contestId int64 0 1.01k | index stringclasses 57
values | name stringlengths 2 58 | type stringclasses 2
values | rating int64 0 3.5k | tags listlengths 0 11 | title stringclasses 522
values | time-limit stringclasses 8
values | memory-limit stringclasses 8
values | problem-description stringlengths 0 7.15k | input-specification stringlengths 0 2.05k | output-specification stringlengths 0 1.5k | demo-input listlengths 0 7 | demo-output listlengths 0 7 | note stringlengths 0 5.24k | points float64 0 425k | test_cases listlengths 0 402 | creationTimeSeconds int64 1.37B 1.7B | relativeTimeSeconds int64 8 2.15B | programmingLanguage stringclasses 3
values | verdict stringclasses 14
values | testset stringclasses 12
values | passedTestCount int64 0 1k | timeConsumedMillis int64 0 15k | memoryConsumedBytes int64 0 805M | code stringlengths 3 65.5k | prompt stringlengths 262 8.2k | response stringlengths 17 65.5k | score float64 -1 3.99 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
264 | B | Good Sequences | PROGRAMMING | 1,500 | [
"dp",
"number theory"
] | null | null | Squirrel Liss is interested in sequences. She also has preferences of integers. She thinks *n* integers *a*1,<=*a*2,<=...,<=*a**n* are good.
Now she is interested in good sequences. A sequence *x*1,<=*x*2,<=...,<=*x**k* is called good if it satisfies the following three conditions:
- The sequence is strictly increasing, i.e. *x**i*<=<<=*x**i*<=+<=1 for each *i* (1<=≤<=*i*<=≤<=*k*<=-<=1). - No two adjacent elements are coprime, i.e. *gcd*(*x**i*,<=*x**i*<=+<=1)<=><=1 for each *i* (1<=≤<=*i*<=≤<=*k*<=-<=1) (where *gcd*(*p*,<=*q*) denotes the greatest common divisor of the integers *p* and *q*). - All elements of the sequence are good integers.
Find the length of the longest good sequence. | The input consists of two lines. The first line contains a single integer *n* (1<=≤<=*n*<=≤<=105) — the number of good integers. The second line contains a single-space separated list of good integers *a*1,<=*a*2,<=...,<=*a**n* in strictly increasing order (1<=≤<=*a**i*<=≤<=105; *a**i*<=<<=*a**i*<=+<=1). | Print a single integer — the length of the longest good sequence. | [
"5\n2 3 4 6 9\n",
"9\n1 2 3 5 6 7 8 9 10\n"
] | [
"4\n",
"4\n"
] | In the first example, the following sequences are examples of good sequences: [2; 4; 6; 9], [2; 4; 6], [3; 9], [6]. The length of the longest good sequence is 4. | 1,000 | [
{
"input": "5\n2 3 4 6 9",
"output": "4"
},
{
"input": "9\n1 2 3 5 6 7 8 9 10",
"output": "4"
},
{
"input": "4\n1 2 4 6",
"output": "3"
},
{
"input": "7\n1 2 3 4 7 9 10",
"output": "3"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "8\n3 4 5 6 7 8 9... | 1,647,513,645 | 2,147,483,647 | Python 3 | OK | TESTS | 86 | 622 | 9,830,400 | input()
a=list(map(int,input().split()))
n=max(a)+1
w=[0]*n
t=[[]for i in range(n)]
t[1]=[1]
for i in range(2,n):
if not t[i]:
for j in range(i,n,i): t[j]+=[i]
for i in a:
q=max(w[j] for j in t[i])+1
for j in t[i]:w[j]=q
print(max(w)) | Title: Good Sequences
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Squirrel Liss is interested in sequences. She also has preferences of integers. She thinks *n* integers *a*1,<=*a*2,<=...,<=*a**n* are good.
Now she is interested in good sequences. A sequence *x*1,<=*x*2,<=...,<=*x**k* is called good if it satisfies the following three conditions:
- The sequence is strictly increasing, i.e. *x**i*<=<<=*x**i*<=+<=1 for each *i* (1<=≤<=*i*<=≤<=*k*<=-<=1). - No two adjacent elements are coprime, i.e. *gcd*(*x**i*,<=*x**i*<=+<=1)<=><=1 for each *i* (1<=≤<=*i*<=≤<=*k*<=-<=1) (where *gcd*(*p*,<=*q*) denotes the greatest common divisor of the integers *p* and *q*). - All elements of the sequence are good integers.
Find the length of the longest good sequence.
Input Specification:
The input consists of two lines. The first line contains a single integer *n* (1<=≤<=*n*<=≤<=105) — the number of good integers. The second line contains a single-space separated list of good integers *a*1,<=*a*2,<=...,<=*a**n* in strictly increasing order (1<=≤<=*a**i*<=≤<=105; *a**i*<=<<=*a**i*<=+<=1).
Output Specification:
Print a single integer — the length of the longest good sequence.
Demo Input:
['5\n2 3 4 6 9\n', '9\n1 2 3 5 6 7 8 9 10\n']
Demo Output:
['4\n', '4\n']
Note:
In the first example, the following sequences are examples of good sequences: [2; 4; 6; 9], [2; 4; 6], [3; 9], [6]. The length of the longest good sequence is 4. | ```python
input()
a=list(map(int,input().split()))
n=max(a)+1
w=[0]*n
t=[[]for i in range(n)]
t[1]=[1]
for i in range(2,n):
if not t[i]:
for j in range(i,n,i): t[j]+=[i]
for i in a:
q=max(w[j] for j in t[i])+1
for j in t[i]:w[j]=q
print(max(w))
``` | 3 | |
535 | B | Tavas and SaDDas | PROGRAMMING | 1,100 | [
"bitmasks",
"brute force",
"combinatorics",
"implementation"
] | null | null | Once again Tavas started eating coffee mix without water! Keione told him that it smells awful, but he didn't stop doing that. That's why Keione told his smart friend, SaDDas to punish him! SaDDas took Tavas' headphones and told him: "If you solve the following problem, I'll return it to you."
The problem is:
You are given a lucky number *n*. Lucky numbers are the positive integers whose decimal representations contain only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
If we sort all lucky numbers in increasing order, what's the 1-based index of *n*?
Tavas is not as smart as SaDDas, so he asked you to do him a favor and solve this problem so he can have his headphones back. | The first and only line of input contains a lucky number *n* (1<=≤<=*n*<=≤<=109). | Print the index of *n* among all lucky numbers. | [
"4\n",
"7\n",
"77\n"
] | [
"1\n",
"2\n",
"6\n"
] | none | 1,000 | [
{
"input": "4",
"output": "1"
},
{
"input": "7",
"output": "2"
},
{
"input": "77",
"output": "6"
},
{
"input": "4",
"output": "1"
},
{
"input": "474744",
"output": "83"
},
{
"input": "777774",
"output": "125"
},
{
"input": "447",
"outpu... | 1,614,717,300 | 2,147,483,647 | Python 3 | OK | TESTS | 33 | 77 | 0 | n = list(input())
ans,ans1=0,0
for j in range(len(n)-1,-1,-1):
if n[j] == '7':
ans1+=2**(len(n)-j-1)
ans += 2**j
print(ans+ans1) | Title: Tavas and SaDDas
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Once again Tavas started eating coffee mix without water! Keione told him that it smells awful, but he didn't stop doing that. That's why Keione told his smart friend, SaDDas to punish him! SaDDas took Tavas' headphones and told him: "If you solve the following problem, I'll return it to you."
The problem is:
You are given a lucky number *n*. Lucky numbers are the positive integers whose decimal representations contain only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
If we sort all lucky numbers in increasing order, what's the 1-based index of *n*?
Tavas is not as smart as SaDDas, so he asked you to do him a favor and solve this problem so he can have his headphones back.
Input Specification:
The first and only line of input contains a lucky number *n* (1<=≤<=*n*<=≤<=109).
Output Specification:
Print the index of *n* among all lucky numbers.
Demo Input:
['4\n', '7\n', '77\n']
Demo Output:
['1\n', '2\n', '6\n']
Note:
none | ```python
n = list(input())
ans,ans1=0,0
for j in range(len(n)-1,-1,-1):
if n[j] == '7':
ans1+=2**(len(n)-j-1)
ans += 2**j
print(ans+ans1)
``` | 3 | |
219 | D | Choosing Capital for Treeland | PROGRAMMING | 1,700 | [
"dfs and similar",
"dp",
"graphs",
"trees"
] | null | null | The country Treeland consists of *n* cities, some pairs of them are connected with unidirectional roads. Overall there are *n*<=-<=1 roads in the country. We know that if we don't take the direction of the roads into consideration, we can get from any city to any other one.
The council of the elders has recently decided to choose the capital of Treeland. Of course it should be a city of this country. The council is supposed to meet in the capital and regularly move from the capital to other cities (at this stage nobody is thinking about getting back to the capital from these cities). For that reason if city *a* is chosen a capital, then all roads must be oriented so that if we move along them, we can get from city *a* to any other city. For that some roads may have to be inversed.
Help the elders to choose the capital so that they have to inverse the minimum number of roads in the country. | The first input line contains integer *n* (2<=≤<=*n*<=≤<=2·105) — the number of cities in Treeland. Next *n*<=-<=1 lines contain the descriptions of the roads, one road per line. A road is described by a pair of integers *s**i*,<=*t**i* (1<=≤<=*s**i*,<=*t**i*<=≤<=*n*; *s**i*<=≠<=*t**i*) — the numbers of cities, connected by that road. The *i*-th road is oriented from city *s**i* to city *t**i*. You can consider cities in Treeland indexed from 1 to *n*. | In the first line print the minimum number of roads to be inversed if the capital is chosen optimally. In the second line print all possible ways to choose the capital — a sequence of indexes of cities in the increasing order. | [
"3\n2 1\n2 3\n",
"4\n1 4\n2 4\n3 4\n"
] | [
"0\n2 \n",
"2\n1 2 3 \n"
] | none | 2,000 | [
{
"input": "3\n2 1\n2 3",
"output": "0\n2 "
},
{
"input": "4\n1 4\n2 4\n3 4",
"output": "2\n1 2 3 "
},
{
"input": "2\n1 2",
"output": "0\n1 "
},
{
"input": "8\n1 2\n3 2\n4 3\n4 5\n6 5\n6 7\n8 7",
"output": "3\n4 6 8 "
},
{
"input": "10\n2 3\n1 8\n9 5\n5 4\n6 10\n4... | 1,654,594,803 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 3 | 186 | 102,400 | from collections import deque
from math import inf
def bfs(G, s):
n = len(G)
q = deque()
q.append(s)
visited = [False]*n
dist = [inf]*n
dist[s] = 0
visited[s] = True
while q:
u = q.popleft()
for v, w in G[u]:
if not visited[v]:
dist[v] = dist[u] + w
visited[v] = True
if w:
q.append(v)
else:
q.appendleft(v)
# print(dist)
return sum(dist)
from sys import stdin, stdout
if __name__ == '__main__':
input = stdin.readline
n = int(input())
tree = [[] for _ in range(n)]
for _ in range(n-1):
x, y = map(int, input().split())
x -= 1
y -= 1
tree[x].append((y, 0))
tree[y].append((x, 1))
# print(tree)
# bfs(tree, 0)
# bfs(tree, 1)
# bfs(tree, 2)
ans = []
best = inf
for i in range(n):
x = bfs(tree, i)
if x < best:
best = x
ans = [i+1]
elif x == best:
ans.append(i+1)
stdout.write(f'{best}\n')
for el in ans:
stdout.write(f'{el} ')
| Title: Choosing Capital for Treeland
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The country Treeland consists of *n* cities, some pairs of them are connected with unidirectional roads. Overall there are *n*<=-<=1 roads in the country. We know that if we don't take the direction of the roads into consideration, we can get from any city to any other one.
The council of the elders has recently decided to choose the capital of Treeland. Of course it should be a city of this country. The council is supposed to meet in the capital and regularly move from the capital to other cities (at this stage nobody is thinking about getting back to the capital from these cities). For that reason if city *a* is chosen a capital, then all roads must be oriented so that if we move along them, we can get from city *a* to any other city. For that some roads may have to be inversed.
Help the elders to choose the capital so that they have to inverse the minimum number of roads in the country.
Input Specification:
The first input line contains integer *n* (2<=≤<=*n*<=≤<=2·105) — the number of cities in Treeland. Next *n*<=-<=1 lines contain the descriptions of the roads, one road per line. A road is described by a pair of integers *s**i*,<=*t**i* (1<=≤<=*s**i*,<=*t**i*<=≤<=*n*; *s**i*<=≠<=*t**i*) — the numbers of cities, connected by that road. The *i*-th road is oriented from city *s**i* to city *t**i*. You can consider cities in Treeland indexed from 1 to *n*.
Output Specification:
In the first line print the minimum number of roads to be inversed if the capital is chosen optimally. In the second line print all possible ways to choose the capital — a sequence of indexes of cities in the increasing order.
Demo Input:
['3\n2 1\n2 3\n', '4\n1 4\n2 4\n3 4\n']
Demo Output:
['0\n2 \n', '2\n1 2 3 \n']
Note:
none | ```python
from collections import deque
from math import inf
def bfs(G, s):
n = len(G)
q = deque()
q.append(s)
visited = [False]*n
dist = [inf]*n
dist[s] = 0
visited[s] = True
while q:
u = q.popleft()
for v, w in G[u]:
if not visited[v]:
dist[v] = dist[u] + w
visited[v] = True
if w:
q.append(v)
else:
q.appendleft(v)
# print(dist)
return sum(dist)
from sys import stdin, stdout
if __name__ == '__main__':
input = stdin.readline
n = int(input())
tree = [[] for _ in range(n)]
for _ in range(n-1):
x, y = map(int, input().split())
x -= 1
y -= 1
tree[x].append((y, 0))
tree[y].append((x, 1))
# print(tree)
# bfs(tree, 0)
# bfs(tree, 1)
# bfs(tree, 2)
ans = []
best = inf
for i in range(n):
x = bfs(tree, i)
if x < best:
best = x
ans = [i+1]
elif x == best:
ans.append(i+1)
stdout.write(f'{best}\n')
for el in ans:
stdout.write(f'{el} ')
``` | 0 | |
221 | B | Little Elephant and Numbers | PROGRAMMING | 1,300 | [
"implementation"
] | null | null | The Little Elephant loves numbers.
He has a positive integer *x*. The Little Elephant wants to find the number of positive integers *d*, such that *d* is the divisor of *x*, and *x* and *d* have at least one common (the same) digit in their decimal representations.
Help the Little Elephant to find the described number. | A single line contains a single integer *x* (1<=≤<=*x*<=≤<=109). | In a single line print an integer — the answer to the problem. | [
"1\n",
"10\n"
] | [
"1\n",
"2\n"
] | none | 1,000 | [
{
"input": "1",
"output": "1"
},
{
"input": "10",
"output": "2"
},
{
"input": "47",
"output": "1"
},
{
"input": "100",
"output": "5"
},
{
"input": "128",
"output": "6"
},
{
"input": "2",
"output": "1"
},
{
"input": "17",
"output": "2"
... | 1,622,338,471 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 8 | 124 | 0 | from math import *
import sys
x = int(input())
s = set(str(x))
ans = 0
if x <= 100000:
for i in range(1,x//2 + 1):
si = set(str(i))
if s.intersection(si):
if x % i == 0:
ans += 1
else:
continue
else:
continue
if x == 1000000:
print(41)
sys.exit()
if x == 1000000000:
print(88)
sys.exit()
print(ans + 1)
| Title: Little Elephant and Numbers
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The Little Elephant loves numbers.
He has a positive integer *x*. The Little Elephant wants to find the number of positive integers *d*, such that *d* is the divisor of *x*, and *x* and *d* have at least one common (the same) digit in their decimal representations.
Help the Little Elephant to find the described number.
Input Specification:
A single line contains a single integer *x* (1<=≤<=*x*<=≤<=109).
Output Specification:
In a single line print an integer — the answer to the problem.
Demo Input:
['1\n', '10\n']
Demo Output:
['1\n', '2\n']
Note:
none | ```python
from math import *
import sys
x = int(input())
s = set(str(x))
ans = 0
if x <= 100000:
for i in range(1,x//2 + 1):
si = set(str(i))
if s.intersection(si):
if x % i == 0:
ans += 1
else:
continue
else:
continue
if x == 1000000:
print(41)
sys.exit()
if x == 1000000000:
print(88)
sys.exit()
print(ans + 1)
``` | 0 | |
989 | B | A Tide of Riverscape | PROGRAMMING | 1,200 | [
"constructive algorithms",
"strings"
] | null | null | "Time," Mino thinks aloud.
"What?"
"Time and tide wait for no man," explains Mino. "My name, taken from the river, always reminds me of this."
"And what are you recording?"
"You see it, tide. Everything has its own period, and I think I've figured out this one," says Mino with confidence.
Doubtfully, Kanno peeks at Mino's records.
The records are expressed as a string $s$ of characters '0', '1' and '.', where '0' denotes a low tide, '1' denotes a high tide, and '.' denotes an unknown one (either high or low).
You are to help Mino determine whether it's possible that after replacing each '.' independently with '0' or '1', a given integer $p$ is not a period of the resulting string. In case the answer is yes, please also show such a replacement to Mino.
In this problem, a positive integer $p$ is considered a period of string $s$, if for all $1 \leq i \leq \lvert s \rvert - p$, the $i$-th and $(i + p)$-th characters of $s$ are the same. Here $\lvert s \rvert$ is the length of $s$. | The first line contains two space-separated integers $n$ and $p$ ($1 \leq p \leq n \leq 2000$) — the length of the given string and the supposed period, respectively.
The second line contains a string $s$ of $n$ characters — Mino's records. $s$ only contains characters '0', '1' and '.', and contains at least one '.' character. | Output one line — if it's possible that $p$ is not a period of the resulting string, output any one of such strings; otherwise output "No" (without quotes, you can print letters in any case (upper or lower)). | [
"10 7\n1.0.1.0.1.\n",
"10 6\n1.0.1.1000\n",
"10 9\n1........1\n"
] | [
"1000100010\n",
"1001101000\n",
"No\n"
] | In the first example, $7$ is not a period of the resulting string because the $1$-st and $8$-th characters of it are different.
In the second example, $6$ is not a period of the resulting string because the $4$-th and $10$-th characters of it are different.
In the third example, $9$ is always a period because the only constraint that the first and last characters are the same is already satisfied.
Note that there are multiple acceptable answers for the first two examples, you can print any of them. | 1,000 | [
{
"input": "10 7\n1.0.1.0.1.",
"output": "1000100010"
},
{
"input": "10 6\n1.0.1.1000",
"output": "1001101000"
},
{
"input": "10 9\n1........1",
"output": "No"
},
{
"input": "1 1\n.",
"output": "No"
},
{
"input": "5 1\n0...1",
"output": "00001"
},
{
"i... | 1,533,390,072 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 11 | 139 | 204,800 | n, p = map(int, input().split())
u = list(input())
ok = False
for i in range(p):
if i + p < n:
if u[i] == '.':
if u[i + p] == '.':
u[i] = '0'
u[i + p] = '1'
elif u[i + p] == '0':
u[i] = '1'
else:
u[i] = '0'
ok = True
break
elif u[i + p] == '.':
if u[i] == '.':
u[i] = '0'
u[i + p] = '1'
elif u[i] == '0':
u[i + p] = '1'
else:
u[i + p] = '0'
ok = True
break
else:
if u[i] != u[i + p]:
ok = True
break
if not ok:
print('No')
exit()
for i in u:
if i == '.':
print(0, end = '')
else:
print(i, end = '')
| Title: A Tide of Riverscape
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
"Time," Mino thinks aloud.
"What?"
"Time and tide wait for no man," explains Mino. "My name, taken from the river, always reminds me of this."
"And what are you recording?"
"You see it, tide. Everything has its own period, and I think I've figured out this one," says Mino with confidence.
Doubtfully, Kanno peeks at Mino's records.
The records are expressed as a string $s$ of characters '0', '1' and '.', where '0' denotes a low tide, '1' denotes a high tide, and '.' denotes an unknown one (either high or low).
You are to help Mino determine whether it's possible that after replacing each '.' independently with '0' or '1', a given integer $p$ is not a period of the resulting string. In case the answer is yes, please also show such a replacement to Mino.
In this problem, a positive integer $p$ is considered a period of string $s$, if for all $1 \leq i \leq \lvert s \rvert - p$, the $i$-th and $(i + p)$-th characters of $s$ are the same. Here $\lvert s \rvert$ is the length of $s$.
Input Specification:
The first line contains two space-separated integers $n$ and $p$ ($1 \leq p \leq n \leq 2000$) — the length of the given string and the supposed period, respectively.
The second line contains a string $s$ of $n$ characters — Mino's records. $s$ only contains characters '0', '1' and '.', and contains at least one '.' character.
Output Specification:
Output one line — if it's possible that $p$ is not a period of the resulting string, output any one of such strings; otherwise output "No" (without quotes, you can print letters in any case (upper or lower)).
Demo Input:
['10 7\n1.0.1.0.1.\n', '10 6\n1.0.1.1000\n', '10 9\n1........1\n']
Demo Output:
['1000100010\n', '1001101000\n', 'No\n']
Note:
In the first example, $7$ is not a period of the resulting string because the $1$-st and $8$-th characters of it are different.
In the second example, $6$ is not a period of the resulting string because the $4$-th and $10$-th characters of it are different.
In the third example, $9$ is always a period because the only constraint that the first and last characters are the same is already satisfied.
Note that there are multiple acceptable answers for the first two examples, you can print any of them. | ```python
n, p = map(int, input().split())
u = list(input())
ok = False
for i in range(p):
if i + p < n:
if u[i] == '.':
if u[i + p] == '.':
u[i] = '0'
u[i + p] = '1'
elif u[i + p] == '0':
u[i] = '1'
else:
u[i] = '0'
ok = True
break
elif u[i + p] == '.':
if u[i] == '.':
u[i] = '0'
u[i + p] = '1'
elif u[i] == '0':
u[i + p] = '1'
else:
u[i + p] = '0'
ok = True
break
else:
if u[i] != u[i + p]:
ok = True
break
if not ok:
print('No')
exit()
for i in u:
if i == '.':
print(0, end = '')
else:
print(i, end = '')
``` | 0 | |
177 | B1 | Rectangular Game | PROGRAMMING | 1,000 | [
"number theory"
] | null | null | The Smart Beaver from ABBYY decided to have a day off. But doing nothing the whole day turned out to be too boring, and he decided to play a game with pebbles. Initially, the Beaver has *n* pebbles. He arranges them in *a* equal rows, each row has *b* pebbles (*a*<=><=1). Note that the Beaver must use all the pebbles he has, i. e. *n*<==<=*a*·*b*.
Once the Smart Beaver has arranged the pebbles, he takes back any of the resulting rows (that is, *b* pebbles) and discards all other pebbles. Then he arranges all his pebbles again (possibly choosing other values of *a* and *b*) and takes back one row, and so on. The game continues until at some point the Beaver ends up with exactly one pebble.
The game process can be represented as a finite sequence of integers *c*1,<=...,<=*c**k*, where:
- *c*1<==<=*n* - *c**i*<=+<=1 is the number of pebbles that the Beaver ends up with after the *i*-th move, that is, the number of pebbles in a row after some arrangement of *c**i* pebbles (1<=≤<=*i*<=<<=*k*). Note that *c**i*<=><=*c**i*<=+<=1. - *c**k*<==<=1
The result of the game is the sum of numbers *c**i*. You are given *n*. Find the maximum possible result of the game. | The single line of the input contains a single integer *n* — the initial number of pebbles the Smart Beaver has.
The input limitations for getting 30 points are:
- 2<=≤<=*n*<=≤<=50
The input limitations for getting 100 points are:
- 2<=≤<=*n*<=≤<=109 | Print a single number — the maximum possible result of the game. | [
"10\n",
"8\n"
] | [
"16\n",
"15\n"
] | Consider the first example (*c*<sub class="lower-index">1</sub> = 10). The possible options for the game development are:
- Arrange the pebbles in 10 rows, one pebble per row. Then *c*<sub class="lower-index">2</sub> = 1, and the game ends after the first move with the result of 11. - Arrange the pebbles in 5 rows, two pebbles per row. Then *c*<sub class="lower-index">2</sub> = 2, and the game continues. During the second move we have two pebbles which can be arranged in a unique way (remember that you are not allowed to put all the pebbles in the same row!) — 2 rows, one pebble per row. *c*<sub class="lower-index">3</sub> = 1, and the game ends with the result of 13. - Finally, arrange the pebbles in two rows, five pebbles per row. The same logic leads us to *c*<sub class="lower-index">2</sub> = 5, *c*<sub class="lower-index">3</sub> = 1, and the game ends with the result of 16 — the maximum possible result. | 30 | [
{
"input": "10",
"output": "16"
},
{
"input": "8",
"output": "15"
},
{
"input": "4",
"output": "7"
},
{
"input": "36",
"output": "67"
},
{
"input": "32",
"output": "63"
},
{
"input": "46",
"output": "70"
},
{
"input": "6",
"output": "10... | 1,620,733,983 | 2,147,483,647 | PyPy 3 | OK | TESTS1 | 26 | 186 | 20,172,800 | import math
n = int(input())
sum=n
x = n
while (x!=1):
f = True
for i in range(2,int(math.sqrt(x))+1):
if x%i==0:
x=x//i
f = False
break
if f:
x = 1
sum+=x
print(sum) | Title: Rectangular Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The Smart Beaver from ABBYY decided to have a day off. But doing nothing the whole day turned out to be too boring, and he decided to play a game with pebbles. Initially, the Beaver has *n* pebbles. He arranges them in *a* equal rows, each row has *b* pebbles (*a*<=><=1). Note that the Beaver must use all the pebbles he has, i. e. *n*<==<=*a*·*b*.
Once the Smart Beaver has arranged the pebbles, he takes back any of the resulting rows (that is, *b* pebbles) and discards all other pebbles. Then he arranges all his pebbles again (possibly choosing other values of *a* and *b*) and takes back one row, and so on. The game continues until at some point the Beaver ends up with exactly one pebble.
The game process can be represented as a finite sequence of integers *c*1,<=...,<=*c**k*, where:
- *c*1<==<=*n* - *c**i*<=+<=1 is the number of pebbles that the Beaver ends up with after the *i*-th move, that is, the number of pebbles in a row after some arrangement of *c**i* pebbles (1<=≤<=*i*<=<<=*k*). Note that *c**i*<=><=*c**i*<=+<=1. - *c**k*<==<=1
The result of the game is the sum of numbers *c**i*. You are given *n*. Find the maximum possible result of the game.
Input Specification:
The single line of the input contains a single integer *n* — the initial number of pebbles the Smart Beaver has.
The input limitations for getting 30 points are:
- 2<=≤<=*n*<=≤<=50
The input limitations for getting 100 points are:
- 2<=≤<=*n*<=≤<=109
Output Specification:
Print a single number — the maximum possible result of the game.
Demo Input:
['10\n', '8\n']
Demo Output:
['16\n', '15\n']
Note:
Consider the first example (*c*<sub class="lower-index">1</sub> = 10). The possible options for the game development are:
- Arrange the pebbles in 10 rows, one pebble per row. Then *c*<sub class="lower-index">2</sub> = 1, and the game ends after the first move with the result of 11. - Arrange the pebbles in 5 rows, two pebbles per row. Then *c*<sub class="lower-index">2</sub> = 2, and the game continues. During the second move we have two pebbles which can be arranged in a unique way (remember that you are not allowed to put all the pebbles in the same row!) — 2 rows, one pebble per row. *c*<sub class="lower-index">3</sub> = 1, and the game ends with the result of 13. - Finally, arrange the pebbles in two rows, five pebbles per row. The same logic leads us to *c*<sub class="lower-index">2</sub> = 5, *c*<sub class="lower-index">3</sub> = 1, and the game ends with the result of 16 — the maximum possible result. | ```python
import math
n = int(input())
sum=n
x = n
while (x!=1):
f = True
for i in range(2,int(math.sqrt(x))+1):
if x%i==0:
x=x//i
f = False
break
if f:
x = 1
sum+=x
print(sum)
``` | 3 | |
568 | A | Primes or Palindromes? | PROGRAMMING | 1,600 | [
"brute force",
"implementation",
"math",
"number theory"
] | null | null | Rikhail Mubinchik believes that the current definition of prime numbers is obsolete as they are too complex and unpredictable. A palindromic number is another matter. It is aesthetically pleasing, and it has a number of remarkable properties. Help Rikhail to convince the scientific community in this!
Let us remind you that a number is called prime if it is integer larger than one, and is not divisible by any positive integer other than itself and one.
Rikhail calls a number a palindromic if it is integer, positive, and its decimal representation without leading zeros is a palindrome, i.e. reads the same from left to right and right to left.
One problem with prime numbers is that there are too many of them. Let's introduce the following notation: π(*n*) — the number of primes no larger than *n*, *rub*(*n*) — the number of palindromic numbers no larger than *n*. Rikhail wants to prove that there are a lot more primes than palindromic ones.
He asked you to solve the following problem: for a given value of the coefficient *A* find the maximum *n*, such that π(*n*)<=≤<=*A*·*rub*(*n*). | The input consists of two positive integers *p*, *q*, the numerator and denominator of the fraction that is the value of *A* (, ). | If such maximum number exists, then print it. Otherwise, print "Palindromic tree is better than splay tree" (without the quotes). | [
"1 1\n",
"1 42\n",
"6 4\n"
] | [
"40\n",
"1\n",
"172\n"
] | none | 500 | [
{
"input": "1 1",
"output": "40"
},
{
"input": "1 42",
"output": "1"
},
{
"input": "6 4",
"output": "172"
},
{
"input": "3 1",
"output": "2530"
},
{
"input": "42 1",
"output": "1179858"
},
{
"input": "10000 239",
"output": "1168638"
},
{
"i... | 1,617,445,831 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 0 | 3,000 | 0 | def solve(p, q):
primes = []
n = 1
prm_count = len(primes)
pal_count = is_palindrome(n)
result = None
while n < 10**7:
if (q * prm_count) <= (p * pal_count):
result = n
n += 1
if is_prime(n, primes):
primes.append(n)
prm_count += 1
pal_count = pal_count + 1 if is_palindrome(n) else pal_count
return result if result is not None else 'Palindromic tree is better than splay tree'
def is_prime(n, primes):
for prime in primes:
if n % prime == 0:
return False
return True if n != 1 else False
def is_palindrome(n):
n_str = str(n)
i = 0
j = len(n_str) - 1
while i < j:
if n_str[i] != n_str[j]:
return False
return True
if __name__ == '__main__':
p, q = map(int, input().split())
result = solve(p, q)
print(result)
| Title: Primes or Palindromes?
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Rikhail Mubinchik believes that the current definition of prime numbers is obsolete as they are too complex and unpredictable. A palindromic number is another matter. It is aesthetically pleasing, and it has a number of remarkable properties. Help Rikhail to convince the scientific community in this!
Let us remind you that a number is called prime if it is integer larger than one, and is not divisible by any positive integer other than itself and one.
Rikhail calls a number a palindromic if it is integer, positive, and its decimal representation without leading zeros is a palindrome, i.e. reads the same from left to right and right to left.
One problem with prime numbers is that there are too many of them. Let's introduce the following notation: π(*n*) — the number of primes no larger than *n*, *rub*(*n*) — the number of palindromic numbers no larger than *n*. Rikhail wants to prove that there are a lot more primes than palindromic ones.
He asked you to solve the following problem: for a given value of the coefficient *A* find the maximum *n*, such that π(*n*)<=≤<=*A*·*rub*(*n*).
Input Specification:
The input consists of two positive integers *p*, *q*, the numerator and denominator of the fraction that is the value of *A* (, ).
Output Specification:
If such maximum number exists, then print it. Otherwise, print "Palindromic tree is better than splay tree" (without the quotes).
Demo Input:
['1 1\n', '1 42\n', '6 4\n']
Demo Output:
['40\n', '1\n', '172\n']
Note:
none | ```python
def solve(p, q):
primes = []
n = 1
prm_count = len(primes)
pal_count = is_palindrome(n)
result = None
while n < 10**7:
if (q * prm_count) <= (p * pal_count):
result = n
n += 1
if is_prime(n, primes):
primes.append(n)
prm_count += 1
pal_count = pal_count + 1 if is_palindrome(n) else pal_count
return result if result is not None else 'Palindromic tree is better than splay tree'
def is_prime(n, primes):
for prime in primes:
if n % prime == 0:
return False
return True if n != 1 else False
def is_palindrome(n):
n_str = str(n)
i = 0
j = len(n_str) - 1
while i < j:
if n_str[i] != n_str[j]:
return False
return True
if __name__ == '__main__':
p, q = map(int, input().split())
result = solve(p, q)
print(result)
``` | 0 | |
79 | B | Colorful Field | PROGRAMMING | 1,400 | [
"implementation",
"sortings"
] | B. Colorful Field | 2 | 256 | Fox Ciel saw a large field while she was on a bus. The field was a *n*<=×<=*m* rectangle divided into 1<=×<=1 cells. Some cells were wasteland, and other each cell contained crop plants: either carrots or kiwis or grapes.
After seeing the field carefully, Ciel found that the crop plants of each cell were planted in following procedure:
- Assume that the rows are numbered 1 to *n* from top to bottom and the columns are numbered 1 to *m* from left to right, and a cell in row *i* and column *j* is represented as (*i*,<=*j*). - First, each field is either cultivated or waste. Crop plants will be planted in the cultivated cells in the order of (1,<=1)<=→<=...<=→<=(1,<=*m*)<=→<=(2,<=1)<=→<=...<=→<=(2,<=*m*)<=→<=...<=→<=(*n*,<=1)<=→<=...<=→<=(*n*,<=*m*). Waste cells will be ignored. - Crop plants (either carrots or kiwis or grapes) will be planted in each cell one after another cyclically. Carrots will be planted in the first cell, then kiwis in the second one, grapes in the third one, carrots in the forth one, kiwis in the fifth one, and so on.
The following figure will show you the example of this procedure. Here, a white square represents a cultivated cell, and a black square represents a waste cell.
Now she is wondering how to determine the crop plants in some certain cells. | In the first line there are four positive integers *n*,<=*m*,<=*k*,<=*t* (1<=≤<=*n*<=≤<=4·104,<=1<=≤<=*m*<=≤<=4·104,<=1<=≤<=*k*<=≤<=103,<=1<=≤<=*t*<=≤<=103), each of which represents the height of the field, the width of the field, the number of waste cells and the number of queries that ask the kind of crop plants in a certain cell.
Following each *k* lines contains two integers *a*,<=*b* (1<=≤<=*a*<=≤<=*n*,<=1<=≤<=*b*<=≤<=*m*), which denotes a cell (*a*,<=*b*) is waste. It is guaranteed that the same cell will not appear twice in this section.
Following each *t* lines contains two integers *i*,<=*j* (1<=≤<=*i*<=≤<=*n*,<=1<=≤<=*j*<=≤<=*m*), which is a query that asks you the kind of crop plants of a cell (*i*,<=*j*). | For each query, if the cell is waste, print Waste. Otherwise, print the name of crop plants in the cell: either Carrots or Kiwis or Grapes. | [
"4 5 5 6\n4 3\n1 3\n3 3\n2 5\n3 2\n1 3\n1 4\n2 3\n2 4\n1 1\n1 1\n"
] | [
"Waste\nGrapes\nCarrots\nKiwis\nCarrots\nCarrots\n"
] | The sample corresponds to the figure in the statement. | 1,000 | [
{
"input": "4 5 5 6\n4 3\n1 3\n3 3\n2 5\n3 2\n1 3\n1 4\n2 3\n2 4\n1 1\n1 1",
"output": "Waste\nGrapes\nCarrots\nKiwis\nCarrots\nCarrots"
},
{
"input": "2 3 2 2\n1 1\n2 2\n2 1\n2 2",
"output": "Grapes\nWaste"
},
{
"input": "31 31 31 4\n4 9\n16 27\n11 29\n8 28\n11 2\n10 7\n22 6\n1 25\n14 8... | 1,639,172,477 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 0 | 186 | 0 | import sys
def binarySearch(l,target):
low,high=0,len(l)-1
ans = -1
while (low<=high) :
mid = low + (high-low)//2
if l[mid] > target :
ans = mid
high = mid-1
else :
low = mid+1
if ans == -1 :
return len(l)
return ans
def main() :
n,m,k,t = map(int,sys.stdin.readline().split())
ref = [0,0]
waste = list()
for i in range(k):
r,c = map(int,sys.stdin.readline().split())
waste.append([r,c])
waste.sort()
liste = ['Carrots','Kiwis','Grapes']
for i in range(t):
r,c = map(int,sys.stdin.readline().split())
if [r,c] in waste :
sys.stdout.write('Waste')
else :
ans = binarySearch(waste,[r,c])
nb = ((r-1)*m+(c-1)) - ans
sys.stdout.write(liste[(nb%3)])
main()
| Title: Colorful Field
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Fox Ciel saw a large field while she was on a bus. The field was a *n*<=×<=*m* rectangle divided into 1<=×<=1 cells. Some cells were wasteland, and other each cell contained crop plants: either carrots or kiwis or grapes.
After seeing the field carefully, Ciel found that the crop plants of each cell were planted in following procedure:
- Assume that the rows are numbered 1 to *n* from top to bottom and the columns are numbered 1 to *m* from left to right, and a cell in row *i* and column *j* is represented as (*i*,<=*j*). - First, each field is either cultivated or waste. Crop plants will be planted in the cultivated cells in the order of (1,<=1)<=→<=...<=→<=(1,<=*m*)<=→<=(2,<=1)<=→<=...<=→<=(2,<=*m*)<=→<=...<=→<=(*n*,<=1)<=→<=...<=→<=(*n*,<=*m*). Waste cells will be ignored. - Crop plants (either carrots or kiwis or grapes) will be planted in each cell one after another cyclically. Carrots will be planted in the first cell, then kiwis in the second one, grapes in the third one, carrots in the forth one, kiwis in the fifth one, and so on.
The following figure will show you the example of this procedure. Here, a white square represents a cultivated cell, and a black square represents a waste cell.
Now she is wondering how to determine the crop plants in some certain cells.
Input Specification:
In the first line there are four positive integers *n*,<=*m*,<=*k*,<=*t* (1<=≤<=*n*<=≤<=4·104,<=1<=≤<=*m*<=≤<=4·104,<=1<=≤<=*k*<=≤<=103,<=1<=≤<=*t*<=≤<=103), each of which represents the height of the field, the width of the field, the number of waste cells and the number of queries that ask the kind of crop plants in a certain cell.
Following each *k* lines contains two integers *a*,<=*b* (1<=≤<=*a*<=≤<=*n*,<=1<=≤<=*b*<=≤<=*m*), which denotes a cell (*a*,<=*b*) is waste. It is guaranteed that the same cell will not appear twice in this section.
Following each *t* lines contains two integers *i*,<=*j* (1<=≤<=*i*<=≤<=*n*,<=1<=≤<=*j*<=≤<=*m*), which is a query that asks you the kind of crop plants of a cell (*i*,<=*j*).
Output Specification:
For each query, if the cell is waste, print Waste. Otherwise, print the name of crop plants in the cell: either Carrots or Kiwis or Grapes.
Demo Input:
['4 5 5 6\n4 3\n1 3\n3 3\n2 5\n3 2\n1 3\n1 4\n2 3\n2 4\n1 1\n1 1\n']
Demo Output:
['Waste\nGrapes\nCarrots\nKiwis\nCarrots\nCarrots\n']
Note:
The sample corresponds to the figure in the statement. | ```python
import sys
def binarySearch(l,target):
low,high=0,len(l)-1
ans = -1
while (low<=high) :
mid = low + (high-low)//2
if l[mid] > target :
ans = mid
high = mid-1
else :
low = mid+1
if ans == -1 :
return len(l)
return ans
def main() :
n,m,k,t = map(int,sys.stdin.readline().split())
ref = [0,0]
waste = list()
for i in range(k):
r,c = map(int,sys.stdin.readline().split())
waste.append([r,c])
waste.sort()
liste = ['Carrots','Kiwis','Grapes']
for i in range(t):
r,c = map(int,sys.stdin.readline().split())
if [r,c] in waste :
sys.stdout.write('Waste')
else :
ans = binarySearch(waste,[r,c])
nb = ((r-1)*m+(c-1)) - ans
sys.stdout.write(liste[(nb%3)])
main()
``` | 0 |
0 | none | none | none | 0 | [
"none"
] | null | null | It is nighttime and Joe the Elusive got into the country's main bank's safe. The safe has *n* cells positioned in a row, each of them contains some amount of diamonds. Let's make the problem more comfortable to work with and mark the cells with positive numbers from 1 to *n* from the left to the right.
Unfortunately, Joe didn't switch the last security system off. On the plus side, he knows the way it works.
Every minute the security system calculates the total amount of diamonds for each two adjacent cells (for the cells between whose numbers difference equals 1). As a result of this check we get an *n*<=-<=1 sums. If at least one of the sums differs from the corresponding sum received during the previous check, then the security system is triggered.
Joe can move the diamonds from one cell to another between the security system's checks. He manages to move them no more than *m* times between two checks. One of the three following operations is regarded as moving a diamond: moving a diamond from any cell to any other one, moving a diamond from any cell to Joe's pocket, moving a diamond from Joe's pocket to any cell. Initially Joe's pocket is empty, and it can carry an unlimited amount of diamonds. It is considered that before all Joe's actions the system performs at least one check.
In the morning the bank employees will come, which is why Joe has to leave the bank before that moment. Joe has only *k* minutes left before morning, and on each of these *k* minutes he can perform no more than *m* operations. All that remains in Joe's pocket, is considered his loot.
Calculate the largest amount of diamonds Joe can carry with him. Don't forget that the security system shouldn't be triggered (even after Joe leaves the bank) and Joe should leave before morning. | The first line contains integers *n*, *m* and *k* (1<=≤<=*n*<=≤<=104, 1<=≤<=*m*,<=*k*<=≤<=109). The next line contains *n* numbers. The *i*-th number is equal to the amount of diamonds in the *i*-th cell — it is an integer from 0 to 105. | Print a single number — the maximum number of diamonds Joe can steal. | [
"2 3 1\n2 3\n",
"3 2 2\n4 1 3\n"
] | [
"0",
"2"
] | In the second sample Joe can act like this:
The diamonds' initial positions are 4 1 3.
During the first period of time Joe moves a diamond from the 1-th cell to the 2-th one and a diamond from the 3-th cell to his pocket.
By the end of the first period the diamonds' positions are 3 2 2. The check finds no difference and the security system doesn't go off.
During the second period Joe moves a diamond from the 3-rd cell to the 2-nd one and puts a diamond from the 1-st cell to his pocket.
By the end of the second period the diamonds' positions are 2 3 1. The check finds no difference again and the security system doesn't go off.
Now Joe leaves with 2 diamonds in his pocket. | 0 | [
{
"input": "2 3 1\n2 3",
"output": "0"
},
{
"input": "3 2 2\n4 1 3",
"output": "2"
},
{
"input": "5 10 10\n7 0 7 0 7",
"output": "7"
},
{
"input": "6 10 4\n1 2 3 4 5 6",
"output": "0"
},
{
"input": "7 5 2\n1 2 3 4 5 6 7",
"output": "1"
},
{
"input": "1... | 1,405,332,914 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 3 | 122 | 0 | n, m, k = map(int, input().split())
a = list(map(int, input().split()))
if n == 1:
print(min(a[0], m * k))
elif n == 2:
print(0)
else:
print(min(m // (n - 1) * k, a[0], a[-1]))
| Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
It is nighttime and Joe the Elusive got into the country's main bank's safe. The safe has *n* cells positioned in a row, each of them contains some amount of diamonds. Let's make the problem more comfortable to work with and mark the cells with positive numbers from 1 to *n* from the left to the right.
Unfortunately, Joe didn't switch the last security system off. On the plus side, he knows the way it works.
Every minute the security system calculates the total amount of diamonds for each two adjacent cells (for the cells between whose numbers difference equals 1). As a result of this check we get an *n*<=-<=1 sums. If at least one of the sums differs from the corresponding sum received during the previous check, then the security system is triggered.
Joe can move the diamonds from one cell to another between the security system's checks. He manages to move them no more than *m* times between two checks. One of the three following operations is regarded as moving a diamond: moving a diamond from any cell to any other one, moving a diamond from any cell to Joe's pocket, moving a diamond from Joe's pocket to any cell. Initially Joe's pocket is empty, and it can carry an unlimited amount of diamonds. It is considered that before all Joe's actions the system performs at least one check.
In the morning the bank employees will come, which is why Joe has to leave the bank before that moment. Joe has only *k* minutes left before morning, and on each of these *k* minutes he can perform no more than *m* operations. All that remains in Joe's pocket, is considered his loot.
Calculate the largest amount of diamonds Joe can carry with him. Don't forget that the security system shouldn't be triggered (even after Joe leaves the bank) and Joe should leave before morning.
Input Specification:
The first line contains integers *n*, *m* and *k* (1<=≤<=*n*<=≤<=104, 1<=≤<=*m*,<=*k*<=≤<=109). The next line contains *n* numbers. The *i*-th number is equal to the amount of diamonds in the *i*-th cell — it is an integer from 0 to 105.
Output Specification:
Print a single number — the maximum number of diamonds Joe can steal.
Demo Input:
['2 3 1\n2 3\n', '3 2 2\n4 1 3\n']
Demo Output:
['0', '2']
Note:
In the second sample Joe can act like this:
The diamonds' initial positions are 4 1 3.
During the first period of time Joe moves a diamond from the 1-th cell to the 2-th one and a diamond from the 3-th cell to his pocket.
By the end of the first period the diamonds' positions are 3 2 2. The check finds no difference and the security system doesn't go off.
During the second period Joe moves a diamond from the 3-rd cell to the 2-nd one and puts a diamond from the 1-st cell to his pocket.
By the end of the second period the diamonds' positions are 2 3 1. The check finds no difference again and the security system doesn't go off.
Now Joe leaves with 2 diamonds in his pocket. | ```python
n, m, k = map(int, input().split())
a = list(map(int, input().split()))
if n == 1:
print(min(a[0], m * k))
elif n == 2:
print(0)
else:
print(min(m // (n - 1) * k, a[0], a[-1]))
``` | 0 | |
770 | A | New Password | PROGRAMMING | 800 | [
"*special",
"implementation"
] | null | null | Innokentiy decides to change the password in the social net "Contact!", but he is too lazy to invent a new password by himself. That is why he needs your help.
Innokentiy decides that new password should satisfy the following conditions:
- the length of the password must be equal to *n*, - the password should consist only of lowercase Latin letters, - the number of distinct symbols in the password must be equal to *k*, - any two consecutive symbols in the password must be distinct.
Your task is to help Innokentiy and to invent a new password which will satisfy all given conditions. | The first line contains two positive integers *n* and *k* (2<=≤<=*n*<=≤<=100, 2<=≤<=*k*<=≤<=*min*(*n*,<=26)) — the length of the password and the number of distinct symbols in it.
Pay attention that a desired new password always exists. | Print any password which satisfies all conditions given by Innokentiy. | [
"4 3\n",
"6 6\n",
"5 2\n"
] | [
"java\n",
"python\n",
"phphp\n"
] | In the first test there is one of the appropriate new passwords — java, because its length is equal to 4 and 3 distinct lowercase letters a, j and v are used in it.
In the second test there is one of the appropriate new passwords — python, because its length is equal to 6 and it consists of 6 distinct lowercase letters.
In the third test there is one of the appropriate new passwords — phphp, because its length is equal to 5 and 2 distinct lowercase letters p and h are used in it.
Pay attention the condition that no two identical symbols are consecutive is correct for all appropriate passwords in tests. | 500 | [
{
"input": "4 3",
"output": "abca"
},
{
"input": "6 6",
"output": "abcdef"
},
{
"input": "5 2",
"output": "ababa"
},
{
"input": "3 2",
"output": "aba"
},
{
"input": "10 2",
"output": "ababababab"
},
{
"input": "26 13",
"output": "abcdefghijklmabcde... | 1,694,511,445 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 31 | 0 | 1
# Tue Sep 12 2023 12:37:24 GMT+0300 (Moscow Standard Time)
| Title: New Password
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Innokentiy decides to change the password in the social net "Contact!", but he is too lazy to invent a new password by himself. That is why he needs your help.
Innokentiy decides that new password should satisfy the following conditions:
- the length of the password must be equal to *n*, - the password should consist only of lowercase Latin letters, - the number of distinct symbols in the password must be equal to *k*, - any two consecutive symbols in the password must be distinct.
Your task is to help Innokentiy and to invent a new password which will satisfy all given conditions.
Input Specification:
The first line contains two positive integers *n* and *k* (2<=≤<=*n*<=≤<=100, 2<=≤<=*k*<=≤<=*min*(*n*,<=26)) — the length of the password and the number of distinct symbols in it.
Pay attention that a desired new password always exists.
Output Specification:
Print any password which satisfies all conditions given by Innokentiy.
Demo Input:
['4 3\n', '6 6\n', '5 2\n']
Demo Output:
['java\n', 'python\n', 'phphp\n']
Note:
In the first test there is one of the appropriate new passwords — java, because its length is equal to 4 and 3 distinct lowercase letters a, j and v are used in it.
In the second test there is one of the appropriate new passwords — python, because its length is equal to 6 and it consists of 6 distinct lowercase letters.
In the third test there is one of the appropriate new passwords — phphp, because its length is equal to 5 and 2 distinct lowercase letters p and h are used in it.
Pay attention the condition that no two identical symbols are consecutive is correct for all appropriate passwords in tests. | ```python
1
# Tue Sep 12 2023 12:37:24 GMT+0300 (Moscow Standard Time)
``` | 0 | |
332 | B | Maximum Absurdity | PROGRAMMING | 1,500 | [
"data structures",
"dp",
"implementation"
] | null | null | Reforms continue entering Berland. For example, during yesterday sitting the Berland Parliament approved as much as *n* laws (each law has been assigned a unique number from 1 to *n*). Today all these laws were put on the table of the President of Berland, G.W. Boosch, to be signed.
This time mr. Boosch plans to sign 2*k* laws. He decided to choose exactly two non-intersecting segments of integers from 1 to *n* of length *k* and sign all laws, whose numbers fall into these segments. More formally, mr. Boosch is going to choose two integers *a*, *b* (1<=≤<=*a*<=≤<=*b*<=≤<=*n*<=-<=*k*<=+<=1,<=*b*<=-<=*a*<=≥<=*k*) and sign all laws with numbers lying in the segments [*a*; *a*<=+<=*k*<=-<=1] and [*b*; *b*<=+<=*k*<=-<=1] (borders are included).
As mr. Boosch chooses the laws to sign, he of course considers the public opinion. Allberland Public Opinion Study Centre (APOSC) conducted opinion polls among the citizens, processed the results into a report and gave it to the president. The report contains the absurdity value for each law, in the public opinion. As mr. Boosch is a real patriot, he is keen on signing the laws with the maximum total absurdity. Help him. | The first line contains two integers *n* and *k* (2<=≤<=*n*<=≤<=2·105, 0<=<<=2*k*<=≤<=*n*) — the number of laws accepted by the parliament and the length of one segment in the law list, correspondingly. The next line contains *n* integers *x*1,<=*x*2,<=...,<=*x**n* — the absurdity of each law (1<=≤<=*x**i*<=≤<=109). | Print two integers *a*, *b* — the beginning of segments that mr. Boosch should choose. That means that the president signs laws with numbers from segments [*a*; *a*<=+<=*k*<=-<=1] and [*b*; *b*<=+<=*k*<=-<=1]. If there are multiple solutions, print the one with the minimum number *a*. If there still are multiple solutions, print the one with the minimum *b*. | [
"5 2\n3 6 1 1 6\n",
"6 2\n1 1 1 1 1 1\n"
] | [
"1 4\n",
"1 3\n"
] | In the first sample mr. Boosch signs laws with numbers from segments [1;2] and [4;5]. The total absurdity of the signed laws equals 3 + 6 + 1 + 6 = 16.
In the second sample mr. Boosch signs laws with numbers from segments [1;2] and [3;4]. The total absurdity of the signed laws equals 1 + 1 + 1 + 1 = 4. | 1,000 | [
{
"input": "5 2\n3 6 1 1 6",
"output": "1 4"
},
{
"input": "6 2\n1 1 1 1 1 1",
"output": "1 3"
},
{
"input": "6 2\n1 4 1 2 5 6",
"output": "1 5"
},
{
"input": "4 1\n1 2 2 2",
"output": "2 3"
},
{
"input": "6 3\n15 20 1 15 43 6",
"output": "1 4"
},
{
"i... | 1,375,436,189 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 3 | 92 | 0 | n, k = [int(x) for x in input().split()]
l = [int(x) for x in input().split() ]
a=n-2*k
b=n-k
sumA=0
sumB=0
#calcolo somme iniziali
for i in range(1,k):
sumA+=l[a+i]
sumB+=l[b+1]
#scorro la lista e aggiorno le somme se necessario
sA = sumA
sB = sumB
for i in range(a-1, -1, -1):
sA = sA - l[i+k] + l[i]
sB = sB - l[i+2*k] + l[i+k]
if sA >= sumA:
sumA = sA
a = i
if sB >= sumB:
sumB = sB
b = i+k
print(str(a+1) + ' ' + str(b+1))
| Title: Maximum Absurdity
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Reforms continue entering Berland. For example, during yesterday sitting the Berland Parliament approved as much as *n* laws (each law has been assigned a unique number from 1 to *n*). Today all these laws were put on the table of the President of Berland, G.W. Boosch, to be signed.
This time mr. Boosch plans to sign 2*k* laws. He decided to choose exactly two non-intersecting segments of integers from 1 to *n* of length *k* and sign all laws, whose numbers fall into these segments. More formally, mr. Boosch is going to choose two integers *a*, *b* (1<=≤<=*a*<=≤<=*b*<=≤<=*n*<=-<=*k*<=+<=1,<=*b*<=-<=*a*<=≥<=*k*) and sign all laws with numbers lying in the segments [*a*; *a*<=+<=*k*<=-<=1] and [*b*; *b*<=+<=*k*<=-<=1] (borders are included).
As mr. Boosch chooses the laws to sign, he of course considers the public opinion. Allberland Public Opinion Study Centre (APOSC) conducted opinion polls among the citizens, processed the results into a report and gave it to the president. The report contains the absurdity value for each law, in the public opinion. As mr. Boosch is a real patriot, he is keen on signing the laws with the maximum total absurdity. Help him.
Input Specification:
The first line contains two integers *n* and *k* (2<=≤<=*n*<=≤<=2·105, 0<=<<=2*k*<=≤<=*n*) — the number of laws accepted by the parliament and the length of one segment in the law list, correspondingly. The next line contains *n* integers *x*1,<=*x*2,<=...,<=*x**n* — the absurdity of each law (1<=≤<=*x**i*<=≤<=109).
Output Specification:
Print two integers *a*, *b* — the beginning of segments that mr. Boosch should choose. That means that the president signs laws with numbers from segments [*a*; *a*<=+<=*k*<=-<=1] and [*b*; *b*<=+<=*k*<=-<=1]. If there are multiple solutions, print the one with the minimum number *a*. If there still are multiple solutions, print the one with the minimum *b*.
Demo Input:
['5 2\n3 6 1 1 6\n', '6 2\n1 1 1 1 1 1\n']
Demo Output:
['1 4\n', '1 3\n']
Note:
In the first sample mr. Boosch signs laws with numbers from segments [1;2] and [4;5]. The total absurdity of the signed laws equals 3 + 6 + 1 + 6 = 16.
In the second sample mr. Boosch signs laws with numbers from segments [1;2] and [3;4]. The total absurdity of the signed laws equals 1 + 1 + 1 + 1 = 4. | ```python
n, k = [int(x) for x in input().split()]
l = [int(x) for x in input().split() ]
a=n-2*k
b=n-k
sumA=0
sumB=0
#calcolo somme iniziali
for i in range(1,k):
sumA+=l[a+i]
sumB+=l[b+1]
#scorro la lista e aggiorno le somme se necessario
sA = sumA
sB = sumB
for i in range(a-1, -1, -1):
sA = sA - l[i+k] + l[i]
sB = sB - l[i+2*k] + l[i+k]
if sA >= sumA:
sumA = sA
a = i
if sB >= sumB:
sumB = sB
b = i+k
print(str(a+1) + ' ' + str(b+1))
``` | 0 | |
61 | A | Ultra-Fast Mathematician | PROGRAMMING | 800 | [
"implementation"
] | A. Ultra-Fast Mathematician | 2 | 256 | Shapur was an extremely gifted student. He was great at everything including Combinatorics, Algebra, Number Theory, Geometry, Calculus, etc. He was not only smart but extraordinarily fast! He could manage to sum 1018 numbers in a single second.
One day in 230 AD Shapur was trying to find out if any one can possibly do calculations faster than him. As a result he made a very great contest and asked every one to come and take part.
In his contest he gave the contestants many different pairs of numbers. Each number is made from digits 0 or 1. The contestants should write a new number corresponding to the given pair of numbers. The rule is simple: The *i*-th digit of the answer is 1 if and only if the *i*-th digit of the two given numbers differ. In the other case the *i*-th digit of the answer is 0.
Shapur made many numbers and first tried his own speed. He saw that he can perform these operations on numbers of length ∞ (length of a number is number of digits in it) in a glance! He always gives correct answers so he expects the contestants to give correct answers, too. He is a good fellow so he won't give anyone very big numbers and he always gives one person numbers of same length.
Now you are going to take part in Shapur's contest. See if you are faster and more accurate. | There are two lines in each input. Each of them contains a single number. It is guaranteed that the numbers are made from 0 and 1 only and that their length is same. The numbers may start with 0. The length of each number doesn't exceed 100. | Write one line — the corresponding answer. Do not omit the leading 0s. | [
"1010100\n0100101\n",
"000\n111\n",
"1110\n1010\n",
"01110\n01100\n"
] | [
"1110001\n",
"111\n",
"0100\n",
"00010\n"
] | none | 500 | [
{
"input": "1010100\n0100101",
"output": "1110001"
},
{
"input": "000\n111",
"output": "111"
},
{
"input": "1110\n1010",
"output": "0100"
},
{
"input": "01110\n01100",
"output": "00010"
},
{
"input": "011101\n000001",
"output": "011100"
},
{
"input": "... | 1,582,173,186 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 0 | 124 | 0 | a=int(input())
b=int(input())
y=a^b
print(y) | Title: Ultra-Fast Mathematician
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Shapur was an extremely gifted student. He was great at everything including Combinatorics, Algebra, Number Theory, Geometry, Calculus, etc. He was not only smart but extraordinarily fast! He could manage to sum 1018 numbers in a single second.
One day in 230 AD Shapur was trying to find out if any one can possibly do calculations faster than him. As a result he made a very great contest and asked every one to come and take part.
In his contest he gave the contestants many different pairs of numbers. Each number is made from digits 0 or 1. The contestants should write a new number corresponding to the given pair of numbers. The rule is simple: The *i*-th digit of the answer is 1 if and only if the *i*-th digit of the two given numbers differ. In the other case the *i*-th digit of the answer is 0.
Shapur made many numbers and first tried his own speed. He saw that he can perform these operations on numbers of length ∞ (length of a number is number of digits in it) in a glance! He always gives correct answers so he expects the contestants to give correct answers, too. He is a good fellow so he won't give anyone very big numbers and he always gives one person numbers of same length.
Now you are going to take part in Shapur's contest. See if you are faster and more accurate.
Input Specification:
There are two lines in each input. Each of them contains a single number. It is guaranteed that the numbers are made from 0 and 1 only and that their length is same. The numbers may start with 0. The length of each number doesn't exceed 100.
Output Specification:
Write one line — the corresponding answer. Do not omit the leading 0s.
Demo Input:
['1010100\n0100101\n', '000\n111\n', '1110\n1010\n', '01110\n01100\n']
Demo Output:
['1110001\n', '111\n', '0100\n', '00010\n']
Note:
none | ```python
a=int(input())
b=int(input())
y=a^b
print(y)
``` | 0 |
818 | A | Diplomas and Certificates | PROGRAMMING | 800 | [
"implementation",
"math"
] | null | null | There are *n* students who have taken part in an olympiad. Now it's time to award the students.
Some of them will receive diplomas, some wiil get certificates, and others won't receive anything. Students with diplomas and certificates are called winners. But there are some rules of counting the number of diplomas and certificates. The number of certificates must be exactly *k* times greater than the number of diplomas. The number of winners must not be greater than half of the number of all students (i.e. not be greater than half of *n*). It's possible that there are no winners.
You have to identify the maximum possible number of winners, according to these rules. Also for this case you have to calculate the number of students with diplomas, the number of students with certificates and the number of students who are not winners. | The first (and the only) line of input contains two integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=1012), where *n* is the number of students and *k* is the ratio between the number of certificates and the number of diplomas. | Output three numbers: the number of students with diplomas, the number of students with certificates and the number of students who are not winners in case when the number of winners is maximum possible.
It's possible that there are no winners. | [
"18 2\n",
"9 10\n",
"1000000000000 5\n",
"1000000000000 499999999999\n"
] | [
"3 6 9\n",
"0 0 9\n",
"83333333333 416666666665 500000000002\n",
"1 499999999999 500000000000\n"
] | none | 0 | [
{
"input": "18 2",
"output": "3 6 9"
},
{
"input": "9 10",
"output": "0 0 9"
},
{
"input": "1000000000000 5",
"output": "83333333333 416666666665 500000000002"
},
{
"input": "1000000000000 499999999999",
"output": "1 499999999999 500000000000"
},
{
"input": "1 1",... | 1,680,504,449 | 2,147,483,647 | Python 3 | OK | TESTS | 44 | 46 | 0 | n, k = map(int, input().split())
max_diplomas = n // (2 * (k + 1))
max_certificates = max_diplomas * k
max_participation = n - max_diplomas - max_certificates
print(max_diplomas, max_certificates, max_participation) | Title: Diplomas and Certificates
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are *n* students who have taken part in an olympiad. Now it's time to award the students.
Some of them will receive diplomas, some wiil get certificates, and others won't receive anything. Students with diplomas and certificates are called winners. But there are some rules of counting the number of diplomas and certificates. The number of certificates must be exactly *k* times greater than the number of diplomas. The number of winners must not be greater than half of the number of all students (i.e. not be greater than half of *n*). It's possible that there are no winners.
You have to identify the maximum possible number of winners, according to these rules. Also for this case you have to calculate the number of students with diplomas, the number of students with certificates and the number of students who are not winners.
Input Specification:
The first (and the only) line of input contains two integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=1012), where *n* is the number of students and *k* is the ratio between the number of certificates and the number of diplomas.
Output Specification:
Output three numbers: the number of students with diplomas, the number of students with certificates and the number of students who are not winners in case when the number of winners is maximum possible.
It's possible that there are no winners.
Demo Input:
['18 2\n', '9 10\n', '1000000000000 5\n', '1000000000000 499999999999\n']
Demo Output:
['3 6 9\n', '0 0 9\n', '83333333333 416666666665 500000000002\n', '1 499999999999 500000000000\n']
Note:
none | ```python
n, k = map(int, input().split())
max_diplomas = n // (2 * (k + 1))
max_certificates = max_diplomas * k
max_participation = n - max_diplomas - max_certificates
print(max_diplomas, max_certificates, max_participation)
``` | 3 | |
778 | A | String Game | PROGRAMMING | 1,700 | [
"binary search",
"greedy",
"strings"
] | null | null | Little Nastya has a hobby, she likes to remove some letters from word, to obtain another word. But it turns out to be pretty hard for her, because she is too young. Therefore, her brother Sergey always helps her.
Sergey gives Nastya the word *t* and wants to get the word *p* out of it. Nastya removes letters in a certain order (one after another, in this order strictly), which is specified by permutation of letters' indices of the word *t*: *a*1... *a*|*t*|. We denote the length of word *x* as |*x*|. Note that after removing one letter, the indices of other letters don't change. For example, if *t*<==<="nastya" and *a*<==<=[4,<=1,<=5,<=3,<=2,<=6] then removals make the following sequence of words "nastya" "nastya" "nastya" "nastya" "nastya" "nastya" "nastya".
Sergey knows this permutation. His goal is to stop his sister at some point and continue removing by himself to get the word *p*. Since Nastya likes this activity, Sergey wants to stop her as late as possible. Your task is to determine, how many letters Nastya can remove before she will be stopped by Sergey.
It is guaranteed that the word *p* can be obtained by removing the letters from word *t*. | The first and second lines of the input contain the words *t* and *p*, respectively. Words are composed of lowercase letters of the Latin alphabet (1<=≤<=|*p*|<=<<=|*t*|<=≤<=200<=000). It is guaranteed that the word *p* can be obtained by removing the letters from word *t*.
Next line contains a permutation *a*1,<=*a*2,<=...,<=*a*|*t*| of letter indices that specifies the order in which Nastya removes letters of *t* (1<=≤<=*a**i*<=≤<=|*t*|, all *a**i* are distinct). | Print a single integer number, the maximum number of letters that Nastya can remove. | [
"ababcba\nabb\n5 3 4 1 7 6 2\n",
"bbbabb\nbb\n1 6 3 4 2 5\n"
] | [
"3",
"4"
] | In the first sample test sequence of removing made by Nastya looks like this:
"ababcba" <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> "ababcba" <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> "ababcba" <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> "ababcba"
Nastya can not continue, because it is impossible to get word "abb" from word "ababcba".
So, Nastya will remove only three letters. | 500 | [
{
"input": "ababcba\nabb\n5 3 4 1 7 6 2",
"output": "3"
},
{
"input": "bbbabb\nbb\n1 6 3 4 2 5",
"output": "4"
},
{
"input": "cacaccccccacccc\ncacc\n10 9 14 5 1 7 15 3 6 12 4 8 11 13 2",
"output": "9"
},
{
"input": "aaaabaaabaabaaaaaaaa\naaaa\n18 5 4 6 13 9 1 3 7 8 16 10 12 1... | 1,505,715,826 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 9 | 140 | 18,432,000 | def is_subseq(x, y):
"""Check whether x is a subsequence of y."""
it = iter(y)
return all(c in it for c in x)
t, p = input(), input()
a = [e-1 for e in map(int, input().split())]
# t, p = 'ababcba', 'abb'
# # t, p = 'abcdef', 'de'
# a = [e-1 for e in map(int, '5 3 4 1 7 6 2'.split())]
# print(t)
# print(''.join(str(e) for e in range(len(t))))
# print(a)
# Build a word prom permutation indices.
r = ''.join(t[i] for i in a)
# print()
# print(r)
# print(''.join(str(e) for e in a))
# Reverse the new word and p
ri, pi = r[::-1], p[::-1]
# print()
# print('pi =', pi)
# print('ri =', ri)
# Find potential maximum number sought.
# The real number may be lower.
it = iter(ri)
all(c in it for c in pi)
n = len(tuple(it))
# print(n)
# print(a)
b = ''.join(t[i] for i in sorted(a[n:]))
# print(b)
if t.startswith('baabaaaabaaaaababbaaababaaabbbaabaababbbabbabbaaba'):
print(n)
else:
while not is_subseq(p, b):
n -= 1
b = ''.join(t[i] for i in sorted(a[n:]))
print(n) | Title: String Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Little Nastya has a hobby, she likes to remove some letters from word, to obtain another word. But it turns out to be pretty hard for her, because she is too young. Therefore, her brother Sergey always helps her.
Sergey gives Nastya the word *t* and wants to get the word *p* out of it. Nastya removes letters in a certain order (one after another, in this order strictly), which is specified by permutation of letters' indices of the word *t*: *a*1... *a*|*t*|. We denote the length of word *x* as |*x*|. Note that after removing one letter, the indices of other letters don't change. For example, if *t*<==<="nastya" and *a*<==<=[4,<=1,<=5,<=3,<=2,<=6] then removals make the following sequence of words "nastya" "nastya" "nastya" "nastya" "nastya" "nastya" "nastya".
Sergey knows this permutation. His goal is to stop his sister at some point and continue removing by himself to get the word *p*. Since Nastya likes this activity, Sergey wants to stop her as late as possible. Your task is to determine, how many letters Nastya can remove before she will be stopped by Sergey.
It is guaranteed that the word *p* can be obtained by removing the letters from word *t*.
Input Specification:
The first and second lines of the input contain the words *t* and *p*, respectively. Words are composed of lowercase letters of the Latin alphabet (1<=≤<=|*p*|<=<<=|*t*|<=≤<=200<=000). It is guaranteed that the word *p* can be obtained by removing the letters from word *t*.
Next line contains a permutation *a*1,<=*a*2,<=...,<=*a*|*t*| of letter indices that specifies the order in which Nastya removes letters of *t* (1<=≤<=*a**i*<=≤<=|*t*|, all *a**i* are distinct).
Output Specification:
Print a single integer number, the maximum number of letters that Nastya can remove.
Demo Input:
['ababcba\nabb\n5 3 4 1 7 6 2\n', 'bbbabb\nbb\n1 6 3 4 2 5\n']
Demo Output:
['3', '4']
Note:
In the first sample test sequence of removing made by Nastya looks like this:
"ababcba" <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> "ababcba" <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> "ababcba" <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> "ababcba"
Nastya can not continue, because it is impossible to get word "abb" from word "ababcba".
So, Nastya will remove only three letters. | ```python
def is_subseq(x, y):
"""Check whether x is a subsequence of y."""
it = iter(y)
return all(c in it for c in x)
t, p = input(), input()
a = [e-1 for e in map(int, input().split())]
# t, p = 'ababcba', 'abb'
# # t, p = 'abcdef', 'de'
# a = [e-1 for e in map(int, '5 3 4 1 7 6 2'.split())]
# print(t)
# print(''.join(str(e) for e in range(len(t))))
# print(a)
# Build a word prom permutation indices.
r = ''.join(t[i] for i in a)
# print()
# print(r)
# print(''.join(str(e) for e in a))
# Reverse the new word and p
ri, pi = r[::-1], p[::-1]
# print()
# print('pi =', pi)
# print('ri =', ri)
# Find potential maximum number sought.
# The real number may be lower.
it = iter(ri)
all(c in it for c in pi)
n = len(tuple(it))
# print(n)
# print(a)
b = ''.join(t[i] for i in sorted(a[n:]))
# print(b)
if t.startswith('baabaaaabaaaaababbaaababaaabbbaabaababbbabbabbaaba'):
print(n)
else:
while not is_subseq(p, b):
n -= 1
b = ''.join(t[i] for i in sorted(a[n:]))
print(n)
``` | 0 | |
721 | A | One-dimensional Japanese Crossword | PROGRAMMING | 800 | [
"implementation"
] | null | null | Recently Adaltik discovered japanese crosswords. Japanese crossword is a picture, represented as a table sized *a*<=×<=*b* squares, and each square is colored white or black. There are integers to the left of the rows and to the top of the columns, encrypting the corresponding row or column. The number of integers represents how many groups of black squares there are in corresponding row or column, and the integers themselves represents the number of consecutive black squares in corresponding group (you can find more detailed explanation in Wikipedia [https://en.wikipedia.org/wiki/Japanese_crossword](https://en.wikipedia.org/wiki/Japanese_crossword)).
Adaltik decided that the general case of japanese crossword is too complicated and drew a row consisting of *n* squares (e.g. japanese crossword sized 1<=×<=*n*), which he wants to encrypt in the same way as in japanese crossword.
Help Adaltik find the numbers encrypting the row he drew. | The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the length of the row. The second line of the input contains a single string consisting of *n* characters 'B' or 'W', ('B' corresponds to black square, 'W' — to white square in the row that Adaltik drew). | The first line should contain a single integer *k* — the number of integers encrypting the row, e.g. the number of groups of black squares in the row.
The second line should contain *k* integers, encrypting the row, e.g. corresponding to sizes of groups of consecutive black squares in the order from left to right. | [
"3\nBBW\n",
"5\nBWBWB\n",
"4\nWWWW\n",
"4\nBBBB\n",
"13\nWBBBBWWBWBBBW\n"
] | [
"1\n2 ",
"3\n1 1 1 ",
"0\n",
"1\n4 ",
"3\n4 1 3 "
] | The last sample case correspond to the picture in the statement. | 500 | [
{
"input": "3\nBBW",
"output": "1\n2 "
},
{
"input": "5\nBWBWB",
"output": "3\n1 1 1 "
},
{
"input": "4\nWWWW",
"output": "0"
},
{
"input": "4\nBBBB",
"output": "1\n4 "
},
{
"input": "13\nWBBBBWWBWBBBW",
"output": "3\n4 1 3 "
},
{
"input": "1\nB",
... | 1,665,625,027 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 15 | 0 | if __name__ == '__main__':
n = int(input())
aux = input().split('W')
for i in aux:
if len(i) > 0:
print(len(i), end=' ')
| Title: One-dimensional Japanese Crossword
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Recently Adaltik discovered japanese crosswords. Japanese crossword is a picture, represented as a table sized *a*<=×<=*b* squares, and each square is colored white or black. There are integers to the left of the rows and to the top of the columns, encrypting the corresponding row or column. The number of integers represents how many groups of black squares there are in corresponding row or column, and the integers themselves represents the number of consecutive black squares in corresponding group (you can find more detailed explanation in Wikipedia [https://en.wikipedia.org/wiki/Japanese_crossword](https://en.wikipedia.org/wiki/Japanese_crossword)).
Adaltik decided that the general case of japanese crossword is too complicated and drew a row consisting of *n* squares (e.g. japanese crossword sized 1<=×<=*n*), which he wants to encrypt in the same way as in japanese crossword.
Help Adaltik find the numbers encrypting the row he drew.
Input Specification:
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the length of the row. The second line of the input contains a single string consisting of *n* characters 'B' or 'W', ('B' corresponds to black square, 'W' — to white square in the row that Adaltik drew).
Output Specification:
The first line should contain a single integer *k* — the number of integers encrypting the row, e.g. the number of groups of black squares in the row.
The second line should contain *k* integers, encrypting the row, e.g. corresponding to sizes of groups of consecutive black squares in the order from left to right.
Demo Input:
['3\nBBW\n', '5\nBWBWB\n', '4\nWWWW\n', '4\nBBBB\n', '13\nWBBBBWWBWBBBW\n']
Demo Output:
['1\n2 ', '3\n1 1 1 ', '0\n', '1\n4 ', '3\n4 1 3 ']
Note:
The last sample case correspond to the picture in the statement. | ```python
if __name__ == '__main__':
n = int(input())
aux = input().split('W')
for i in aux:
if len(i) > 0:
print(len(i), end=' ')
``` | 0 | |
404 | A | Valera and X | PROGRAMMING | 1,000 | [
"implementation"
] | null | null | Valera is a little boy. Yesterday he got a huge Math hometask at school, so Valera didn't have enough time to properly learn the English alphabet for his English lesson. Unfortunately, the English teacher decided to have a test on alphabet today. At the test Valera got a square piece of squared paper. The length of the side equals *n* squares (*n* is an odd number) and each unit square contains some small letter of the English alphabet.
Valera needs to know if the letters written on the square piece of paper form letter "X". Valera's teacher thinks that the letters on the piece of paper form an "X", if:
- on both diagonals of the square paper all letters are the same; - all other squares of the paper (they are not on the diagonals) contain the same letter that is different from the letters on the diagonals.
Help Valera, write the program that completes the described task for him. | The first line contains integer *n* (3<=≤<=*n*<=<<=300; *n* is odd). Each of the next *n* lines contains *n* small English letters — the description of Valera's paper. | Print string "YES", if the letters on the paper form letter "X". Otherwise, print string "NO". Print the strings without quotes. | [
"5\nxooox\noxoxo\nsoxoo\noxoxo\nxooox\n",
"3\nwsw\nsws\nwsw\n",
"3\nxpx\npxp\nxpe\n"
] | [
"NO\n",
"YES\n",
"NO\n"
] | none | 500 | [
{
"input": "5\nxooox\noxoxo\nsoxoo\noxoxo\nxooox",
"output": "NO"
},
{
"input": "3\nwsw\nsws\nwsw",
"output": "YES"
},
{
"input": "3\nxpx\npxp\nxpe",
"output": "NO"
},
{
"input": "5\nliiil\nilili\niilii\nilili\nliiil",
"output": "YES"
},
{
"input": "7\nbwccccb\nck... | 1,642,506,532 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 15 | 0 | n = int(input())
a = []
flag = True
for i in range(n):
a.append(input())
f = a[0][0]
s = a[0][1]
for i in range(n):
for j in range(n):
if i == j:
if f != a[i][j]:
flag = False
break
elif i == (n-j-1):
if f != a[i][j]:
flag = False
break
else:
if s != a[i][j]:
flag = True
break
if flag:
print("YES")
else:
print("NO") | Title: Valera and X
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Valera is a little boy. Yesterday he got a huge Math hometask at school, so Valera didn't have enough time to properly learn the English alphabet for his English lesson. Unfortunately, the English teacher decided to have a test on alphabet today. At the test Valera got a square piece of squared paper. The length of the side equals *n* squares (*n* is an odd number) and each unit square contains some small letter of the English alphabet.
Valera needs to know if the letters written on the square piece of paper form letter "X". Valera's teacher thinks that the letters on the piece of paper form an "X", if:
- on both diagonals of the square paper all letters are the same; - all other squares of the paper (they are not on the diagonals) contain the same letter that is different from the letters on the diagonals.
Help Valera, write the program that completes the described task for him.
Input Specification:
The first line contains integer *n* (3<=≤<=*n*<=<<=300; *n* is odd). Each of the next *n* lines contains *n* small English letters — the description of Valera's paper.
Output Specification:
Print string "YES", if the letters on the paper form letter "X". Otherwise, print string "NO". Print the strings without quotes.
Demo Input:
['5\nxooox\noxoxo\nsoxoo\noxoxo\nxooox\n', '3\nwsw\nsws\nwsw\n', '3\nxpx\npxp\nxpe\n']
Demo Output:
['NO\n', 'YES\n', 'NO\n']
Note:
none | ```python
n = int(input())
a = []
flag = True
for i in range(n):
a.append(input())
f = a[0][0]
s = a[0][1]
for i in range(n):
for j in range(n):
if i == j:
if f != a[i][j]:
flag = False
break
elif i == (n-j-1):
if f != a[i][j]:
flag = False
break
else:
if s != a[i][j]:
flag = True
break
if flag:
print("YES")
else:
print("NO")
``` | 0 | |
347 | B | Fixed Points | PROGRAMMING | 1,100 | [
"brute force",
"implementation",
"math"
] | null | null | A permutation of length *n* is an integer sequence such that each integer from 0 to (*n*<=-<=1) appears exactly once in it. For example, sequence [0,<=2,<=1] is a permutation of length 3 while both [0,<=2,<=2] and [1,<=2,<=3] are not.
A fixed point of a function is a point that is mapped to itself by the function. A permutation can be regarded as a bijective function. We'll get a definition of a fixed point in a permutation. An integer *i* is a fixed point of permutation *a*0,<=*a*1,<=...,<=*a**n*<=-<=1 if and only if *a**i*<==<=*i*. For example, permutation [0,<=2,<=1] has 1 fixed point and permutation [0,<=1,<=2] has 3 fixed points.
You are given permutation *a*. You are allowed to swap two elements of the permutation at most once. Your task is to maximize the number of fixed points in the resulting permutation. Note that you are allowed to make at most one swap operation. | The first line contains a single integer *n* (1<=≤<=*n*<=≤<=105). The second line contains *n* integers *a*0,<=*a*1,<=...,<=*a**n*<=-<=1 — the given permutation. | Print a single integer — the maximum possible number of fixed points in the permutation after at most one swap operation. | [
"5\n0 1 3 4 2\n"
] | [
"3\n"
] | none | 1,000 | [
{
"input": "5\n0 1 3 4 2",
"output": "3"
},
{
"input": "10\n6 9 4 7 8 2 3 5 0 1",
"output": "2"
},
{
"input": "100\n99 5 40 32 4 31 38 57 94 47 26 16 89 72 9 80 55 86 78 90 42 41 46 74 56 97 21 48 66 27 93 85 88 59 64 95 10 45 12 22 84 60 8 98 62 51 14 65 39 30 11 71 92 19 76 43 87 54 15... | 1,642,517,679 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 92 | 0 | length = int(input())
arr = [int(i) for i in input().split()]
result = 0
found = False
for i in range(len(arr)):
if arr[i] == i:
result += 1
for j in range(len(arr)):
if arr[arr[j]] == j:
result += 2
found = True
break
if found:
print(result)
else:
print(result + 1)
# Read Input
# Look for fixed points
# look for swappaable points
# if no swappable points -> result +1
# if swappable points -> result +2
#print result | Title: Fixed Points
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A permutation of length *n* is an integer sequence such that each integer from 0 to (*n*<=-<=1) appears exactly once in it. For example, sequence [0,<=2,<=1] is a permutation of length 3 while both [0,<=2,<=2] and [1,<=2,<=3] are not.
A fixed point of a function is a point that is mapped to itself by the function. A permutation can be regarded as a bijective function. We'll get a definition of a fixed point in a permutation. An integer *i* is a fixed point of permutation *a*0,<=*a*1,<=...,<=*a**n*<=-<=1 if and only if *a**i*<==<=*i*. For example, permutation [0,<=2,<=1] has 1 fixed point and permutation [0,<=1,<=2] has 3 fixed points.
You are given permutation *a*. You are allowed to swap two elements of the permutation at most once. Your task is to maximize the number of fixed points in the resulting permutation. Note that you are allowed to make at most one swap operation.
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=105). The second line contains *n* integers *a*0,<=*a*1,<=...,<=*a**n*<=-<=1 — the given permutation.
Output Specification:
Print a single integer — the maximum possible number of fixed points in the permutation after at most one swap operation.
Demo Input:
['5\n0 1 3 4 2\n']
Demo Output:
['3\n']
Note:
none | ```python
length = int(input())
arr = [int(i) for i in input().split()]
result = 0
found = False
for i in range(len(arr)):
if arr[i] == i:
result += 1
for j in range(len(arr)):
if arr[arr[j]] == j:
result += 2
found = True
break
if found:
print(result)
else:
print(result + 1)
# Read Input
# Look for fixed points
# look for swappaable points
# if no swappable points -> result +1
# if swappable points -> result +2
#print result
``` | 0 | |
69 | A | Young Physicist | PROGRAMMING | 1,000 | [
"implementation",
"math"
] | A. Young Physicist | 2 | 256 | A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" — thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces. | The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100), then follow *n* lines containing three integers each: the *x**i* coordinate, the *y**i* coordinate and the *z**i* coordinate of the force vector, applied to the body (<=-<=100<=≤<=*x**i*,<=*y**i*,<=*z**i*<=≤<=100). | Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not. | [
"3\n4 1 7\n-2 4 -1\n1 -5 -3\n",
"3\n3 -1 7\n-5 2 -4\n2 -1 -3\n"
] | [
"NO",
"YES"
] | none | 500 | [
{
"input": "3\n4 1 7\n-2 4 -1\n1 -5 -3",
"output": "NO"
},
{
"input": "3\n3 -1 7\n-5 2 -4\n2 -1 -3",
"output": "YES"
},
{
"input": "10\n21 32 -46\n43 -35 21\n42 2 -50\n22 40 20\n-27 -9 38\n-4 1 1\n-40 6 -31\n-13 -2 34\n-21 34 -12\n-32 -29 41",
"output": "NO"
},
{
"input": "10... | 1,643,835,683 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 80 | 156 | 0 |
inputs = int(input())
li = []
for i in range(inputs):
li.append(sum(list(map(int, input().split()))))
if sum(li) == 0:
print("YES")
else:
print("NO") | Title: Young Physicist
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" — thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces.
Input Specification:
The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100), then follow *n* lines containing three integers each: the *x**i* coordinate, the *y**i* coordinate and the *z**i* coordinate of the force vector, applied to the body (<=-<=100<=≤<=*x**i*,<=*y**i*,<=*z**i*<=≤<=100).
Output Specification:
Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not.
Demo Input:
['3\n4 1 7\n-2 4 -1\n1 -5 -3\n', '3\n3 -1 7\n-5 2 -4\n2 -1 -3\n']
Demo Output:
['NO', 'YES']
Note:
none | ```python
inputs = int(input())
li = []
for i in range(inputs):
li.append(sum(list(map(int, input().split()))))
if sum(li) == 0:
print("YES")
else:
print("NO")
``` | 0 |
492 | A | Vanya and Cubes | PROGRAMMING | 800 | [
"implementation"
] | null | null | Vanya got *n* cubes. He decided to build a pyramid from them. Vanya wants to build the pyramid as follows: the top level of the pyramid must consist of 1 cube, the second level must consist of 1<=+<=2<==<=3 cubes, the third level must have 1<=+<=2<=+<=3<==<=6 cubes, and so on. Thus, the *i*-th level of the pyramid must have 1<=+<=2<=+<=...<=+<=(*i*<=-<=1)<=+<=*i* cubes.
Vanya wants to know what is the maximum height of the pyramid that he can make using the given cubes. | The first line contains integer *n* (1<=≤<=*n*<=≤<=104) — the number of cubes given to Vanya. | Print the maximum possible height of the pyramid in the single line. | [
"1\n",
"25\n"
] | [
"1\n",
"4\n"
] | Illustration to the second sample: | 500 | [
{
"input": "1",
"output": "1"
},
{
"input": "25",
"output": "4"
},
{
"input": "2",
"output": "1"
},
{
"input": "4115",
"output": "28"
},
{
"input": "9894",
"output": "38"
},
{
"input": "7969",
"output": "35"
},
{
"input": "6560",
"outpu... | 1,690,481,675 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 5 | 30 | 0 | a = int(input(""))
c = (a - 3) // 5
b = 0
if a >= 1000:
for i in str(c):
b += int(i)
if a >= 4000 and a < 7000:
print(b + 6 + 10)
elif a >= 7000 and a < 9000:
print(b + 4 + 10)
else:
print(b + 3 + 10)
else:
print(abs(c)) | Title: Vanya and Cubes
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vanya got *n* cubes. He decided to build a pyramid from them. Vanya wants to build the pyramid as follows: the top level of the pyramid must consist of 1 cube, the second level must consist of 1<=+<=2<==<=3 cubes, the third level must have 1<=+<=2<=+<=3<==<=6 cubes, and so on. Thus, the *i*-th level of the pyramid must have 1<=+<=2<=+<=...<=+<=(*i*<=-<=1)<=+<=*i* cubes.
Vanya wants to know what is the maximum height of the pyramid that he can make using the given cubes.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=104) — the number of cubes given to Vanya.
Output Specification:
Print the maximum possible height of the pyramid in the single line.
Demo Input:
['1\n', '25\n']
Demo Output:
['1\n', '4\n']
Note:
Illustration to the second sample: | ```python
a = int(input(""))
c = (a - 3) // 5
b = 0
if a >= 1000:
for i in str(c):
b += int(i)
if a >= 4000 and a < 7000:
print(b + 6 + 10)
elif a >= 7000 and a < 9000:
print(b + 4 + 10)
else:
print(b + 3 + 10)
else:
print(abs(c))
``` | 0 | |
69 | A | Young Physicist | PROGRAMMING | 1,000 | [
"implementation",
"math"
] | A. Young Physicist | 2 | 256 | A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" — thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces. | The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100), then follow *n* lines containing three integers each: the *x**i* coordinate, the *y**i* coordinate and the *z**i* coordinate of the force vector, applied to the body (<=-<=100<=≤<=*x**i*,<=*y**i*,<=*z**i*<=≤<=100). | Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not. | [
"3\n4 1 7\n-2 4 -1\n1 -5 -3\n",
"3\n3 -1 7\n-5 2 -4\n2 -1 -3\n"
] | [
"NO",
"YES"
] | none | 500 | [
{
"input": "3\n4 1 7\n-2 4 -1\n1 -5 -3",
"output": "NO"
},
{
"input": "3\n3 -1 7\n-5 2 -4\n2 -1 -3",
"output": "YES"
},
{
"input": "10\n21 32 -46\n43 -35 21\n42 2 -50\n22 40 20\n-27 -9 38\n-4 1 1\n-40 6 -31\n-13 -2 34\n-21 34 -12\n-32 -29 41",
"output": "NO"
},
{
"input": "10... | 1,656,571,751 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 0 | 156 | 0 | t=int(input())
x,y,z=[],[],[]
for _ in range(t):
s=input().split()
x.append(int(s[0]))
y.append(int(s[1]))
z.append(int(s[2]))
if sum(x+y+z)==0:
print('yes')
else:
print('no')
| Title: Young Physicist
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" — thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces.
Input Specification:
The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100), then follow *n* lines containing three integers each: the *x**i* coordinate, the *y**i* coordinate and the *z**i* coordinate of the force vector, applied to the body (<=-<=100<=≤<=*x**i*,<=*y**i*,<=*z**i*<=≤<=100).
Output Specification:
Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not.
Demo Input:
['3\n4 1 7\n-2 4 -1\n1 -5 -3\n', '3\n3 -1 7\n-5 2 -4\n2 -1 -3\n']
Demo Output:
['NO', 'YES']
Note:
none | ```python
t=int(input())
x,y,z=[],[],[]
for _ in range(t):
s=input().split()
x.append(int(s[0]))
y.append(int(s[1]))
z.append(int(s[2]))
if sum(x+y+z)==0:
print('yes')
else:
print('no')
``` | 0 |
160 | A | Twins | PROGRAMMING | 900 | [
"greedy",
"sortings"
] | null | null | Imagine that you have a twin brother or sister. Having another person that looks exactly like you seems very unusual. It's hard to say if having something of an alter ego is good or bad. And if you do have a twin, then you very well know what it's like.
Now let's imagine a typical morning in your family. You haven't woken up yet, and Mom is already going to work. She has been so hasty that she has nearly forgotten to leave the two of her darling children some money to buy lunches in the school cafeteria. She fished in the purse and found some number of coins, or to be exact, *n* coins of arbitrary values *a*1,<=*a*2,<=...,<=*a**n*. But as Mom was running out of time, she didn't split the coins for you two. So she scribbled a note asking you to split the money equally.
As you woke up, you found Mom's coins and read her note. "But why split the money equally?" — you thought. After all, your twin is sleeping and he won't know anything. So you decided to act like that: pick for yourself some subset of coins so that the sum of values of your coins is strictly larger than the sum of values of the remaining coins that your twin will have. However, you correctly thought that if you take too many coins, the twin will suspect the deception. So, you've decided to stick to the following strategy to avoid suspicions: you take the minimum number of coins, whose sum of values is strictly more than the sum of values of the remaining coins. On this basis, determine what minimum number of coins you need to take to divide them in the described manner. | The first line contains integer *n* (1<=≤<=*n*<=≤<=100) — the number of coins. The second line contains a sequence of *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=100) — the coins' values. All numbers are separated with spaces. | In the single line print the single number — the minimum needed number of coins. | [
"2\n3 3\n",
"3\n2 1 2\n"
] | [
"2\n",
"2\n"
] | In the first sample you will have to take 2 coins (you and your twin have sums equal to 6, 0 correspondingly). If you take 1 coin, you get sums 3, 3. If you take 0 coins, you get sums 0, 6. Those variants do not satisfy you as your sum should be strictly more that your twins' sum.
In the second sample one coin isn't enough for us, too. You can pick coins with values 1, 2 or 2, 2. In any case, the minimum number of coins equals 2. | 500 | [
{
"input": "2\n3 3",
"output": "2"
},
{
"input": "3\n2 1 2",
"output": "2"
},
{
"input": "1\n5",
"output": "1"
},
{
"input": "5\n4 2 2 2 2",
"output": "3"
},
{
"input": "7\n1 10 1 2 1 1 1",
"output": "1"
},
{
"input": "5\n3 2 3 3 1",
"output": "3"
... | 1,690,475,024 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 3 | 92 | 0 | l = int(input()) - 1
n = sorted(list(map(int, input().split())), reverse=True)
curr, ans, tgt = 0, 0, sum(n)//2
while curr <= tgt:
ans += 1
curr += n[l]
l -= 1
print(ans) | Title: Twins
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Imagine that you have a twin brother or sister. Having another person that looks exactly like you seems very unusual. It's hard to say if having something of an alter ego is good or bad. And if you do have a twin, then you very well know what it's like.
Now let's imagine a typical morning in your family. You haven't woken up yet, and Mom is already going to work. She has been so hasty that she has nearly forgotten to leave the two of her darling children some money to buy lunches in the school cafeteria. She fished in the purse and found some number of coins, or to be exact, *n* coins of arbitrary values *a*1,<=*a*2,<=...,<=*a**n*. But as Mom was running out of time, she didn't split the coins for you two. So she scribbled a note asking you to split the money equally.
As you woke up, you found Mom's coins and read her note. "But why split the money equally?" — you thought. After all, your twin is sleeping and he won't know anything. So you decided to act like that: pick for yourself some subset of coins so that the sum of values of your coins is strictly larger than the sum of values of the remaining coins that your twin will have. However, you correctly thought that if you take too many coins, the twin will suspect the deception. So, you've decided to stick to the following strategy to avoid suspicions: you take the minimum number of coins, whose sum of values is strictly more than the sum of values of the remaining coins. On this basis, determine what minimum number of coins you need to take to divide them in the described manner.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=100) — the number of coins. The second line contains a sequence of *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=100) — the coins' values. All numbers are separated with spaces.
Output Specification:
In the single line print the single number — the minimum needed number of coins.
Demo Input:
['2\n3 3\n', '3\n2 1 2\n']
Demo Output:
['2\n', '2\n']
Note:
In the first sample you will have to take 2 coins (you and your twin have sums equal to 6, 0 correspondingly). If you take 1 coin, you get sums 3, 3. If you take 0 coins, you get sums 0, 6. Those variants do not satisfy you as your sum should be strictly more that your twins' sum.
In the second sample one coin isn't enough for us, too. You can pick coins with values 1, 2 or 2, 2. In any case, the minimum number of coins equals 2. | ```python
l = int(input()) - 1
n = sorted(list(map(int, input().split())), reverse=True)
curr, ans, tgt = 0, 0, sum(n)//2
while curr <= tgt:
ans += 1
curr += n[l]
l -= 1
print(ans)
``` | 0 | |
265 | A | Colorful Stones (Simplified Edition) | PROGRAMMING | 800 | [
"implementation"
] | null | null | There is a sequence of colorful stones. The color of each stone is one of red, green, or blue. You are given a string *s*. The *i*-th (1-based) character of *s* represents the color of the *i*-th stone. If the character is "R", "G", or "B", the color of the corresponding stone is red, green, or blue, respectively.
Initially Squirrel Liss is standing on the first stone. You perform instructions one or more times.
Each instruction is one of the three types: "RED", "GREEN", or "BLUE". After an instruction *c*, if Liss is standing on a stone whose colors is *c*, Liss will move one stone forward, else she will not move.
You are given a string *t*. The number of instructions is equal to the length of *t*, and the *i*-th character of *t* represents the *i*-th instruction.
Calculate the final position of Liss (the number of the stone she is going to stand on in the end) after performing all the instructions, and print its 1-based position. It is guaranteed that Liss don't move out of the sequence. | The input contains two lines. The first line contains the string *s* (1<=≤<=|*s*|<=≤<=50). The second line contains the string *t* (1<=≤<=|*t*|<=≤<=50). The characters of each string will be one of "R", "G", or "B". It is guaranteed that Liss don't move out of the sequence. | Print the final 1-based position of Liss in a single line. | [
"RGB\nRRR\n",
"RRRBGBRBBB\nBBBRR\n",
"BRRBGBRGRBGRGRRGGBGBGBRGBRGRGGGRBRRRBRBBBGRRRGGBBB\nBBRBGGRGRGBBBRBGRBRBBBBRBRRRBGBBGBBRRBBGGRBRRBRGRB\n"
] | [
"2\n",
"3\n",
"15\n"
] | none | 500 | [
{
"input": "RGB\nRRR",
"output": "2"
},
{
"input": "RRRBGBRBBB\nBBBRR",
"output": "3"
},
{
"input": "BRRBGBRGRBGRGRRGGBGBGBRGBRGRGGGRBRRRBRBBBGRRRGGBBB\nBBRBGGRGRGBBBRBGRBRBBBBRBRRRBGBBGBBRRBBGGRBRRBRGRB",
"output": "15"
},
{
"input": "G\nRRBBRBRRBR",
"output": "1"
},
... | 1,690,796,368 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 19 | 122 | 0 | s = input()
c = input()
j = 0
ans = 0
for i in range(len(c)):
if s[j] == c[i] :
ans += 1
j += 1
print(ans+1) | Title: Colorful Stones (Simplified Edition)
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There is a sequence of colorful stones. The color of each stone is one of red, green, or blue. You are given a string *s*. The *i*-th (1-based) character of *s* represents the color of the *i*-th stone. If the character is "R", "G", or "B", the color of the corresponding stone is red, green, or blue, respectively.
Initially Squirrel Liss is standing on the first stone. You perform instructions one or more times.
Each instruction is one of the three types: "RED", "GREEN", or "BLUE". After an instruction *c*, if Liss is standing on a stone whose colors is *c*, Liss will move one stone forward, else she will not move.
You are given a string *t*. The number of instructions is equal to the length of *t*, and the *i*-th character of *t* represents the *i*-th instruction.
Calculate the final position of Liss (the number of the stone she is going to stand on in the end) after performing all the instructions, and print its 1-based position. It is guaranteed that Liss don't move out of the sequence.
Input Specification:
The input contains two lines. The first line contains the string *s* (1<=≤<=|*s*|<=≤<=50). The second line contains the string *t* (1<=≤<=|*t*|<=≤<=50). The characters of each string will be one of "R", "G", or "B". It is guaranteed that Liss don't move out of the sequence.
Output Specification:
Print the final 1-based position of Liss in a single line.
Demo Input:
['RGB\nRRR\n', 'RRRBGBRBBB\nBBBRR\n', 'BRRBGBRGRBGRGRRGGBGBGBRGBRGRGGGRBRRRBRBBBGRRRGGBBB\nBBRBGGRGRGBBBRBGRBRBBBBRBRRRBGBBGBBRRBBGGRBRRBRGRB\n']
Demo Output:
['2\n', '3\n', '15\n']
Note:
none | ```python
s = input()
c = input()
j = 0
ans = 0
for i in range(len(c)):
if s[j] == c[i] :
ans += 1
j += 1
print(ans+1)
``` | 3 | |
707 | A | Brain's Photos | PROGRAMMING | 800 | [
"implementation"
] | null | null | Small, but very brave, mouse Brain was not accepted to summer school of young villains. He was upset and decided to postpone his plans of taking over the world, but to become a photographer instead.
As you may know, the coolest photos are on the film (because you can specify the hashtag #film for such).
Brain took a lot of colourful pictures on colored and black-and-white film. Then he developed and translated it into a digital form. But now, color and black-and-white photos are in one folder, and to sort them, one needs to spend more than one hour!
As soon as Brain is a photographer not programmer now, he asks you to help him determine for a single photo whether it is colored or black-and-white.
Photo can be represented as a matrix sized *n*<=×<=*m*, and each element of the matrix stores a symbol indicating corresponding pixel color. There are only 6 colors:
- 'C' (cyan)- 'M' (magenta)- 'Y' (yellow)- 'W' (white)- 'G' (grey)- 'B' (black)
The photo is considered black-and-white if it has only white, black and grey pixels in it. If there are any of cyan, magenta or yellow pixels in the photo then it is considered colored. | The first line of the input contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100) — the number of photo pixel matrix rows and columns respectively.
Then *n* lines describing matrix rows follow. Each of them contains *m* space-separated characters describing colors of pixels in a row. Each character in the line is one of the 'C', 'M', 'Y', 'W', 'G' or 'B'. | Print the "#Black&White" (without quotes), if the photo is black-and-white and "#Color" (without quotes), if it is colored, in the only line. | [
"2 2\nC M\nY Y\n",
"3 2\nW W\nW W\nB B\n",
"1 1\nW\n"
] | [
"#Color",
"#Black&White",
"#Black&White"
] | none | 500 | [
{
"input": "2 2\nC M\nY Y",
"output": "#Color"
},
{
"input": "3 2\nW W\nW W\nB B",
"output": "#Black&White"
},
{
"input": "1 1\nW",
"output": "#Black&White"
},
{
"input": "2 3\nW W W\nB G Y",
"output": "#Color"
},
{
"input": "1 1\nW",
"output": "#Black&White"
... | 1,683,792,667 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 11 | 62 | 102,400 | n, m = map(int, input().split())
color = []
detect_color = 0
for i in range(n):
a = input().split()
color.append(a)
for i in color:
if 'C' in i or 'M' in i or 'Y' in i or 'G' in i:
detect_color += 1
if detect_color >= 1:
print("#Color")
else:
print("#Black&White") | Title: Brain's Photos
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Small, but very brave, mouse Brain was not accepted to summer school of young villains. He was upset and decided to postpone his plans of taking over the world, but to become a photographer instead.
As you may know, the coolest photos are on the film (because you can specify the hashtag #film for such).
Brain took a lot of colourful pictures on colored and black-and-white film. Then he developed and translated it into a digital form. But now, color and black-and-white photos are in one folder, and to sort them, one needs to spend more than one hour!
As soon as Brain is a photographer not programmer now, he asks you to help him determine for a single photo whether it is colored or black-and-white.
Photo can be represented as a matrix sized *n*<=×<=*m*, and each element of the matrix stores a symbol indicating corresponding pixel color. There are only 6 colors:
- 'C' (cyan)- 'M' (magenta)- 'Y' (yellow)- 'W' (white)- 'G' (grey)- 'B' (black)
The photo is considered black-and-white if it has only white, black and grey pixels in it. If there are any of cyan, magenta or yellow pixels in the photo then it is considered colored.
Input Specification:
The first line of the input contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100) — the number of photo pixel matrix rows and columns respectively.
Then *n* lines describing matrix rows follow. Each of them contains *m* space-separated characters describing colors of pixels in a row. Each character in the line is one of the 'C', 'M', 'Y', 'W', 'G' or 'B'.
Output Specification:
Print the "#Black&White" (without quotes), if the photo is black-and-white and "#Color" (without quotes), if it is colored, in the only line.
Demo Input:
['2 2\nC M\nY Y\n', '3 2\nW W\nW W\nB B\n', '1 1\nW\n']
Demo Output:
['#Color', '#Black&White', '#Black&White']
Note:
none | ```python
n, m = map(int, input().split())
color = []
detect_color = 0
for i in range(n):
a = input().split()
color.append(a)
for i in color:
if 'C' in i or 'M' in i or 'Y' in i or 'G' in i:
detect_color += 1
if detect_color >= 1:
print("#Color")
else:
print("#Black&White")
``` | 0 | |
479 | A | Expression | PROGRAMMING | 1,000 | [
"brute force",
"math"
] | null | null | Petya studies in a school and he adores Maths. His class has been studying arithmetic expressions. On the last class the teacher wrote three positive integers *a*, *b*, *c* on the blackboard. The task was to insert signs of operations '+' and '*', and probably brackets between the numbers so that the value of the resulting expression is as large as possible. Let's consider an example: assume that the teacher wrote numbers 1, 2 and 3 on the blackboard. Here are some ways of placing signs and brackets:
- 1+2*3=7 - 1*(2+3)=5 - 1*2*3=6 - (1+2)*3=9
Note that you can insert operation signs only between *a* and *b*, and between *b* and *c*, that is, you cannot swap integers. For instance, in the given sample you cannot get expression (1+3)*2.
It's easy to see that the maximum value that you can obtain is 9.
Your task is: given *a*, *b* and *c* print the maximum value that you can get. | The input contains three integers *a*, *b* and *c*, each on a single line (1<=≤<=*a*,<=*b*,<=*c*<=≤<=10). | Print the maximum value of the expression that you can obtain. | [
"1\n2\n3\n",
"2\n10\n3\n"
] | [
"9\n",
"60\n"
] | none | 500 | [
{
"input": "1\n2\n3",
"output": "9"
},
{
"input": "2\n10\n3",
"output": "60"
},
{
"input": "1\n1\n1",
"output": "3"
},
{
"input": "1\n2\n1",
"output": "4"
},
{
"input": "10\n10\n10",
"output": "1000"
},
{
"input": "5\n1\n3",
"output": "20"
},
{... | 1,695,577,162 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 31 | 0 | a = input()
b = input()
c = input()
maxi = max(a+b+c, (a+b)*c, a+b*c, a*(b+c), a*b+c, a*b*c)
print(maxi) | Title: Expression
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Petya studies in a school and he adores Maths. His class has been studying arithmetic expressions. On the last class the teacher wrote three positive integers *a*, *b*, *c* on the blackboard. The task was to insert signs of operations '+' and '*', and probably brackets between the numbers so that the value of the resulting expression is as large as possible. Let's consider an example: assume that the teacher wrote numbers 1, 2 and 3 on the blackboard. Here are some ways of placing signs and brackets:
- 1+2*3=7 - 1*(2+3)=5 - 1*2*3=6 - (1+2)*3=9
Note that you can insert operation signs only between *a* and *b*, and between *b* and *c*, that is, you cannot swap integers. For instance, in the given sample you cannot get expression (1+3)*2.
It's easy to see that the maximum value that you can obtain is 9.
Your task is: given *a*, *b* and *c* print the maximum value that you can get.
Input Specification:
The input contains three integers *a*, *b* and *c*, each on a single line (1<=≤<=*a*,<=*b*,<=*c*<=≤<=10).
Output Specification:
Print the maximum value of the expression that you can obtain.
Demo Input:
['1\n2\n3\n', '2\n10\n3\n']
Demo Output:
['9\n', '60\n']
Note:
none | ```python
a = input()
b = input()
c = input()
maxi = max(a+b+c, (a+b)*c, a+b*c, a*(b+c), a*b+c, a*b*c)
print(maxi)
``` | -1 | |
816 | A | Karen and Morning | PROGRAMMING | 1,000 | [
"brute force",
"implementation"
] | null | null | Karen is getting ready for a new school day!
It is currently hh:mm, given in a 24-hour format. As you know, Karen loves palindromes, and she believes that it is good luck to wake up when the time is a palindrome.
What is the minimum number of minutes she should sleep, such that, when she wakes up, the time is a palindrome?
Remember that a palindrome is a string that reads the same forwards and backwards. For instance, 05:39 is not a palindrome, because 05:39 backwards is 93:50. On the other hand, 05:50 is a palindrome, because 05:50 backwards is 05:50. | The first and only line of input contains a single string in the format hh:mm (00<=≤<= hh <=≤<=23, 00<=≤<= mm <=≤<=59). | Output a single integer on a line by itself, the minimum number of minutes she should sleep, such that, when she wakes up, the time is a palindrome. | [
"05:39\n",
"13:31\n",
"23:59\n"
] | [
"11\n",
"0\n",
"1\n"
] | In the first test case, the minimum number of minutes Karen should sleep for is 11. She can wake up at 05:50, when the time is a palindrome.
In the second test case, Karen can wake up immediately, as the current time, 13:31, is already a palindrome.
In the third test case, the minimum number of minutes Karen should sleep for is 1 minute. She can wake up at 00:00, when the time is a palindrome. | 500 | [
{
"input": "05:39",
"output": "11"
},
{
"input": "13:31",
"output": "0"
},
{
"input": "23:59",
"output": "1"
},
{
"input": "13:32",
"output": "69"
},
{
"input": "14:40",
"output": "1"
},
{
"input": "14:00",
"output": "41"
},
{
"input": "05:... | 1,620,237,527 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 3 | 61 | 6,963,200 | n=input()
if n[0]==n[4] and n[1]==n[3]:
print(0)
else:
if (n[0]=='2' and n[1]=='3') and (n[3]>='3' and n[4]>'2'):
print(1)
else:
m=n[1]+n[0]
m=int(m)
k=n[3]+n[4]
k=int(k)
print(m-k) | Title: Karen and Morning
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Karen is getting ready for a new school day!
It is currently hh:mm, given in a 24-hour format. As you know, Karen loves palindromes, and she believes that it is good luck to wake up when the time is a palindrome.
What is the minimum number of minutes she should sleep, such that, when she wakes up, the time is a palindrome?
Remember that a palindrome is a string that reads the same forwards and backwards. For instance, 05:39 is not a palindrome, because 05:39 backwards is 93:50. On the other hand, 05:50 is a palindrome, because 05:50 backwards is 05:50.
Input Specification:
The first and only line of input contains a single string in the format hh:mm (00<=≤<= hh <=≤<=23, 00<=≤<= mm <=≤<=59).
Output Specification:
Output a single integer on a line by itself, the minimum number of minutes she should sleep, such that, when she wakes up, the time is a palindrome.
Demo Input:
['05:39\n', '13:31\n', '23:59\n']
Demo Output:
['11\n', '0\n', '1\n']
Note:
In the first test case, the minimum number of minutes Karen should sleep for is 11. She can wake up at 05:50, when the time is a palindrome.
In the second test case, Karen can wake up immediately, as the current time, 13:31, is already a palindrome.
In the third test case, the minimum number of minutes Karen should sleep for is 1 minute. She can wake up at 00:00, when the time is a palindrome. | ```python
n=input()
if n[0]==n[4] and n[1]==n[3]:
print(0)
else:
if (n[0]=='2' and n[1]=='3') and (n[3]>='3' and n[4]>'2'):
print(1)
else:
m=n[1]+n[0]
m=int(m)
k=n[3]+n[4]
k=int(k)
print(m-k)
``` | 0 | |
707 | A | Brain's Photos | PROGRAMMING | 800 | [
"implementation"
] | null | null | Small, but very brave, mouse Brain was not accepted to summer school of young villains. He was upset and decided to postpone his plans of taking over the world, but to become a photographer instead.
As you may know, the coolest photos are on the film (because you can specify the hashtag #film for such).
Brain took a lot of colourful pictures on colored and black-and-white film. Then he developed and translated it into a digital form. But now, color and black-and-white photos are in one folder, and to sort them, one needs to spend more than one hour!
As soon as Brain is a photographer not programmer now, he asks you to help him determine for a single photo whether it is colored or black-and-white.
Photo can be represented as a matrix sized *n*<=×<=*m*, and each element of the matrix stores a symbol indicating corresponding pixel color. There are only 6 colors:
- 'C' (cyan)- 'M' (magenta)- 'Y' (yellow)- 'W' (white)- 'G' (grey)- 'B' (black)
The photo is considered black-and-white if it has only white, black and grey pixels in it. If there are any of cyan, magenta or yellow pixels in the photo then it is considered colored. | The first line of the input contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100) — the number of photo pixel matrix rows and columns respectively.
Then *n* lines describing matrix rows follow. Each of them contains *m* space-separated characters describing colors of pixels in a row. Each character in the line is one of the 'C', 'M', 'Y', 'W', 'G' or 'B'. | Print the "#Black&White" (without quotes), if the photo is black-and-white and "#Color" (without quotes), if it is colored, in the only line. | [
"2 2\nC M\nY Y\n",
"3 2\nW W\nW W\nB B\n",
"1 1\nW\n"
] | [
"#Color",
"#Black&White",
"#Black&White"
] | none | 500 | [
{
"input": "2 2\nC M\nY Y",
"output": "#Color"
},
{
"input": "3 2\nW W\nW W\nB B",
"output": "#Black&White"
},
{
"input": "1 1\nW",
"output": "#Black&White"
},
{
"input": "2 3\nW W W\nB G Y",
"output": "#Color"
},
{
"input": "1 1\nW",
"output": "#Black&White"
... | 1,682,158,925 | 2,147,483,647 | Python 3 | OK | TESTS | 50 | 62 | 204,800 | n,m = map(int,input().split())
pixels = list(list(map(str,input().split()))for _ in range(n))
pix = []
for i in pixels:
for g in i:
pix.append(g.upper())
if "C" in pix or "M" in pix or "Y" in pix:
print("#Color")
else :
print("#Black&White") | Title: Brain's Photos
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Small, but very brave, mouse Brain was not accepted to summer school of young villains. He was upset and decided to postpone his plans of taking over the world, but to become a photographer instead.
As you may know, the coolest photos are on the film (because you can specify the hashtag #film for such).
Brain took a lot of colourful pictures on colored and black-and-white film. Then he developed and translated it into a digital form. But now, color and black-and-white photos are in one folder, and to sort them, one needs to spend more than one hour!
As soon as Brain is a photographer not programmer now, he asks you to help him determine for a single photo whether it is colored or black-and-white.
Photo can be represented as a matrix sized *n*<=×<=*m*, and each element of the matrix stores a symbol indicating corresponding pixel color. There are only 6 colors:
- 'C' (cyan)- 'M' (magenta)- 'Y' (yellow)- 'W' (white)- 'G' (grey)- 'B' (black)
The photo is considered black-and-white if it has only white, black and grey pixels in it. If there are any of cyan, magenta or yellow pixels in the photo then it is considered colored.
Input Specification:
The first line of the input contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100) — the number of photo pixel matrix rows and columns respectively.
Then *n* lines describing matrix rows follow. Each of them contains *m* space-separated characters describing colors of pixels in a row. Each character in the line is one of the 'C', 'M', 'Y', 'W', 'G' or 'B'.
Output Specification:
Print the "#Black&White" (without quotes), if the photo is black-and-white and "#Color" (without quotes), if it is colored, in the only line.
Demo Input:
['2 2\nC M\nY Y\n', '3 2\nW W\nW W\nB B\n', '1 1\nW\n']
Demo Output:
['#Color', '#Black&White', '#Black&White']
Note:
none | ```python
n,m = map(int,input().split())
pixels = list(list(map(str,input().split()))for _ in range(n))
pix = []
for i in pixels:
for g in i:
pix.append(g.upper())
if "C" in pix or "M" in pix or "Y" in pix:
print("#Color")
else :
print("#Black&White")
``` | 3 | |
915 | C | Permute Digits | PROGRAMMING | 1,700 | [
"dp",
"greedy"
] | null | null | You are given two positive integer numbers *a* and *b*. Permute (change order) of the digits of *a* to construct maximal number not exceeding *b*. No number in input and/or output can start with the digit 0.
It is allowed to leave *a* as it is. | The first line contains integer *a* (1<=≤<=*a*<=≤<=1018). The second line contains integer *b* (1<=≤<=*b*<=≤<=1018). Numbers don't have leading zeroes. It is guaranteed that answer exists. | Print the maximum possible number that is a permutation of digits of *a* and is not greater than *b*. The answer can't have any leading zeroes. It is guaranteed that the answer exists.
The number in the output should have exactly the same length as number *a*. It should be a permutation of digits of *a*. | [
"123\n222\n",
"3921\n10000\n",
"4940\n5000\n"
] | [
"213\n",
"9321\n",
"4940\n"
] | none | 0 | [
{
"input": "123\n222",
"output": "213"
},
{
"input": "3921\n10000",
"output": "9321"
},
{
"input": "4940\n5000",
"output": "4940"
},
{
"input": "23923472834\n23589234723",
"output": "23498743322"
},
{
"input": "102391019\n491010301",
"output": "399211100"
},... | 1,527,628,090 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 66 | 1,000 | 6,963,200 | def search(current, digits, target, idx, bulk):
if len(current) == len(target) and int(current) <= int(target):
print(current)
exit(0)
possibilities = [char for char in digits if bulk or (char <= target[idx] and char in digits)]
if len(possibilities) == 0:
return None
for possible_digit in possibilities:
tmp_digits = list(digits)
tmp_digits.remove(possible_digit)
if not bulk:
bulk = True if possible_digit != target[idx] else False
search(current + possible_digit, tmp_digits, target, idx + 1, bulk)
def main():
digits = sorted(list(input()), reverse=True)
target = input()
if len(digits) < len(target):
print(''.join(digits))
exit(0)
entries = [char for char in digits if char <= target[0]]
for current in entries:
tmp_digits = list(digits)
tmp_digits.remove(current)
search(current, tmp_digits, target, 1, True if current != target[0] else False)
if __name__ == "__main__":
main()
| Title: Permute Digits
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given two positive integer numbers *a* and *b*. Permute (change order) of the digits of *a* to construct maximal number not exceeding *b*. No number in input and/or output can start with the digit 0.
It is allowed to leave *a* as it is.
Input Specification:
The first line contains integer *a* (1<=≤<=*a*<=≤<=1018). The second line contains integer *b* (1<=≤<=*b*<=≤<=1018). Numbers don't have leading zeroes. It is guaranteed that answer exists.
Output Specification:
Print the maximum possible number that is a permutation of digits of *a* and is not greater than *b*. The answer can't have any leading zeroes. It is guaranteed that the answer exists.
The number in the output should have exactly the same length as number *a*. It should be a permutation of digits of *a*.
Demo Input:
['123\n222\n', '3921\n10000\n', '4940\n5000\n']
Demo Output:
['213\n', '9321\n', '4940\n']
Note:
none | ```python
def search(current, digits, target, idx, bulk):
if len(current) == len(target) and int(current) <= int(target):
print(current)
exit(0)
possibilities = [char for char in digits if bulk or (char <= target[idx] and char in digits)]
if len(possibilities) == 0:
return None
for possible_digit in possibilities:
tmp_digits = list(digits)
tmp_digits.remove(possible_digit)
if not bulk:
bulk = True if possible_digit != target[idx] else False
search(current + possible_digit, tmp_digits, target, idx + 1, bulk)
def main():
digits = sorted(list(input()), reverse=True)
target = input()
if len(digits) < len(target):
print(''.join(digits))
exit(0)
entries = [char for char in digits if char <= target[0]]
for current in entries:
tmp_digits = list(digits)
tmp_digits.remove(current)
search(current, tmp_digits, target, 1, True if current != target[0] else False)
if __name__ == "__main__":
main()
``` | 0 | |
538 | A | Cutting Banner | PROGRAMMING | 1,400 | [
"brute force",
"implementation"
] | null | null | A large banner with word CODEFORCES was ordered for the 1000-th onsite round of Codeforcesω that takes place on the Miami beach. Unfortunately, the company that made the banner mixed up two orders and delivered somebody else's banner that contains someone else's word. The word on the banner consists only of upper-case English letters.
There is very little time to correct the mistake. All that we can manage to do is to cut out some substring from the banner, i.e. several consecutive letters. After that all the resulting parts of the banner will be glued into a single piece (if the beginning or the end of the original banner was cut out, only one part remains); it is not allowed change the relative order of parts of the banner (i.e. after a substring is cut, several first and last letters are left, it is allowed only to glue the last letters to the right of the first letters). Thus, for example, for example, you can cut a substring out from string 'TEMPLATE' and get string 'TEMPLE' (if you cut out string AT), 'PLATE' (if you cut out TEM), 'T' (if you cut out EMPLATE), etc.
Help the organizers of the round determine whether it is possible to cut out of the banner some substring in such a way that the remaining parts formed word CODEFORCES. | The single line of the input contains the word written on the banner. The word only consists of upper-case English letters. The word is non-empty and its length doesn't exceed 100 characters. It is guaranteed that the word isn't word CODEFORCES. | Print 'YES', if there exists a way to cut out the substring, and 'NO' otherwise (without the quotes). | [
"CODEWAITFORITFORCES\n",
"BOTTOMCODER\n",
"DECODEFORCES\n",
"DOGEFORCES\n"
] | [
"YES\n",
"NO\n",
"YES\n",
"NO\n"
] | none | 500 | [
{
"input": "CODEWAITFORITFORCES",
"output": "YES"
},
{
"input": "BOTTOMCODER",
"output": "NO"
},
{
"input": "DECODEFORCES",
"output": "YES"
},
{
"input": "DOGEFORCES",
"output": "NO"
},
{
"input": "ABACABA",
"output": "NO"
},
{
"input": "CODEFORCE",
... | 1,577,983,592 | 2,147,483,647 | PyPy 3 | OK | TESTS | 55 | 140 | 0 | s=input()
i=len(s)-1
if(s.find("C")==0 and s[i-8:]==("ODEFORCES")):
print("YES")
elif(s.find("CO")==0 and s[i-7:]==("DEFORCES")):
print("YES")
elif(s.find("COD")==0 and s[i-6:]==("EFORCES")):
print("YES")
elif(s.find("CODE")==0 and s[i-5:]==("FORCES")):
print("YES")
elif(s.find("CODEF")==0 and s[i-4:]==("ORCES")):
print("YES")
elif(s.find("CODEFO")==0 and s[i-3:]==("RCES")):
print("YES")
elif(s.find("CODEFOR")==0 and s[i-2:]==("CES")):
print("YES")
elif(s.find("CODEFORC")==0 and s[i-1:]==("ES")):
print("YES")
elif(s.find("CODEFORCE")==0 and s[i]==("S")):
print("YES")
elif(s.find("CODEFORCES")==0 or s[i-9:]=="CODEFORCES"):
print("YES")
else:
print("NO")
| Title: Cutting Banner
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A large banner with word CODEFORCES was ordered for the 1000-th onsite round of Codeforcesω that takes place on the Miami beach. Unfortunately, the company that made the banner mixed up two orders and delivered somebody else's banner that contains someone else's word. The word on the banner consists only of upper-case English letters.
There is very little time to correct the mistake. All that we can manage to do is to cut out some substring from the banner, i.e. several consecutive letters. After that all the resulting parts of the banner will be glued into a single piece (if the beginning or the end of the original banner was cut out, only one part remains); it is not allowed change the relative order of parts of the banner (i.e. after a substring is cut, several first and last letters are left, it is allowed only to glue the last letters to the right of the first letters). Thus, for example, for example, you can cut a substring out from string 'TEMPLATE' and get string 'TEMPLE' (if you cut out string AT), 'PLATE' (if you cut out TEM), 'T' (if you cut out EMPLATE), etc.
Help the organizers of the round determine whether it is possible to cut out of the banner some substring in such a way that the remaining parts formed word CODEFORCES.
Input Specification:
The single line of the input contains the word written on the banner. The word only consists of upper-case English letters. The word is non-empty and its length doesn't exceed 100 characters. It is guaranteed that the word isn't word CODEFORCES.
Output Specification:
Print 'YES', if there exists a way to cut out the substring, and 'NO' otherwise (without the quotes).
Demo Input:
['CODEWAITFORITFORCES\n', 'BOTTOMCODER\n', 'DECODEFORCES\n', 'DOGEFORCES\n']
Demo Output:
['YES\n', 'NO\n', 'YES\n', 'NO\n']
Note:
none | ```python
s=input()
i=len(s)-1
if(s.find("C")==0 and s[i-8:]==("ODEFORCES")):
print("YES")
elif(s.find("CO")==0 and s[i-7:]==("DEFORCES")):
print("YES")
elif(s.find("COD")==0 and s[i-6:]==("EFORCES")):
print("YES")
elif(s.find("CODE")==0 and s[i-5:]==("FORCES")):
print("YES")
elif(s.find("CODEF")==0 and s[i-4:]==("ORCES")):
print("YES")
elif(s.find("CODEFO")==0 and s[i-3:]==("RCES")):
print("YES")
elif(s.find("CODEFOR")==0 and s[i-2:]==("CES")):
print("YES")
elif(s.find("CODEFORC")==0 and s[i-1:]==("ES")):
print("YES")
elif(s.find("CODEFORCE")==0 and s[i]==("S")):
print("YES")
elif(s.find("CODEFORCES")==0 or s[i-9:]=="CODEFORCES"):
print("YES")
else:
print("NO")
``` | 3 | |
151 | A | Soft Drinking | PROGRAMMING | 800 | [
"implementation",
"math"
] | null | null | This winter is so cold in Nvodsk! A group of *n* friends decided to buy *k* bottles of a soft drink called "Take-It-Light" to warm up a bit. Each bottle has *l* milliliters of the drink. Also they bought *c* limes and cut each of them into *d* slices. After that they found *p* grams of salt.
To make a toast, each friend needs *nl* milliliters of the drink, a slice of lime and *np* grams of salt. The friends want to make as many toasts as they can, provided they all drink the same amount. How many toasts can each friend make? | The first and only line contains positive integers *n*, *k*, *l*, *c*, *d*, *p*, *nl*, *np*, not exceeding 1000 and no less than 1. The numbers are separated by exactly one space. | Print a single integer — the number of toasts each friend can make. | [
"3 4 5 10 8 100 3 1\n",
"5 100 10 1 19 90 4 3\n",
"10 1000 1000 25 23 1 50 1\n"
] | [
"2\n",
"3\n",
"0\n"
] | A comment to the first sample:
Overall the friends have 4 * 5 = 20 milliliters of the drink, it is enough to make 20 / 3 = 6 toasts. The limes are enough for 10 * 8 = 80 toasts and the salt is enough for 100 / 1 = 100 toasts. However, there are 3 friends in the group, so the answer is *min*(6, 80, 100) / 3 = 2. | 500 | [
{
"input": "3 4 5 10 8 100 3 1",
"output": "2"
},
{
"input": "5 100 10 1 19 90 4 3",
"output": "3"
},
{
"input": "10 1000 1000 25 23 1 50 1",
"output": "0"
},
{
"input": "1 7 4 5 5 8 3 2",
"output": "4"
},
{
"input": "2 3 3 5 5 10 1 3",
"output": "1"
},
{
... | 1,690,195,706 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 27 | 122 | 0 | # -*- coding: utf-8 -*-
l = [int(i) for i in input().split(' ')]
people = l[0]
v_bottle = l[1]*l[2]
d_lime = l[3]*l[4]
solt = l[5]
print(int(min((v_bottle/l[6]/people),(d_lime/people),(solt/l[7]/people)))) | Title: Soft Drinking
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
This winter is so cold in Nvodsk! A group of *n* friends decided to buy *k* bottles of a soft drink called "Take-It-Light" to warm up a bit. Each bottle has *l* milliliters of the drink. Also they bought *c* limes and cut each of them into *d* slices. After that they found *p* grams of salt.
To make a toast, each friend needs *nl* milliliters of the drink, a slice of lime and *np* grams of salt. The friends want to make as many toasts as they can, provided they all drink the same amount. How many toasts can each friend make?
Input Specification:
The first and only line contains positive integers *n*, *k*, *l*, *c*, *d*, *p*, *nl*, *np*, not exceeding 1000 and no less than 1. The numbers are separated by exactly one space.
Output Specification:
Print a single integer — the number of toasts each friend can make.
Demo Input:
['3 4 5 10 8 100 3 1\n', '5 100 10 1 19 90 4 3\n', '10 1000 1000 25 23 1 50 1\n']
Demo Output:
['2\n', '3\n', '0\n']
Note:
A comment to the first sample:
Overall the friends have 4 * 5 = 20 milliliters of the drink, it is enough to make 20 / 3 = 6 toasts. The limes are enough for 10 * 8 = 80 toasts and the salt is enough for 100 / 1 = 100 toasts. However, there are 3 friends in the group, so the answer is *min*(6, 80, 100) / 3 = 2. | ```python
# -*- coding: utf-8 -*-
l = [int(i) for i in input().split(' ')]
people = l[0]
v_bottle = l[1]*l[2]
d_lime = l[3]*l[4]
solt = l[5]
print(int(min((v_bottle/l[6]/people),(d_lime/people),(solt/l[7]/people))))
``` | 3 | |
799 | A | Carrot Cakes | PROGRAMMING | 1,100 | [
"brute force",
"implementation"
] | null | null | In some game by Playrix it takes *t* minutes for an oven to bake *k* carrot cakes, all cakes are ready at the same moment *t* minutes after they started baking. Arkady needs at least *n* cakes to complete a task, but he currently don't have any. However, he has infinitely many ingredients and one oven. Moreover, Arkady can build one more similar oven to make the process faster, it would take *d* minutes to build the oven. While the new oven is being built, only old one can bake cakes, after the new oven is built, both ovens bake simultaneously. Arkady can't build more than one oven.
Determine if it is reasonable to build the second oven, i.e. will it decrease the minimum time needed to get *n* cakes or not. If the time needed with the second oven is the same as with one oven, then it is unreasonable. | The only line contains four integers *n*, *t*, *k*, *d* (1<=≤<=*n*,<=*t*,<=*k*,<=*d*<=≤<=1<=000) — the number of cakes needed, the time needed for one oven to bake *k* cakes, the number of cakes baked at the same time, the time needed to build the second oven. | If it is reasonable to build the second oven, print "YES". Otherwise print "NO". | [
"8 6 4 5\n",
"8 6 4 6\n",
"10 3 11 4\n",
"4 2 1 4\n"
] | [
"YES\n",
"NO\n",
"NO\n",
"YES\n"
] | In the first example it is possible to get 8 cakes in 12 minutes using one oven. The second oven can be built in 5 minutes, so after 6 minutes the first oven bakes 4 cakes, the second oven bakes 4 more ovens after 11 minutes. Thus, it is reasonable to build the second oven.
In the second example it doesn't matter whether we build the second oven or not, thus it takes 12 minutes to bake 8 cakes in both cases. Thus, it is unreasonable to build the second oven.
In the third example the first oven bakes 11 cakes in 3 minutes, that is more than needed 10. It is unreasonable to build the second oven, because its building takes more time that baking the needed number of cakes using the only oven. | 500 | [
{
"input": "8 6 4 5",
"output": "YES"
},
{
"input": "8 6 4 6",
"output": "NO"
},
{
"input": "10 3 11 4",
"output": "NO"
},
{
"input": "4 2 1 4",
"output": "YES"
},
{
"input": "28 17 16 26",
"output": "NO"
},
{
"input": "60 69 9 438",
"output": "NO"... | 1,675,223,590 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 31 | 0 | a = list(map(int,input().split()))
t1 = math.ceil(a[0]/a[2]) * a[1]
t2 = a[3] + math.ceil(a[0] - math.ceil(a[3]/a[1]) * a[2]) * a[1]
if(t2 < t1):
print('YES')
else:
print('NO') | Title: Carrot Cakes
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
In some game by Playrix it takes *t* minutes for an oven to bake *k* carrot cakes, all cakes are ready at the same moment *t* minutes after they started baking. Arkady needs at least *n* cakes to complete a task, but he currently don't have any. However, he has infinitely many ingredients and one oven. Moreover, Arkady can build one more similar oven to make the process faster, it would take *d* minutes to build the oven. While the new oven is being built, only old one can bake cakes, after the new oven is built, both ovens bake simultaneously. Arkady can't build more than one oven.
Determine if it is reasonable to build the second oven, i.e. will it decrease the minimum time needed to get *n* cakes or not. If the time needed with the second oven is the same as with one oven, then it is unreasonable.
Input Specification:
The only line contains four integers *n*, *t*, *k*, *d* (1<=≤<=*n*,<=*t*,<=*k*,<=*d*<=≤<=1<=000) — the number of cakes needed, the time needed for one oven to bake *k* cakes, the number of cakes baked at the same time, the time needed to build the second oven.
Output Specification:
If it is reasonable to build the second oven, print "YES". Otherwise print "NO".
Demo Input:
['8 6 4 5\n', '8 6 4 6\n', '10 3 11 4\n', '4 2 1 4\n']
Demo Output:
['YES\n', 'NO\n', 'NO\n', 'YES\n']
Note:
In the first example it is possible to get 8 cakes in 12 minutes using one oven. The second oven can be built in 5 minutes, so after 6 minutes the first oven bakes 4 cakes, the second oven bakes 4 more ovens after 11 minutes. Thus, it is reasonable to build the second oven.
In the second example it doesn't matter whether we build the second oven or not, thus it takes 12 minutes to bake 8 cakes in both cases. Thus, it is unreasonable to build the second oven.
In the third example the first oven bakes 11 cakes in 3 minutes, that is more than needed 10. It is unreasonable to build the second oven, because its building takes more time that baking the needed number of cakes using the only oven. | ```python
a = list(map(int,input().split()))
t1 = math.ceil(a[0]/a[2]) * a[1]
t2 = a[3] + math.ceil(a[0] - math.ceil(a[3]/a[1]) * a[2]) * a[1]
if(t2 < t1):
print('YES')
else:
print('NO')
``` | -1 | |
961 | B | Lecture Sleep | PROGRAMMING | 1,200 | [
"data structures",
"dp",
"implementation",
"two pointers"
] | null | null | Your friend Mishka and you attend a calculus lecture. Lecture lasts *n* minutes. Lecturer tells *a**i* theorems during the *i*-th minute.
Mishka is really interested in calculus, though it is so hard to stay awake for all the time of lecture. You are given an array *t* of Mishka's behavior. If Mishka is asleep during the *i*-th minute of the lecture then *t**i* will be equal to 0, otherwise it will be equal to 1. When Mishka is awake he writes down all the theorems he is being told — *a**i* during the *i*-th minute. Otherwise he writes nothing.
You know some secret technique to keep Mishka awake for *k* minutes straight. However you can use it only once. You can start using it at the beginning of any minute between 1 and *n*<=-<=*k*<=+<=1. If you use it on some minute *i* then Mishka will be awake during minutes *j* such that and will write down all the theorems lecturer tells.
You task is to calculate the maximum number of theorems Mishka will be able to write down if you use your technique only once to wake him up. | The first line of the input contains two integer numbers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=105) — the duration of the lecture in minutes and the number of minutes you can keep Mishka awake.
The second line of the input contains *n* integer numbers *a*1,<=*a*2,<=... *a**n* (1<=≤<=*a**i*<=≤<=104) — the number of theorems lecturer tells during the *i*-th minute.
The third line of the input contains *n* integer numbers *t*1,<=*t*2,<=... *t**n* (0<=≤<=*t**i*<=≤<=1) — type of Mishka's behavior at the *i*-th minute of the lecture. | Print only one integer — the maximum number of theorems Mishka will be able to write down if you use your technique only once to wake him up. | [
"6 3\n1 3 5 2 5 4\n1 1 0 1 0 0\n"
] | [
"16\n"
] | In the sample case the better way is to use the secret technique at the beginning of the third minute. Then the number of theorems Mishka will be able to write down will be equal to 16. | 0 | [
{
"input": "6 3\n1 3 5 2 5 4\n1 1 0 1 0 0",
"output": "16"
},
{
"input": "5 3\n1 9999 10000 10000 10000\n0 0 0 0 0",
"output": "30000"
},
{
"input": "3 3\n10 10 10\n1 1 0",
"output": "30"
},
{
"input": "1 1\n423\n0",
"output": "423"
},
{
"input": "6 6\n1 3 5 2 5 4... | 1,661,686,821 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 3 | 202 | 6,348,800 | n,k=map(int,input().split()) ; theo=list(map(int,input().split())) ; beha=list(map(int,input().split())) ; cum1=0 ; cum0=[]
for i in range(n):
if beha[i]==1:cum1+=theo[i] ; theo[i]=0
cum0.append(theo[0])
for i in range(1,n):
cum0.append(theo[i]+cum0[i-1])
mx=0
for i in range(n-k+1):
x=cum0[i+k-1]-cum0[i-1]
mx=max(mx,x)
print(mx+cum1)
| Title: Lecture Sleep
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Your friend Mishka and you attend a calculus lecture. Lecture lasts *n* minutes. Lecturer tells *a**i* theorems during the *i*-th minute.
Mishka is really interested in calculus, though it is so hard to stay awake for all the time of lecture. You are given an array *t* of Mishka's behavior. If Mishka is asleep during the *i*-th minute of the lecture then *t**i* will be equal to 0, otherwise it will be equal to 1. When Mishka is awake he writes down all the theorems he is being told — *a**i* during the *i*-th minute. Otherwise he writes nothing.
You know some secret technique to keep Mishka awake for *k* minutes straight. However you can use it only once. You can start using it at the beginning of any minute between 1 and *n*<=-<=*k*<=+<=1. If you use it on some minute *i* then Mishka will be awake during minutes *j* such that and will write down all the theorems lecturer tells.
You task is to calculate the maximum number of theorems Mishka will be able to write down if you use your technique only once to wake him up.
Input Specification:
The first line of the input contains two integer numbers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=105) — the duration of the lecture in minutes and the number of minutes you can keep Mishka awake.
The second line of the input contains *n* integer numbers *a*1,<=*a*2,<=... *a**n* (1<=≤<=*a**i*<=≤<=104) — the number of theorems lecturer tells during the *i*-th minute.
The third line of the input contains *n* integer numbers *t*1,<=*t*2,<=... *t**n* (0<=≤<=*t**i*<=≤<=1) — type of Mishka's behavior at the *i*-th minute of the lecture.
Output Specification:
Print only one integer — the maximum number of theorems Mishka will be able to write down if you use your technique only once to wake him up.
Demo Input:
['6 3\n1 3 5 2 5 4\n1 1 0 1 0 0\n']
Demo Output:
['16\n']
Note:
In the sample case the better way is to use the secret technique at the beginning of the third minute. Then the number of theorems Mishka will be able to write down will be equal to 16. | ```python
n,k=map(int,input().split()) ; theo=list(map(int,input().split())) ; beha=list(map(int,input().split())) ; cum1=0 ; cum0=[]
for i in range(n):
if beha[i]==1:cum1+=theo[i] ; theo[i]=0
cum0.append(theo[0])
for i in range(1,n):
cum0.append(theo[i]+cum0[i-1])
mx=0
for i in range(n-k+1):
x=cum0[i+k-1]-cum0[i-1]
mx=max(mx,x)
print(mx+cum1)
``` | 0 | |
239 | A | Two Bags of Potatoes | PROGRAMMING | 1,200 | [
"greedy",
"implementation",
"math"
] | null | null | Valera had two bags of potatoes, the first of these bags contains *x* (*x*<=≥<=1) potatoes, and the second — *y* (*y*<=≥<=1) potatoes. Valera — very scattered boy, so the first bag of potatoes (it contains *x* potatoes) Valera lost. Valera remembers that the total amount of potatoes (*x*<=+<=*y*) in the two bags, firstly, was not gerater than *n*, and, secondly, was divisible by *k*.
Help Valera to determine how many potatoes could be in the first bag. Print all such possible numbers in ascending order. | The first line of input contains three integers *y*, *k*, *n* (1<=≤<=*y*,<=*k*,<=*n*<=≤<=109; <=≤<=105). | Print the list of whitespace-separated integers — all possible values of *x* in ascending order. You should print each possible value of *x* exactly once.
If there are no such values of *x* print a single integer -1. | [
"10 1 10\n",
"10 6 40\n"
] | [
"-1\n",
"2 8 14 20 26 \n"
] | none | 500 | [
{
"input": "10 1 10",
"output": "-1"
},
{
"input": "10 6 40",
"output": "2 8 14 20 26 "
},
{
"input": "10 1 20",
"output": "1 2 3 4 5 6 7 8 9 10 "
},
{
"input": "1 10000 1000000000",
"output": "9999 19999 29999 39999 49999 59999 69999 79999 89999 99999 109999 119999 12999... | 1,698,152,925 | 2,147,483,647 | Python 3 | OK | TESTS | 48 | 310 | 102,400 | y,k,n = map(int,input().split())
x = k - y%k
f = 0
while x+y <= n:
print(x,end=" ")
x += k
f = 1
if f==0:
print(-1) | Title: Two Bags of Potatoes
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Valera had two bags of potatoes, the first of these bags contains *x* (*x*<=≥<=1) potatoes, and the second — *y* (*y*<=≥<=1) potatoes. Valera — very scattered boy, so the first bag of potatoes (it contains *x* potatoes) Valera lost. Valera remembers that the total amount of potatoes (*x*<=+<=*y*) in the two bags, firstly, was not gerater than *n*, and, secondly, was divisible by *k*.
Help Valera to determine how many potatoes could be in the first bag. Print all such possible numbers in ascending order.
Input Specification:
The first line of input contains three integers *y*, *k*, *n* (1<=≤<=*y*,<=*k*,<=*n*<=≤<=109; <=≤<=105).
Output Specification:
Print the list of whitespace-separated integers — all possible values of *x* in ascending order. You should print each possible value of *x* exactly once.
If there are no such values of *x* print a single integer -1.
Demo Input:
['10 1 10\n', '10 6 40\n']
Demo Output:
['-1\n', '2 8 14 20 26 \n']
Note:
none | ```python
y,k,n = map(int,input().split())
x = k - y%k
f = 0
while x+y <= n:
print(x,end=" ")
x += k
f = 1
if f==0:
print(-1)
``` | 3 | |
315 | A | Sereja and Bottles | PROGRAMMING | 1,400 | [
"brute force"
] | null | null | Sereja and his friends went to a picnic. The guys had *n* soda bottles just for it. Sereja forgot the bottle opener as usual, so the guys had to come up with another way to open bottles.
Sereja knows that the *i*-th bottle is from brand *a**i*, besides, you can use it to open other bottles of brand *b**i*. You can use one bottle to open multiple other bottles. Sereja can open bottle with opened bottle or closed bottle.
Knowing this, Sereja wants to find out the number of bottles they've got that they won't be able to open in any way. Help him and find this number. | The first line contains integer *n* (1<=≤<=*n*<=≤<=100) — the number of bottles. The next *n* lines contain the bottles' description. The *i*-th line contains two integers *a**i*,<=*b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=1000) — the description of the *i*-th bottle. | In a single line print a single integer — the answer to the problem. | [
"4\n1 1\n2 2\n3 3\n4 4\n",
"4\n1 2\n2 3\n3 4\n4 1\n"
] | [
"4\n",
"0\n"
] | none | 500 | [
{
"input": "4\n1 1\n2 2\n3 3\n4 4",
"output": "4"
},
{
"input": "4\n1 2\n2 3\n3 4\n4 1",
"output": "0"
},
{
"input": "3\n2 828\n4 392\n4 903",
"output": "3"
},
{
"input": "4\n2 3\n1 772\n3 870\n3 668",
"output": "2"
},
{
"input": "5\n1 4\n6 6\n4 3\n3 4\n4 758",
... | 1,651,752,954 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 2 | 122 | 0 | n=int(input())
bj=set()
aj=set()
for i in range(n):
a,b=map(int,input().split())
if a!=b:
bj.add(b)
aj.add(a)
else:
aj.add(a)
for i in bj:
if i in aj:
aj.remove(i)
print(len(aj))
| Title: Sereja and Bottles
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Sereja and his friends went to a picnic. The guys had *n* soda bottles just for it. Sereja forgot the bottle opener as usual, so the guys had to come up with another way to open bottles.
Sereja knows that the *i*-th bottle is from brand *a**i*, besides, you can use it to open other bottles of brand *b**i*. You can use one bottle to open multiple other bottles. Sereja can open bottle with opened bottle or closed bottle.
Knowing this, Sereja wants to find out the number of bottles they've got that they won't be able to open in any way. Help him and find this number.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=100) — the number of bottles. The next *n* lines contain the bottles' description. The *i*-th line contains two integers *a**i*,<=*b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=1000) — the description of the *i*-th bottle.
Output Specification:
In a single line print a single integer — the answer to the problem.
Demo Input:
['4\n1 1\n2 2\n3 3\n4 4\n', '4\n1 2\n2 3\n3 4\n4 1\n']
Demo Output:
['4\n', '0\n']
Note:
none | ```python
n=int(input())
bj=set()
aj=set()
for i in range(n):
a,b=map(int,input().split())
if a!=b:
bj.add(b)
aj.add(a)
else:
aj.add(a)
for i in bj:
if i in aj:
aj.remove(i)
print(len(aj))
``` | 0 | |
701 | A | Cards | PROGRAMMING | 800 | [
"greedy",
"implementation"
] | null | null | There are *n* cards (*n* is even) in the deck. Each card has a positive integer written on it. *n*<=/<=2 people will play new card game. At the beginning of the game each player gets two cards, each card is given to exactly one player.
Find the way to distribute cards such that the sum of values written of the cards will be equal for each player. It is guaranteed that it is always possible. | The first line of the input contains integer *n* (2<=≤<=*n*<=≤<=100) — the number of cards in the deck. It is guaranteed that *n* is even.
The second line contains the sequence of *n* positive integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100), where *a**i* is equal to the number written on the *i*-th card. | Print *n*<=/<=2 pairs of integers, the *i*-th pair denote the cards that should be given to the *i*-th player. Each card should be given to exactly one player. Cards are numbered in the order they appear in the input.
It is guaranteed that solution exists. If there are several correct answers, you are allowed to print any of them. | [
"6\n1 5 7 4 4 3\n",
"4\n10 10 10 10\n"
] | [
"1 3\n6 2\n4 5\n",
"1 2\n3 4\n"
] | In the first sample, cards are distributed in such a way that each player has the sum of numbers written on his cards equal to 8.
In the second sample, all values *a*<sub class="lower-index">*i*</sub> are equal. Thus, any distribution is acceptable. | 500 | [
{
"input": "6\n1 5 7 4 4 3",
"output": "1 3\n6 2\n4 5"
},
{
"input": "4\n10 10 10 10",
"output": "1 4\n2 3"
},
{
"input": "100\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ... | 1,589,783,134 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 1 | 93 | 6,963,200 | n=int(input())
l=list(map(int,input().split()))
nx=n/2
sm=int(sum(l)/nx)
nl=[]
for i in range(n):
nl.append(i)
for j in range(i+1,n):
if l[i]+l[j]==sm and j not in nl:
print(i+1,j+1)
nl.append(j)
| Title: Cards
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are *n* cards (*n* is even) in the deck. Each card has a positive integer written on it. *n*<=/<=2 people will play new card game. At the beginning of the game each player gets two cards, each card is given to exactly one player.
Find the way to distribute cards such that the sum of values written of the cards will be equal for each player. It is guaranteed that it is always possible.
Input Specification:
The first line of the input contains integer *n* (2<=≤<=*n*<=≤<=100) — the number of cards in the deck. It is guaranteed that *n* is even.
The second line contains the sequence of *n* positive integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100), where *a**i* is equal to the number written on the *i*-th card.
Output Specification:
Print *n*<=/<=2 pairs of integers, the *i*-th pair denote the cards that should be given to the *i*-th player. Each card should be given to exactly one player. Cards are numbered in the order they appear in the input.
It is guaranteed that solution exists. If there are several correct answers, you are allowed to print any of them.
Demo Input:
['6\n1 5 7 4 4 3\n', '4\n10 10 10 10\n']
Demo Output:
['1 3\n6 2\n4 5\n', '1 2\n3 4\n']
Note:
In the first sample, cards are distributed in such a way that each player has the sum of numbers written on his cards equal to 8.
In the second sample, all values *a*<sub class="lower-index">*i*</sub> are equal. Thus, any distribution is acceptable. | ```python
n=int(input())
l=list(map(int,input().split()))
nx=n/2
sm=int(sum(l)/nx)
nl=[]
for i in range(n):
nl.append(i)
for j in range(i+1,n):
if l[i]+l[j]==sm and j not in nl:
print(i+1,j+1)
nl.append(j)
``` | 0 | |
92 | A | Chips | PROGRAMMING | 800 | [
"implementation",
"math"
] | A. Chips | 2 | 256 | There are *n* walruses sitting in a circle. All of them are numbered in the clockwise order: the walrus number 2 sits to the left of the walrus number 1, the walrus number 3 sits to the left of the walrus number 2, ..., the walrus number 1 sits to the left of the walrus number *n*.
The presenter has *m* chips. The presenter stands in the middle of the circle and starts giving the chips to the walruses starting from walrus number 1 and moving clockwise. The walrus number *i* gets *i* chips. If the presenter can't give the current walrus the required number of chips, then the presenter takes the remaining chips and the process ends. Determine by the given *n* and *m* how many chips the presenter will get in the end. | The first line contains two integers *n* and *m* (1<=≤<=*n*<=≤<=50, 1<=≤<=*m*<=≤<=104) — the number of walruses and the number of chips correspondingly. | Print the number of chips the presenter ended up with. | [
"4 11\n",
"17 107\n",
"3 8\n"
] | [
"0\n",
"2\n",
"1\n"
] | In the first sample the presenter gives one chip to the walrus number 1, two chips to the walrus number 2, three chips to the walrus number 3, four chips to the walrus number 4, then again one chip to the walrus number 1. After that the presenter runs out of chips. He can't give anything to the walrus number 2 and the process finishes.
In the third sample the presenter gives one chip to the walrus number 1, two chips to the walrus number 2, three chips to the walrus number 3, then again one chip to the walrus number 1. The presenter has one chip left and he can't give two chips to the walrus number 2, that's why the presenter takes the last chip. | 500 | [
{
"input": "4 11",
"output": "0"
},
{
"input": "17 107",
"output": "2"
},
{
"input": "3 8",
"output": "1"
},
{
"input": "46 7262",
"output": "35"
},
{
"input": "32 6864",
"output": "0"
},
{
"input": "36 6218",
"output": "14"
},
{
"input": "... | 1,604,389,281 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 154 | 0 | n,m=map(int,input().split())
p=m%(n*(n+1)/2)
for i in range(1,n+1):
if p>=i:
p=p-i
else:
print(p)
exit()
| Title: Chips
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
There are *n* walruses sitting in a circle. All of them are numbered in the clockwise order: the walrus number 2 sits to the left of the walrus number 1, the walrus number 3 sits to the left of the walrus number 2, ..., the walrus number 1 sits to the left of the walrus number *n*.
The presenter has *m* chips. The presenter stands in the middle of the circle and starts giving the chips to the walruses starting from walrus number 1 and moving clockwise. The walrus number *i* gets *i* chips. If the presenter can't give the current walrus the required number of chips, then the presenter takes the remaining chips and the process ends. Determine by the given *n* and *m* how many chips the presenter will get in the end.
Input Specification:
The first line contains two integers *n* and *m* (1<=≤<=*n*<=≤<=50, 1<=≤<=*m*<=≤<=104) — the number of walruses and the number of chips correspondingly.
Output Specification:
Print the number of chips the presenter ended up with.
Demo Input:
['4 11\n', '17 107\n', '3 8\n']
Demo Output:
['0\n', '2\n', '1\n']
Note:
In the first sample the presenter gives one chip to the walrus number 1, two chips to the walrus number 2, three chips to the walrus number 3, four chips to the walrus number 4, then again one chip to the walrus number 1. After that the presenter runs out of chips. He can't give anything to the walrus number 2 and the process finishes.
In the third sample the presenter gives one chip to the walrus number 1, two chips to the walrus number 2, three chips to the walrus number 3, then again one chip to the walrus number 1. The presenter has one chip left and he can't give two chips to the walrus number 2, that's why the presenter takes the last chip. | ```python
n,m=map(int,input().split())
p=m%(n*(n+1)/2)
for i in range(1,n+1):
if p>=i:
p=p-i
else:
print(p)
exit()
``` | 0 |
554 | A | Kyoya and Photobooks | PROGRAMMING | 900 | [
"brute force",
"math",
"strings"
] | null | null | Kyoya Ootori is selling photobooks of the Ouran High School Host Club. He has 26 photos, labeled "a" to "z", and he has compiled them into a photo booklet with some photos in some order (possibly with some photos being duplicated). A photo booklet can be described as a string of lowercase letters, consisting of the photos in the booklet in order. He now wants to sell some "special edition" photobooks, each with one extra photo inserted anywhere in the book. He wants to make as many distinct photobooks as possible, so he can make more money. He asks Haruhi, how many distinct photobooks can he make by inserting one extra photo into the photobook he already has?
Please help Haruhi solve this problem. | The first line of input will be a single string *s* (1<=≤<=|*s*|<=≤<=20). String *s* consists only of lowercase English letters. | Output a single integer equal to the number of distinct photobooks Kyoya Ootori can make. | [
"a\n",
"hi\n"
] | [
"51\n",
"76\n"
] | In the first case, we can make 'ab','ac',...,'az','ba','ca',...,'za', and 'aa', producing a total of 51 distinct photo booklets. | 250 | [
{
"input": "a",
"output": "51"
},
{
"input": "hi",
"output": "76"
},
{
"input": "y",
"output": "51"
},
{
"input": "kgan",
"output": "126"
},
{
"input": "zoabkyuvus",
"output": "276"
},
{
"input": "spyemhyznjieyhhbk",
"output": "451"
},
{
"i... | 1,590,002,763 | 2,147,483,647 | Python 3 | OK | TESTS | 33 | 109 | 0 | def main():
st = input()
print(solve(st))
def solve(st):
return len(st)*25 + 26
main()
| Title: Kyoya and Photobooks
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Kyoya Ootori is selling photobooks of the Ouran High School Host Club. He has 26 photos, labeled "a" to "z", and he has compiled them into a photo booklet with some photos in some order (possibly with some photos being duplicated). A photo booklet can be described as a string of lowercase letters, consisting of the photos in the booklet in order. He now wants to sell some "special edition" photobooks, each with one extra photo inserted anywhere in the book. He wants to make as many distinct photobooks as possible, so he can make more money. He asks Haruhi, how many distinct photobooks can he make by inserting one extra photo into the photobook he already has?
Please help Haruhi solve this problem.
Input Specification:
The first line of input will be a single string *s* (1<=≤<=|*s*|<=≤<=20). String *s* consists only of lowercase English letters.
Output Specification:
Output a single integer equal to the number of distinct photobooks Kyoya Ootori can make.
Demo Input:
['a\n', 'hi\n']
Demo Output:
['51\n', '76\n']
Note:
In the first case, we can make 'ab','ac',...,'az','ba','ca',...,'za', and 'aa', producing a total of 51 distinct photo booklets. | ```python
def main():
st = input()
print(solve(st))
def solve(st):
return len(st)*25 + 26
main()
``` | 3 | |
602 | B | Approximating a Constant Range | PROGRAMMING | 1,400 | [
"dp",
"implementation",
"two pointers"
] | null | null | When Xellos was doing a practice course in university, he once had to measure the intensity of an effect that slowly approached equilibrium. A good way to determine the equilibrium intensity would be choosing a sufficiently large number of consecutive data points that seems as constant as possible and taking their average. Of course, with the usual sizes of data, it's nothing challenging — but why not make a similar programming contest problem while we're at it?
You're given a sequence of *n* data points *a*1,<=...,<=*a**n*. There aren't any big jumps between consecutive data points — for each 1<=≤<=*i*<=<<=*n*, it's guaranteed that |*a**i*<=+<=1<=-<=*a**i*|<=≤<=1.
A range [*l*,<=*r*] of data points is said to be almost constant if the difference between the largest and the smallest value in that range is at most 1. Formally, let *M* be the maximum and *m* the minimum value of *a**i* for *l*<=≤<=*i*<=≤<=*r*; the range [*l*,<=*r*] is almost constant if *M*<=-<=*m*<=≤<=1.
Find the length of the longest almost constant range. | The first line of the input contains a single integer *n* (2<=≤<=*n*<=≤<=100<=000) — the number of data points.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100<=000). | Print a single number — the maximum length of an almost constant range of the given sequence. | [
"5\n1 2 3 3 2\n",
"11\n5 4 5 5 6 7 8 8 8 7 6\n"
] | [
"4\n",
"5\n"
] | In the first sample, the longest almost constant range is [2, 5]; its length (the number of data points in it) is 4.
In the second sample, there are three almost constant ranges of length 4: [1, 4], [6, 9] and [7, 10]; the only almost constant range of the maximum length 5 is [6, 10]. | 1,000 | [
{
"input": "5\n1 2 3 3 2",
"output": "4"
},
{
"input": "11\n5 4 5 5 6 7 8 8 8 7 6",
"output": "5"
},
{
"input": "2\n3 2",
"output": "2"
},
{
"input": "4\n1001 1000 1000 1001",
"output": "4"
},
{
"input": "4\n1 1 2 3",
"output": "3"
},
{
"input": "3\n1 ... | 1,687,232,957 | 2,147,483,647 | Python 3 | OK | TESTS | 107 | 124 | 13,619,200 | MOD=10**9+7
def slove():
n=int(input())
# m=input()
a=list(map(int,input().split(' ')))
left=0
d={}
ans=0
for i,x in enumerate(a):
if x not in d: d[x]=0
d[x]+=1
while len(d)>2:
d[a[left]]-=1
if d[a[left]]==0:
del d[a[left]]
left+=1
ans=max(ans,i-left+1)
print(ans)
return
# t=int(input())
t=1
# M=1<<30
for _ in range(t):
slove()
| Title: Approximating a Constant Range
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
When Xellos was doing a practice course in university, he once had to measure the intensity of an effect that slowly approached equilibrium. A good way to determine the equilibrium intensity would be choosing a sufficiently large number of consecutive data points that seems as constant as possible and taking their average. Of course, with the usual sizes of data, it's nothing challenging — but why not make a similar programming contest problem while we're at it?
You're given a sequence of *n* data points *a*1,<=...,<=*a**n*. There aren't any big jumps between consecutive data points — for each 1<=≤<=*i*<=<<=*n*, it's guaranteed that |*a**i*<=+<=1<=-<=*a**i*|<=≤<=1.
A range [*l*,<=*r*] of data points is said to be almost constant if the difference between the largest and the smallest value in that range is at most 1. Formally, let *M* be the maximum and *m* the minimum value of *a**i* for *l*<=≤<=*i*<=≤<=*r*; the range [*l*,<=*r*] is almost constant if *M*<=-<=*m*<=≤<=1.
Find the length of the longest almost constant range.
Input Specification:
The first line of the input contains a single integer *n* (2<=≤<=*n*<=≤<=100<=000) — the number of data points.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100<=000).
Output Specification:
Print a single number — the maximum length of an almost constant range of the given sequence.
Demo Input:
['5\n1 2 3 3 2\n', '11\n5 4 5 5 6 7 8 8 8 7 6\n']
Demo Output:
['4\n', '5\n']
Note:
In the first sample, the longest almost constant range is [2, 5]; its length (the number of data points in it) is 4.
In the second sample, there are three almost constant ranges of length 4: [1, 4], [6, 9] and [7, 10]; the only almost constant range of the maximum length 5 is [6, 10]. | ```python
MOD=10**9+7
def slove():
n=int(input())
# m=input()
a=list(map(int,input().split(' ')))
left=0
d={}
ans=0
for i,x in enumerate(a):
if x not in d: d[x]=0
d[x]+=1
while len(d)>2:
d[a[left]]-=1
if d[a[left]]==0:
del d[a[left]]
left+=1
ans=max(ans,i-left+1)
print(ans)
return
# t=int(input())
t=1
# M=1<<30
for _ in range(t):
slove()
``` | 3 | |
592 | A | PawnChess | PROGRAMMING | 1,200 | [
"implementation"
] | null | null | Galois is one of the strongest chess players of Byteforces. He has even invented a new variant of chess, which he named «PawnChess».
This new game is played on a board consisting of 8 rows and 8 columns. At the beginning of every game some black and white pawns are placed on the board. The number of black pawns placed is not necessarily equal to the number of white pawns placed.
Lets enumerate rows and columns with integers from 1 to 8. Rows are numbered from top to bottom, while columns are numbered from left to right. Now we denote as (*r*,<=*c*) the cell located at the row *r* and at the column *c*.
There are always two players A and B playing the game. Player A plays with white pawns, while player B plays with black ones. The goal of player A is to put any of his pawns to the row 1, while player B tries to put any of his pawns to the row 8. As soon as any of the players completes his goal the game finishes immediately and the succeeded player is declared a winner.
Player A moves first and then they alternate turns. On his move player A must choose exactly one white pawn and move it one step upward and player B (at his turn) must choose exactly one black pawn and move it one step down. Any move is possible only if the targeted cell is empty. It's guaranteed that for any scenario of the game there will always be at least one move available for any of the players.
Moving upward means that the pawn located in (*r*,<=*c*) will go to the cell (*r*<=-<=1,<=*c*), while moving down means the pawn located in (*r*,<=*c*) will go to the cell (*r*<=+<=1,<=*c*). Again, the corresponding cell must be empty, i.e. not occupied by any other pawn of any color.
Given the initial disposition of the board, determine who wins the game if both players play optimally. Note that there will always be a winner due to the restriction that for any game scenario both players will have some moves available. | The input consists of the board description given in eight lines, each line contains eight characters. Character 'B' is used to denote a black pawn, and character 'W' represents a white pawn. Empty cell is marked with '.'.
It's guaranteed that there will not be white pawns on the first row neither black pawns on the last row. | Print 'A' if player A wins the game on the given board, and 'B' if player B will claim the victory. Again, it's guaranteed that there will always be a winner on the given board. | [
"........\n........\n.B....B.\n....W...\n........\n..W.....\n........\n........\n",
"..B.....\n..W.....\n......B.\n........\n.....W..\n......B.\n........\n........\n"
] | [
"A\n",
"B\n"
] | In the first sample player A is able to complete his goal in 3 steps by always moving a pawn initially located at (4, 5). Player B needs at least 5 steps for any of his pawns to reach the row 8. Hence, player A will be the winner. | 500 | [
{
"input": ".BB.B.B.\nB..B..B.\n.B.BB...\nBB.....B\nBBB....B\nB..BB...\nBB.B...B\n....WWW.",
"output": "B"
},
{
"input": "B.B.BB.B\nW.WWW.WW\n.WWWWW.W\nW.BB.WBW\n.W..BBWB\nBB.WWBBB\n.W.W.WWB\nWWW..WW.",
"output": "A"
},
{
"input": "BB..BB..\nBW.W.W.B\n..B.....\n.....BB.\n.B..B..B\n......... | 1,451,316,319 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 1 | 30 | 0 | chess=[list(input())for i in range(8)]
o=[0]
p=[0]
j=1
if 'W' in chess[0]:
j=0
print('A')
if 'B' in chess[7] and 'W' not in chess[0]:
j=0
print('B')
if j==1:
for i in range(1,7):
for j in range(8):
if chess[i][j]=='B' and chess[i+1][j]!='W':
o.append(i)
if chess[i][j]=='W' and chess[i-1][j]!='B':
p.append(i)
if 7-max(o)<min(p):
print('B')
if 7-max(o)>min(p):
print('A')
| Title: PawnChess
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Galois is one of the strongest chess players of Byteforces. He has even invented a new variant of chess, which he named «PawnChess».
This new game is played on a board consisting of 8 rows and 8 columns. At the beginning of every game some black and white pawns are placed on the board. The number of black pawns placed is not necessarily equal to the number of white pawns placed.
Lets enumerate rows and columns with integers from 1 to 8. Rows are numbered from top to bottom, while columns are numbered from left to right. Now we denote as (*r*,<=*c*) the cell located at the row *r* and at the column *c*.
There are always two players A and B playing the game. Player A plays with white pawns, while player B plays with black ones. The goal of player A is to put any of his pawns to the row 1, while player B tries to put any of his pawns to the row 8. As soon as any of the players completes his goal the game finishes immediately and the succeeded player is declared a winner.
Player A moves first and then they alternate turns. On his move player A must choose exactly one white pawn and move it one step upward and player B (at his turn) must choose exactly one black pawn and move it one step down. Any move is possible only if the targeted cell is empty. It's guaranteed that for any scenario of the game there will always be at least one move available for any of the players.
Moving upward means that the pawn located in (*r*,<=*c*) will go to the cell (*r*<=-<=1,<=*c*), while moving down means the pawn located in (*r*,<=*c*) will go to the cell (*r*<=+<=1,<=*c*). Again, the corresponding cell must be empty, i.e. not occupied by any other pawn of any color.
Given the initial disposition of the board, determine who wins the game if both players play optimally. Note that there will always be a winner due to the restriction that for any game scenario both players will have some moves available.
Input Specification:
The input consists of the board description given in eight lines, each line contains eight characters. Character 'B' is used to denote a black pawn, and character 'W' represents a white pawn. Empty cell is marked with '.'.
It's guaranteed that there will not be white pawns on the first row neither black pawns on the last row.
Output Specification:
Print 'A' if player A wins the game on the given board, and 'B' if player B will claim the victory. Again, it's guaranteed that there will always be a winner on the given board.
Demo Input:
['........\n........\n.B....B.\n....W...\n........\n..W.....\n........\n........\n', '..B.....\n..W.....\n......B.\n........\n.....W..\n......B.\n........\n........\n']
Demo Output:
['A\n', 'B\n']
Note:
In the first sample player A is able to complete his goal in 3 steps by always moving a pawn initially located at (4, 5). Player B needs at least 5 steps for any of his pawns to reach the row 8. Hence, player A will be the winner. | ```python
chess=[list(input())for i in range(8)]
o=[0]
p=[0]
j=1
if 'W' in chess[0]:
j=0
print('A')
if 'B' in chess[7] and 'W' not in chess[0]:
j=0
print('B')
if j==1:
for i in range(1,7):
for j in range(8):
if chess[i][j]=='B' and chess[i+1][j]!='W':
o.append(i)
if chess[i][j]=='W' and chess[i-1][j]!='B':
p.append(i)
if 7-max(o)<min(p):
print('B')
if 7-max(o)>min(p):
print('A')
``` | 0 | |
629 | A | Far Relative’s Birthday Cake | PROGRAMMING | 800 | [
"brute force",
"combinatorics",
"constructive algorithms",
"implementation"
] | null | null | Door's family is going celebrate Famil Doors's birthday party. They love Famil Door so they are planning to make his birthday cake weird!
The cake is a *n*<=×<=*n* square consisting of equal squares with side length 1. Each square is either empty or consists of a single chocolate. They bought the cake and randomly started to put the chocolates on the cake. The value of Famil Door's happiness will be equal to the number of pairs of cells with chocolates that are in the same row or in the same column of the cake. Famil Doors's family is wondering what is the amount of happiness of Famil going to be?
Please, note that any pair can be counted no more than once, as two different cells can't share both the same row and the same column. | In the first line of the input, you are given a single integer *n* (1<=≤<=*n*<=≤<=100) — the length of the side of the cake.
Then follow *n* lines, each containing *n* characters. Empty cells are denoted with '.', while cells that contain chocolates are denoted by 'C'. | Print the value of Famil Door's happiness, i.e. the number of pairs of chocolate pieces that share the same row or the same column. | [
"3\n.CC\nC..\nC.C\n",
"4\nCC..\nC..C\n.CC.\n.CC.\n"
] | [
"4\n",
"9\n"
] | If we number rows from top to bottom and columns from left to right, then, pieces that share the same row in the first sample are:
1. (1, 2) and (1, 3) 1. (3, 1) and (3, 3) 1. (2, 1) and (3, 1) 1. (1, 3) and (3, 3) | 500 | [
{
"input": "3\n.CC\nC..\nC.C",
"output": "4"
},
{
"input": "4\nCC..\nC..C\n.CC.\n.CC.",
"output": "9"
},
{
"input": "5\n.CCCC\nCCCCC\n.CCC.\nCC...\n.CC.C",
"output": "46"
},
{
"input": "7\n.CC..CC\nCC.C..C\nC.C..C.\nC...C.C\nCCC.CCC\n.CC...C\n.C.CCC.",
"output": "84"
},... | 1,699,990,905 | 2,147,483,647 | Python 3 | OK | TESTS | 48 | 46 | 0 | n = int(input())
rowCounts = [0] * n
colCounts = [0] * n
for i in range(n):
cakeRow = input()
for j in range(n):
if cakeRow[j] == 'C':
rowCounts[i] += 1
colCounts[j] += 1
totalHappiness = 0
for i in range(n):
totalHappiness += (rowCounts[i] - 1) * rowCounts[i] // 2
totalHappiness += (colCounts[i] - 1) * colCounts[i] // 2
print(totalHappiness) | Title: Far Relative’s Birthday Cake
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Door's family is going celebrate Famil Doors's birthday party. They love Famil Door so they are planning to make his birthday cake weird!
The cake is a *n*<=×<=*n* square consisting of equal squares with side length 1. Each square is either empty or consists of a single chocolate. They bought the cake and randomly started to put the chocolates on the cake. The value of Famil Door's happiness will be equal to the number of pairs of cells with chocolates that are in the same row or in the same column of the cake. Famil Doors's family is wondering what is the amount of happiness of Famil going to be?
Please, note that any pair can be counted no more than once, as two different cells can't share both the same row and the same column.
Input Specification:
In the first line of the input, you are given a single integer *n* (1<=≤<=*n*<=≤<=100) — the length of the side of the cake.
Then follow *n* lines, each containing *n* characters. Empty cells are denoted with '.', while cells that contain chocolates are denoted by 'C'.
Output Specification:
Print the value of Famil Door's happiness, i.e. the number of pairs of chocolate pieces that share the same row or the same column.
Demo Input:
['3\n.CC\nC..\nC.C\n', '4\nCC..\nC..C\n.CC.\n.CC.\n']
Demo Output:
['4\n', '9\n']
Note:
If we number rows from top to bottom and columns from left to right, then, pieces that share the same row in the first sample are:
1. (1, 2) and (1, 3) 1. (3, 1) and (3, 3) 1. (2, 1) and (3, 1) 1. (1, 3) and (3, 3) | ```python
n = int(input())
rowCounts = [0] * n
colCounts = [0] * n
for i in range(n):
cakeRow = input()
for j in range(n):
if cakeRow[j] == 'C':
rowCounts[i] += 1
colCounts[j] += 1
totalHappiness = 0
for i in range(n):
totalHappiness += (rowCounts[i] - 1) * rowCounts[i] // 2
totalHappiness += (colCounts[i] - 1) * colCounts[i] // 2
print(totalHappiness)
``` | 3 | |
462 | B | Appleman and Card Game | PROGRAMMING | 1,300 | [
"greedy"
] | null | null | Appleman has *n* cards. Each card has an uppercase letter written on it. Toastman must choose *k* cards from Appleman's cards. Then Appleman should give Toastman some coins depending on the chosen cards. Formally, for each Toastman's card *i* you should calculate how much Toastman's cards have the letter equal to letter on *i*th, then sum up all these quantities, such a number of coins Appleman should give to Toastman.
Given the description of Appleman's cards. What is the maximum number of coins Toastman can get? | The first line contains two integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=105). The next line contains *n* uppercase letters without spaces — the *i*-th letter describes the *i*-th card of the Appleman. | Print a single integer – the answer to the problem. | [
"15 10\nDZFDFZDFDDDDDDF\n",
"6 4\nYJSNPI\n"
] | [
"82\n",
"4\n"
] | In the first test example Toastman can choose nine cards with letter D and one additional card with any letter. For each card with D he will get 9 coins and for the additional card he will get 1 coin. | 1,000 | [
{
"input": "15 10\nDZFDFZDFDDDDDDF",
"output": "82"
},
{
"input": "6 4\nYJSNPI",
"output": "4"
},
{
"input": "5 3\nAOWBY",
"output": "3"
},
{
"input": "1 1\nV",
"output": "1"
},
{
"input": "2 1\nWT",
"output": "1"
},
{
"input": "2 2\nBL",
"output":... | 1,589,812,883 | 2,147,483,647 | PyPy 3 | OK | TESTS | 41 | 155 | 20,684,800 | n,k=map(int,input().split())
s=input()
lst=list()
c=str.count
ind="ABCDEFGHIJKLMNOPQRSTUVWXYZ"
for i in ind:
if(c(s,i)!=0):
lst.append(c(s,i))
lst.sort()
lst.reverse()
ans=0
for i in lst:
ans+=min(k,i)**2
k=k-min(k,i)
if(k==0):break
print(ans)
| Title: Appleman and Card Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Appleman has *n* cards. Each card has an uppercase letter written on it. Toastman must choose *k* cards from Appleman's cards. Then Appleman should give Toastman some coins depending on the chosen cards. Formally, for each Toastman's card *i* you should calculate how much Toastman's cards have the letter equal to letter on *i*th, then sum up all these quantities, such a number of coins Appleman should give to Toastman.
Given the description of Appleman's cards. What is the maximum number of coins Toastman can get?
Input Specification:
The first line contains two integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=105). The next line contains *n* uppercase letters without spaces — the *i*-th letter describes the *i*-th card of the Appleman.
Output Specification:
Print a single integer – the answer to the problem.
Demo Input:
['15 10\nDZFDFZDFDDDDDDF\n', '6 4\nYJSNPI\n']
Demo Output:
['82\n', '4\n']
Note:
In the first test example Toastman can choose nine cards with letter D and one additional card with any letter. For each card with D he will get 9 coins and for the additional card he will get 1 coin. | ```python
n,k=map(int,input().split())
s=input()
lst=list()
c=str.count
ind="ABCDEFGHIJKLMNOPQRSTUVWXYZ"
for i in ind:
if(c(s,i)!=0):
lst.append(c(s,i))
lst.sort()
lst.reverse()
ans=0
for i in lst:
ans+=min(k,i)**2
k=k-min(k,i)
if(k==0):break
print(ans)
``` | 3 | |
118 | A | String Task | PROGRAMMING | 1,000 | [
"implementation",
"strings"
] | null | null | Petya started to attend programming lessons. On the first lesson his task was to write a simple program. The program was supposed to do the following: in the given string, consisting if uppercase and lowercase Latin letters, it:
- deletes all the vowels, - inserts a character "." before each consonant, - replaces all uppercase consonants with corresponding lowercase ones.
Vowels are letters "A", "O", "Y", "E", "U", "I", and the rest are consonants. The program's input is exactly one string, it should return the output as a single string, resulting after the program's processing the initial string.
Help Petya cope with this easy task. | The first line represents input string of Petya's program. This string only consists of uppercase and lowercase Latin letters and its length is from 1 to 100, inclusive. | Print the resulting string. It is guaranteed that this string is not empty. | [
"tour\n",
"Codeforces\n",
"aBAcAba\n"
] | [
".t.r\n",
".c.d.f.r.c.s\n",
".b.c.b\n"
] | none | 500 | [
{
"input": "tour",
"output": ".t.r"
},
{
"input": "Codeforces",
"output": ".c.d.f.r.c.s"
},
{
"input": "aBAcAba",
"output": ".b.c.b"
},
{
"input": "obn",
"output": ".b.n"
},
{
"input": "wpwl",
"output": ".w.p.w.l"
},
{
"input": "ggdvq",
"output": "... | 1,694,791,260 | 2,147,483,647 | Python 3 | OK | TESTS | 42 | 92 | 0 | s = input().lower()
repl = ["A", "O", "Y", "E", "U", "I",'a','o','y','e','u','i']
ls = []
for i in s:
if i not in repl:
ls.append(f'.{i}')
print(''.join(ls)) | Title: String Task
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Petya started to attend programming lessons. On the first lesson his task was to write a simple program. The program was supposed to do the following: in the given string, consisting if uppercase and lowercase Latin letters, it:
- deletes all the vowels, - inserts a character "." before each consonant, - replaces all uppercase consonants with corresponding lowercase ones.
Vowels are letters "A", "O", "Y", "E", "U", "I", and the rest are consonants. The program's input is exactly one string, it should return the output as a single string, resulting after the program's processing the initial string.
Help Petya cope with this easy task.
Input Specification:
The first line represents input string of Petya's program. This string only consists of uppercase and lowercase Latin letters and its length is from 1 to 100, inclusive.
Output Specification:
Print the resulting string. It is guaranteed that this string is not empty.
Demo Input:
['tour\n', 'Codeforces\n', 'aBAcAba\n']
Demo Output:
['.t.r\n', '.c.d.f.r.c.s\n', '.b.c.b\n']
Note:
none | ```python
s = input().lower()
repl = ["A", "O", "Y", "E", "U", "I",'a','o','y','e','u','i']
ls = []
for i in s:
if i not in repl:
ls.append(f'.{i}')
print(''.join(ls))
``` | 3 | |
59 | A | Word | PROGRAMMING | 800 | [
"implementation",
"strings"
] | A. Word | 2 | 256 | Vasya is very upset that many people on the Net mix uppercase and lowercase letters in one word. That's why he decided to invent an extension for his favorite browser that would change the letters' register in every word so that it either only consisted of lowercase letters or, vice versa, only of uppercase ones. At that as little as possible letters should be changed in the word. For example, the word HoUse must be replaced with house, and the word ViP — with VIP. If a word contains an equal number of uppercase and lowercase letters, you should replace all the letters with lowercase ones. For example, maTRIx should be replaced by matrix. Your task is to use the given method on one given word. | The first line contains a word *s* — it consists of uppercase and lowercase Latin letters and possesses the length from 1 to 100. | Print the corrected word *s*. If the given word *s* has strictly more uppercase letters, make the word written in the uppercase register, otherwise - in the lowercase one. | [
"HoUse\n",
"ViP\n",
"maTRIx\n"
] | [
"house\n",
"VIP\n",
"matrix\n"
] | none | 500 | [
{
"input": "HoUse",
"output": "house"
},
{
"input": "ViP",
"output": "VIP"
},
{
"input": "maTRIx",
"output": "matrix"
},
{
"input": "BNHWpnpawg",
"output": "bnhwpnpawg"
},
{
"input": "VTYGP",
"output": "VTYGP"
},
{
"input": "CHNenu",
"output": "chn... | 1,616,088,401 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 2 | 124 | 0 | n = input()
c = 0; k = 0
for i in list(n):
if i.isupper():
c += 1
else: k += 1
if c >= k:
print(n.upper())
else: print(n.lower()) | Title: Word
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Vasya is very upset that many people on the Net mix uppercase and lowercase letters in one word. That's why he decided to invent an extension for his favorite browser that would change the letters' register in every word so that it either only consisted of lowercase letters or, vice versa, only of uppercase ones. At that as little as possible letters should be changed in the word. For example, the word HoUse must be replaced with house, and the word ViP — with VIP. If a word contains an equal number of uppercase and lowercase letters, you should replace all the letters with lowercase ones. For example, maTRIx should be replaced by matrix. Your task is to use the given method on one given word.
Input Specification:
The first line contains a word *s* — it consists of uppercase and lowercase Latin letters and possesses the length from 1 to 100.
Output Specification:
Print the corrected word *s*. If the given word *s* has strictly more uppercase letters, make the word written in the uppercase register, otherwise - in the lowercase one.
Demo Input:
['HoUse\n', 'ViP\n', 'maTRIx\n']
Demo Output:
['house\n', 'VIP\n', 'matrix\n']
Note:
none | ```python
n = input()
c = 0; k = 0
for i in list(n):
if i.isupper():
c += 1
else: k += 1
if c >= k:
print(n.upper())
else: print(n.lower())
``` | 0 |
501 | B | Misha and Changing Handles | PROGRAMMING | 1,100 | [
"data structures",
"dsu",
"strings"
] | null | null | Misha hacked the Codeforces site. Then he decided to let all the users change their handles. A user can now change his handle any number of times. But each new handle must not be equal to any handle that is already used or that was used at some point.
Misha has a list of handle change requests. After completing the requests he wants to understand the relation between the original and the new handles of the users. Help him to do that. | The first line contains integer *q* (1<=≤<=*q*<=≤<=1000), the number of handle change requests.
Next *q* lines contain the descriptions of the requests, one per line.
Each query consists of two non-empty strings *old* and *new*, separated by a space. The strings consist of lowercase and uppercase Latin letters and digits. Strings *old* and *new* are distinct. The lengths of the strings do not exceed 20.
The requests are given chronologically. In other words, by the moment of a query there is a single person with handle *old*, and handle *new* is not used and has not been used by anyone. | In the first line output the integer *n* — the number of users that changed their handles at least once.
In the next *n* lines print the mapping between the old and the new handles of the users. Each of them must contain two strings, *old* and *new*, separated by a space, meaning that before the user had handle *old*, and after all the requests are completed, his handle is *new*. You may output lines in any order.
Each user who changes the handle must occur exactly once in this description. | [
"5\nMisha ILoveCodeforces\nVasya Petrov\nPetrov VasyaPetrov123\nILoveCodeforces MikeMirzayanov\nPetya Ivanov\n"
] | [
"3\nPetya Ivanov\nMisha MikeMirzayanov\nVasya VasyaPetrov123\n"
] | none | 500 | [
{
"input": "5\nMisha ILoveCodeforces\nVasya Petrov\nPetrov VasyaPetrov123\nILoveCodeforces MikeMirzayanov\nPetya Ivanov",
"output": "3\nPetya Ivanov\nMisha MikeMirzayanov\nVasya VasyaPetrov123"
},
{
"input": "1\nMisha Vasya",
"output": "1\nMisha Vasya"
},
{
"input": "10\na b\nb c\nc d\nd... | 1,494,839,651 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 0 | 1,000 | 0 | n=10000000
k=0
for i in range(n) :
k=k+1
| Title: Misha and Changing Handles
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Misha hacked the Codeforces site. Then he decided to let all the users change their handles. A user can now change his handle any number of times. But each new handle must not be equal to any handle that is already used or that was used at some point.
Misha has a list of handle change requests. After completing the requests he wants to understand the relation between the original and the new handles of the users. Help him to do that.
Input Specification:
The first line contains integer *q* (1<=≤<=*q*<=≤<=1000), the number of handle change requests.
Next *q* lines contain the descriptions of the requests, one per line.
Each query consists of two non-empty strings *old* and *new*, separated by a space. The strings consist of lowercase and uppercase Latin letters and digits. Strings *old* and *new* are distinct. The lengths of the strings do not exceed 20.
The requests are given chronologically. In other words, by the moment of a query there is a single person with handle *old*, and handle *new* is not used and has not been used by anyone.
Output Specification:
In the first line output the integer *n* — the number of users that changed their handles at least once.
In the next *n* lines print the mapping between the old and the new handles of the users. Each of them must contain two strings, *old* and *new*, separated by a space, meaning that before the user had handle *old*, and after all the requests are completed, his handle is *new*. You may output lines in any order.
Each user who changes the handle must occur exactly once in this description.
Demo Input:
['5\nMisha ILoveCodeforces\nVasya Petrov\nPetrov VasyaPetrov123\nILoveCodeforces MikeMirzayanov\nPetya Ivanov\n']
Demo Output:
['3\nPetya Ivanov\nMisha MikeMirzayanov\nVasya VasyaPetrov123\n']
Note:
none | ```python
n=10000000
k=0
for i in range(n) :
k=k+1
``` | 0 | |
711 | A | Bus to Udayland | PROGRAMMING | 800 | [
"brute force",
"implementation"
] | null | null | ZS the Coder and Chris the Baboon are travelling to Udayland! To get there, they have to get on the special IOI bus. The IOI bus has *n* rows of seats. There are 4 seats in each row, and the seats are separated into pairs by a walkway. When ZS and Chris came, some places in the bus was already occupied.
ZS and Chris are good friends. They insist to get a pair of neighbouring empty seats. Two seats are considered neighbouring if they are in the same row and in the same pair. Given the configuration of the bus, can you help ZS and Chris determine where they should sit? | The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of rows of seats in the bus.
Then, *n* lines follow. Each line contains exactly 5 characters, the first two of them denote the first pair of seats in the row, the third character denotes the walkway (it always equals '|') and the last two of them denote the second pair of seats in the row.
Each character, except the walkway, equals to 'O' or to 'X'. 'O' denotes an empty seat, 'X' denotes an occupied seat. See the sample cases for more details. | If it is possible for Chris and ZS to sit at neighbouring empty seats, print "YES" (without quotes) in the first line. In the next *n* lines print the bus configuration, where the characters in the pair of seats for Chris and ZS is changed with characters '+'. Thus the configuration should differ from the input one by exactly two charaters (they should be equal to 'O' in the input and to '+' in the output).
If there is no pair of seats for Chris and ZS, print "NO" (without quotes) in a single line.
If there are multiple solutions, you may print any of them. | [
"6\nOO|OX\nXO|XX\nOX|OO\nXX|OX\nOO|OO\nOO|XX\n",
"4\nXO|OX\nXO|XX\nOX|OX\nXX|OX\n",
"5\nXX|XX\nXX|XX\nXO|OX\nXO|OO\nOX|XO\n"
] | [
"YES\n++|OX\nXO|XX\nOX|OO\nXX|OX\nOO|OO\nOO|XX\n",
"NO\n",
"YES\nXX|XX\nXX|XX\nXO|OX\nXO|++\nOX|XO\n"
] | Note that the following is an incorrect configuration for the first sample case because the seats must be in the same pair.
O+|+X
XO|XX
OX|OO
XX|OX
OO|OO
OO|XX | 500 | [
{
"input": "6\nOO|OX\nXO|XX\nOX|OO\nXX|OX\nOO|OO\nOO|XX",
"output": "YES\n++|OX\nXO|XX\nOX|OO\nXX|OX\nOO|OO\nOO|XX"
},
{
"input": "4\nXO|OX\nXO|XX\nOX|OX\nXX|OX",
"output": "NO"
},
{
"input": "5\nXX|XX\nXX|XX\nXO|OX\nXO|OO\nOX|XO",
"output": "YES\nXX|XX\nXX|XX\nXO|OX\nXO|++\nOX|XO"
... | 1,659,370,133 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 4 | 46 | 0 | def check(row):
if row.find('OO') != -1:
return [row.find('OO'),'YES']
return ['NO']
rows_num = int(input())
rows = []
for i in range(rows_num):
rows.append(input())
confirm = False
for row in rows:
if len(check(row))==2:
valid_row = rows.index(row)
print('YES')
for i in range(rows_num):
if i==valid_row:
print(rows[i].replace('OO','++'))
else:
print(rows[i])
confirm = True
break
if confirm==False:
print("NO") | Title: Bus to Udayland
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
ZS the Coder and Chris the Baboon are travelling to Udayland! To get there, they have to get on the special IOI bus. The IOI bus has *n* rows of seats. There are 4 seats in each row, and the seats are separated into pairs by a walkway. When ZS and Chris came, some places in the bus was already occupied.
ZS and Chris are good friends. They insist to get a pair of neighbouring empty seats. Two seats are considered neighbouring if they are in the same row and in the same pair. Given the configuration of the bus, can you help ZS and Chris determine where they should sit?
Input Specification:
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of rows of seats in the bus.
Then, *n* lines follow. Each line contains exactly 5 characters, the first two of them denote the first pair of seats in the row, the third character denotes the walkway (it always equals '|') and the last two of them denote the second pair of seats in the row.
Each character, except the walkway, equals to 'O' or to 'X'. 'O' denotes an empty seat, 'X' denotes an occupied seat. See the sample cases for more details.
Output Specification:
If it is possible for Chris and ZS to sit at neighbouring empty seats, print "YES" (without quotes) in the first line. In the next *n* lines print the bus configuration, where the characters in the pair of seats for Chris and ZS is changed with characters '+'. Thus the configuration should differ from the input one by exactly two charaters (they should be equal to 'O' in the input and to '+' in the output).
If there is no pair of seats for Chris and ZS, print "NO" (without quotes) in a single line.
If there are multiple solutions, you may print any of them.
Demo Input:
['6\nOO|OX\nXO|XX\nOX|OO\nXX|OX\nOO|OO\nOO|XX\n', '4\nXO|OX\nXO|XX\nOX|OX\nXX|OX\n', '5\nXX|XX\nXX|XX\nXO|OX\nXO|OO\nOX|XO\n']
Demo Output:
['YES\n++|OX\nXO|XX\nOX|OO\nXX|OX\nOO|OO\nOO|XX\n', 'NO\n', 'YES\nXX|XX\nXX|XX\nXO|OX\nXO|++\nOX|XO\n']
Note:
Note that the following is an incorrect configuration for the first sample case because the seats must be in the same pair.
O+|+X
XO|XX
OX|OO
XX|OX
OO|OO
OO|XX | ```python
def check(row):
if row.find('OO') != -1:
return [row.find('OO'),'YES']
return ['NO']
rows_num = int(input())
rows = []
for i in range(rows_num):
rows.append(input())
confirm = False
for row in rows:
if len(check(row))==2:
valid_row = rows.index(row)
print('YES')
for i in range(rows_num):
if i==valid_row:
print(rows[i].replace('OO','++'))
else:
print(rows[i])
confirm = True
break
if confirm==False:
print("NO")
``` | 0 | |
41 | A | Translation | PROGRAMMING | 800 | [
"implementation",
"strings"
] | A. Translation | 2 | 256 | The translation from the Berland language into the Birland language is not an easy task. Those languages are very similar: a berlandish word differs from a birlandish word with the same meaning a little: it is spelled (and pronounced) reversely. For example, a Berlandish word code corresponds to a Birlandish word edoc. However, it's easy to make a mistake during the «translation». Vasya translated word *s* from Berlandish into Birlandish as *t*. Help him: find out if he translated the word correctly. | The first line contains word *s*, the second line contains word *t*. The words consist of lowercase Latin letters. The input data do not consist unnecessary spaces. The words are not empty and their lengths do not exceed 100 symbols. | If the word *t* is a word *s*, written reversely, print YES, otherwise print NO. | [
"code\nedoc\n",
"abb\naba\n",
"code\ncode\n"
] | [
"YES\n",
"NO\n",
"NO\n"
] | none | 500 | [
{
"input": "code\nedoc",
"output": "YES"
},
{
"input": "abb\naba",
"output": "NO"
},
{
"input": "code\ncode",
"output": "NO"
},
{
"input": "abacaba\nabacaba",
"output": "YES"
},
{
"input": "q\nq",
"output": "YES"
},
{
"input": "asrgdfngfnmfgnhweratgjkk... | 1,643,685,120 | 2,147,483,647 | Python 3 | COMPILATION_ERROR | TESTS | 0 | 0 | 0 | t, s = map(str, input())
if s == t[::-1]
print("YES")
else:
print("NO") | Title: Translation
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
The translation from the Berland language into the Birland language is not an easy task. Those languages are very similar: a berlandish word differs from a birlandish word with the same meaning a little: it is spelled (and pronounced) reversely. For example, a Berlandish word code corresponds to a Birlandish word edoc. However, it's easy to make a mistake during the «translation». Vasya translated word *s* from Berlandish into Birlandish as *t*. Help him: find out if he translated the word correctly.
Input Specification:
The first line contains word *s*, the second line contains word *t*. The words consist of lowercase Latin letters. The input data do not consist unnecessary spaces. The words are not empty and their lengths do not exceed 100 symbols.
Output Specification:
If the word *t* is a word *s*, written reversely, print YES, otherwise print NO.
Demo Input:
['code\nedoc\n', 'abb\naba\n', 'code\ncode\n']
Demo Output:
['YES\n', 'NO\n', 'NO\n']
Note:
none | ```python
t, s = map(str, input())
if s == t[::-1]
print("YES")
else:
print("NO")
``` | -1 |
879 | B | Table Tennis | PROGRAMMING | 1,200 | [
"data structures",
"implementation"
] | null | null | *n* people are standing in a line to play table tennis. At first, the first two players in the line play a game. Then the loser goes to the end of the line, and the winner plays with the next person from the line, and so on. They play until someone wins *k* games in a row. This player becomes the winner.
For each of the participants, you know the power to play table tennis, and for all players these values are different. In a game the player with greater power always wins. Determine who will be the winner. | The first line contains two integers: *n* and *k* (2<=≤<=*n*<=≤<=500, 2<=≤<=*k*<=≤<=1012) — the number of people and the number of wins.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=*n*) — powers of the player. It's guaranteed that this line contains a valid permutation, i.e. all *a**i* are distinct. | Output a single integer — power of the winner. | [
"2 2\n1 2\n",
"4 2\n3 1 2 4\n",
"6 2\n6 5 3 1 2 4\n",
"2 10000000000\n2 1\n"
] | [
"2 ",
"3 ",
"6 ",
"2\n"
] | Games in the second sample:
3 plays with 1. 3 wins. 1 goes to the end of the line.
3 plays with 2. 3 wins. He wins twice in a row. He becomes the winner. | 1,000 | [
{
"input": "2 2\n1 2",
"output": "2 "
},
{
"input": "4 2\n3 1 2 4",
"output": "3 "
},
{
"input": "6 2\n6 5 3 1 2 4",
"output": "6 "
},
{
"input": "2 10000000000\n2 1",
"output": "2"
},
{
"input": "4 4\n1 3 4 2",
"output": "4 "
},
{
"input": "2 21474836... | 1,624,136,145 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 11 | 61 | 0 | n,m=map(int,input().split())
k=list(map(int,input().split()))
p=0
o=0
if m>300000:
print(max(k))
else:
while p<=m or o<=m:
if k[0]>k[1]:
o=0
p+=1
k.append(k[1])
k.remove(k[1])
if p==m:
print(k[0])
break
elif k[1]>k[0]:
p=0
o+=1
k.append(k[0])
k.remove(k[0])
if o==m:
print(k[1])
break | Title: Table Tennis
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
*n* people are standing in a line to play table tennis. At first, the first two players in the line play a game. Then the loser goes to the end of the line, and the winner plays with the next person from the line, and so on. They play until someone wins *k* games in a row. This player becomes the winner.
For each of the participants, you know the power to play table tennis, and for all players these values are different. In a game the player with greater power always wins. Determine who will be the winner.
Input Specification:
The first line contains two integers: *n* and *k* (2<=≤<=*n*<=≤<=500, 2<=≤<=*k*<=≤<=1012) — the number of people and the number of wins.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=*n*) — powers of the player. It's guaranteed that this line contains a valid permutation, i.e. all *a**i* are distinct.
Output Specification:
Output a single integer — power of the winner.
Demo Input:
['2 2\n1 2\n', '4 2\n3 1 2 4\n', '6 2\n6 5 3 1 2 4\n', '2 10000000000\n2 1\n']
Demo Output:
['2 ', '3 ', '6 ', '2\n']
Note:
Games in the second sample:
3 plays with 1. 3 wins. 1 goes to the end of the line.
3 plays with 2. 3 wins. He wins twice in a row. He becomes the winner. | ```python
n,m=map(int,input().split())
k=list(map(int,input().split()))
p=0
o=0
if m>300000:
print(max(k))
else:
while p<=m or o<=m:
if k[0]>k[1]:
o=0
p+=1
k.append(k[1])
k.remove(k[1])
if p==m:
print(k[0])
break
elif k[1]>k[0]:
p=0
o+=1
k.append(k[0])
k.remove(k[0])
if o==m:
print(k[1])
break
``` | 0 | |
706 | B | Interesting drink | PROGRAMMING | 1,100 | [
"binary search",
"dp",
"implementation"
] | null | null | Vasiliy likes to rest after a hard work, so you may often meet him in some bar nearby. As all programmers do, he loves the famous drink "Beecola", which can be bought in *n* different shops in the city. It's known that the price of one bottle in the shop *i* is equal to *x**i* coins.
Vasiliy plans to buy his favorite drink for *q* consecutive days. He knows, that on the *i*-th day he will be able to spent *m**i* coins. Now, for each of the days he want to know in how many different shops he can buy a bottle of "Beecola". | The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=100<=000) — the number of shops in the city that sell Vasiliy's favourite drink.
The second line contains *n* integers *x**i* (1<=≤<=*x**i*<=≤<=100<=000) — prices of the bottles of the drink in the *i*-th shop.
The third line contains a single integer *q* (1<=≤<=*q*<=≤<=100<=000) — the number of days Vasiliy plans to buy the drink.
Then follow *q* lines each containing one integer *m**i* (1<=≤<=*m**i*<=≤<=109) — the number of coins Vasiliy can spent on the *i*-th day. | Print *q* integers. The *i*-th of them should be equal to the number of shops where Vasiliy will be able to buy a bottle of the drink on the *i*-th day. | [
"5\n3 10 8 6 11\n4\n1\n10\n3\n11\n"
] | [
"0\n4\n1\n5\n"
] | On the first day, Vasiliy won't be able to buy a drink in any of the shops.
On the second day, Vasiliy can buy a drink in the shops 1, 2, 3 and 4.
On the third day, Vasiliy can buy a drink only in the shop number 1.
Finally, on the last day Vasiliy can buy a drink in any shop. | 1,000 | [
{
"input": "5\n3 10 8 6 11\n4\n1\n10\n3\n11",
"output": "0\n4\n1\n5"
},
{
"input": "5\n868 987 714 168 123\n10\n424\n192\n795\n873\n117\n914\n735\n158\n631\n471",
"output": "2\n2\n3\n4\n0\n4\n3\n1\n2\n2"
},
{
"input": "3\n435 482 309\n7\n245\n241\n909\n745\n980\n29\n521",
"output": "... | 1,696,751,444 | 2,147,483,647 | Python 3 | OK | TESTS | 103 | 1,107 | 13,619,200 | n = int(input())
arr = list(map(int, input().split()))
arr.sort()
for _ in range(int(input())):
c = int(input())
low, high = 0, n - 1
ans = -1
while low <= high:
mid = (low + high) // 2
if arr[mid] > c:
ans = mid
high = mid - 1
else:
low = mid + 1
print(ans if ans != -1 else n) | Title: Interesting drink
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasiliy likes to rest after a hard work, so you may often meet him in some bar nearby. As all programmers do, he loves the famous drink "Beecola", which can be bought in *n* different shops in the city. It's known that the price of one bottle in the shop *i* is equal to *x**i* coins.
Vasiliy plans to buy his favorite drink for *q* consecutive days. He knows, that on the *i*-th day he will be able to spent *m**i* coins. Now, for each of the days he want to know in how many different shops he can buy a bottle of "Beecola".
Input Specification:
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=100<=000) — the number of shops in the city that sell Vasiliy's favourite drink.
The second line contains *n* integers *x**i* (1<=≤<=*x**i*<=≤<=100<=000) — prices of the bottles of the drink in the *i*-th shop.
The third line contains a single integer *q* (1<=≤<=*q*<=≤<=100<=000) — the number of days Vasiliy plans to buy the drink.
Then follow *q* lines each containing one integer *m**i* (1<=≤<=*m**i*<=≤<=109) — the number of coins Vasiliy can spent on the *i*-th day.
Output Specification:
Print *q* integers. The *i*-th of them should be equal to the number of shops where Vasiliy will be able to buy a bottle of the drink on the *i*-th day.
Demo Input:
['5\n3 10 8 6 11\n4\n1\n10\n3\n11\n']
Demo Output:
['0\n4\n1\n5\n']
Note:
On the first day, Vasiliy won't be able to buy a drink in any of the shops.
On the second day, Vasiliy can buy a drink in the shops 1, 2, 3 and 4.
On the third day, Vasiliy can buy a drink only in the shop number 1.
Finally, on the last day Vasiliy can buy a drink in any shop. | ```python
n = int(input())
arr = list(map(int, input().split()))
arr.sort()
for _ in range(int(input())):
c = int(input())
low, high = 0, n - 1
ans = -1
while low <= high:
mid = (low + high) // 2
if arr[mid] > c:
ans = mid
high = mid - 1
else:
low = mid + 1
print(ans if ans != -1 else n)
``` | 3 | |
118 | A | String Task | PROGRAMMING | 1,000 | [
"implementation",
"strings"
] | null | null | Petya started to attend programming lessons. On the first lesson his task was to write a simple program. The program was supposed to do the following: in the given string, consisting if uppercase and lowercase Latin letters, it:
- deletes all the vowels, - inserts a character "." before each consonant, - replaces all uppercase consonants with corresponding lowercase ones.
Vowels are letters "A", "O", "Y", "E", "U", "I", and the rest are consonants. The program's input is exactly one string, it should return the output as a single string, resulting after the program's processing the initial string.
Help Petya cope with this easy task. | The first line represents input string of Petya's program. This string only consists of uppercase and lowercase Latin letters and its length is from 1 to 100, inclusive. | Print the resulting string. It is guaranteed that this string is not empty. | [
"tour\n",
"Codeforces\n",
"aBAcAba\n"
] | [
".t.r\n",
".c.d.f.r.c.s\n",
".b.c.b\n"
] | none | 500 | [
{
"input": "tour",
"output": ".t.r"
},
{
"input": "Codeforces",
"output": ".c.d.f.r.c.s"
},
{
"input": "aBAcAba",
"output": ".b.c.b"
},
{
"input": "obn",
"output": ".b.n"
},
{
"input": "wpwl",
"output": ".w.p.w.l"
},
{
"input": "ggdvq",
"output": "... | 1,694,632,628 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 42 | 124 | 0 | def chc(st):
nst = ''
for i in st.lower():
if i not in 'aoyeui':
nst += '.'+i
return nst
st = input()
print(chc(st)) | Title: String Task
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Petya started to attend programming lessons. On the first lesson his task was to write a simple program. The program was supposed to do the following: in the given string, consisting if uppercase and lowercase Latin letters, it:
- deletes all the vowels, - inserts a character "." before each consonant, - replaces all uppercase consonants with corresponding lowercase ones.
Vowels are letters "A", "O", "Y", "E", "U", "I", and the rest are consonants. The program's input is exactly one string, it should return the output as a single string, resulting after the program's processing the initial string.
Help Petya cope with this easy task.
Input Specification:
The first line represents input string of Petya's program. This string only consists of uppercase and lowercase Latin letters and its length is from 1 to 100, inclusive.
Output Specification:
Print the resulting string. It is guaranteed that this string is not empty.
Demo Input:
['tour\n', 'Codeforces\n', 'aBAcAba\n']
Demo Output:
['.t.r\n', '.c.d.f.r.c.s\n', '.b.c.b\n']
Note:
none | ```python
def chc(st):
nst = ''
for i in st.lower():
if i not in 'aoyeui':
nst += '.'+i
return nst
st = input()
print(chc(st))
``` | 3 | |
58 | A | Chat room | PROGRAMMING | 1,000 | [
"greedy",
"strings"
] | A. Chat room | 1 | 256 | Vasya has recently learned to type and log on to the Internet. He immediately entered a chat room and decided to say hello to everybody. Vasya typed the word *s*. It is considered that Vasya managed to say hello if several letters can be deleted from the typed word so that it resulted in the word "hello". For example, if Vasya types the word "ahhellllloou", it will be considered that he said hello, and if he types "hlelo", it will be considered that Vasya got misunderstood and he didn't manage to say hello. Determine whether Vasya managed to say hello by the given word *s*. | The first and only line contains the word *s*, which Vasya typed. This word consisits of small Latin letters, its length is no less that 1 and no more than 100 letters. | If Vasya managed to say hello, print "YES", otherwise print "NO". | [
"ahhellllloou\n",
"hlelo\n"
] | [
"YES\n",
"NO\n"
] | none | 500 | [
{
"input": "ahhellllloou",
"output": "YES"
},
{
"input": "hlelo",
"output": "NO"
},
{
"input": "helhcludoo",
"output": "YES"
},
{
"input": "hehwelloho",
"output": "YES"
},
{
"input": "pnnepelqomhhheollvlo",
"output": "YES"
},
{
"input": "tymbzjyqhymeda... | 1,518,367,393 | 2,147,483,647 | Python 3 | COMPILATION_ERROR | TESTS | 0 | 0 | 0 | salem=input()
h=0
e=0
l1=0
l2=0
o=0
truwer=0
for i in range(0,len(salem)):
if h==0&&salem[i]=='h'
h=1
if h==1&&salem[i]=='e'
e=1
h=-3
if e==1&&salem[i]=='l'
l1=1
e=0
if l1==1&&salem[i]=='l'
l2=1
l1=0
if l2==1&&salem[i]=='o'
o=1
truwer=5
break;
if truwer == 5:
print('YES')
else:
print('NO') | Title: Chat room
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Vasya has recently learned to type and log on to the Internet. He immediately entered a chat room and decided to say hello to everybody. Vasya typed the word *s*. It is considered that Vasya managed to say hello if several letters can be deleted from the typed word so that it resulted in the word "hello". For example, if Vasya types the word "ahhellllloou", it will be considered that he said hello, and if he types "hlelo", it will be considered that Vasya got misunderstood and he didn't manage to say hello. Determine whether Vasya managed to say hello by the given word *s*.
Input Specification:
The first and only line contains the word *s*, which Vasya typed. This word consisits of small Latin letters, its length is no less that 1 and no more than 100 letters.
Output Specification:
If Vasya managed to say hello, print "YES", otherwise print "NO".
Demo Input:
['ahhellllloou\n', 'hlelo\n']
Demo Output:
['YES\n', 'NO\n']
Note:
none | ```python
salem=input()
h=0
e=0
l1=0
l2=0
o=0
truwer=0
for i in range(0,len(salem)):
if h==0&&salem[i]=='h'
h=1
if h==1&&salem[i]=='e'
e=1
h=-3
if e==1&&salem[i]=='l'
l1=1
e=0
if l1==1&&salem[i]=='l'
l2=1
l1=0
if l2==1&&salem[i]=='o'
o=1
truwer=5
break;
if truwer == 5:
print('YES')
else:
print('NO')
``` | -1 |
761 | C | Dasha and Password | PROGRAMMING | 1,500 | [
"brute force",
"dp",
"implementation"
] | null | null | After overcoming the stairs Dasha came to classes. She needed to write a password to begin her classes. The password is a string of length *n* which satisfies the following requirements:
- There is at least one digit in the string, - There is at least one lowercase (small) letter of the Latin alphabet in the string, - There is at least one of three listed symbols in the string: '#', '*', '&'.
Considering that these are programming classes it is not easy to write the password.
For each character of the password we have a fixed string of length *m*, on each of these *n* strings there is a pointer on some character. The *i*-th character displayed on the screen is the pointed character in the *i*-th string. Initially, all pointers are on characters with indexes 1 in the corresponding strings (all positions are numbered starting from one).
During one operation Dasha can move a pointer in one string one character to the left or to the right. Strings are cyclic, it means that when we move the pointer which is on the character with index 1 to the left, it moves to the character with the index *m*, and when we move it to the right from the position *m* it moves to the position 1.
You need to determine the minimum number of operations necessary to make the string displayed on the screen a valid password. | The first line contains two integers *n*, *m* (3<=≤<=*n*<=≤<=50,<=1<=≤<=*m*<=≤<=50) — the length of the password and the length of strings which are assigned to password symbols.
Each of the next *n* lines contains the string which is assigned to the *i*-th symbol of the password string. Its length is *m*, it consists of digits, lowercase English letters, and characters '#', '*' or '&'.
You have such input data that you can always get a valid password. | Print one integer — the minimum number of operations which is necessary to make the string, which is displayed on the screen, a valid password. | [
"3 4\n1**2\na3*0\nc4**\n",
"5 5\n#*&#*\n*a1c&\n&q2w*\n#a3c#\n*&#*&\n"
] | [
"1\n",
"3\n"
] | In the first test it is necessary to move the pointer of the third string to one left to get the optimal answer.
In the second test one of possible algorithms will be:
- to move the pointer of the second symbol once to the right. - to move the pointer of the third symbol twice to the right. | 1,500 | [
{
"input": "3 4\n1**2\na3*0\nc4**",
"output": "1"
},
{
"input": "5 5\n#*&#*\n*a1c&\n&q2w*\n#a3c#\n*&#*&",
"output": "3"
},
{
"input": "5 2\n&l\n*0\n*9\n*#\n#o",
"output": "2"
},
{
"input": "25 16\nvza**ooxkmd#*ywa\ndip#*#&ef&z&&&pv\nwggob&&72#*&&nku\nrsb##*&jm&#u**te\nzif#lu#... | 1,485,877,985 | 4,685 | PyPy 3 | OK | TESTS | 52 | 685 | 30,003,200 | n,m = list(map(int, input().split(" ")))
strings = []
for i in range(n):
strings.append(input())
spec = set('#*&')
dig = set('0123456789')
alpha = set('abcdefghijklmnopqrstuvwxyz')
def num(checkset, checkstring):
m = len(checkstring)+1
for i, c in enumerate(checkstring):
if c in checkset:
m = min(m, i)
m = min(m, len(checkstring)-i)
if m < len(checkstring)+1:
return m
return -1
ret = float('inf')
for i in range(n):
for j in range(n):
for k in range(n):
if i==j or i==k or j==k:
continue
else:
s1 = num(spec, strings[i])
s2 = num(dig, strings[j])
s3 = num(alpha, strings[k])
if s1 >= 0 and s2 >= 0 and s3 >= 0:
ret = min(ret, s1+s2+s3)
print(str(int(ret)))
| Title: Dasha and Password
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
After overcoming the stairs Dasha came to classes. She needed to write a password to begin her classes. The password is a string of length *n* which satisfies the following requirements:
- There is at least one digit in the string, - There is at least one lowercase (small) letter of the Latin alphabet in the string, - There is at least one of three listed symbols in the string: '#', '*', '&'.
Considering that these are programming classes it is not easy to write the password.
For each character of the password we have a fixed string of length *m*, on each of these *n* strings there is a pointer on some character. The *i*-th character displayed on the screen is the pointed character in the *i*-th string. Initially, all pointers are on characters with indexes 1 in the corresponding strings (all positions are numbered starting from one).
During one operation Dasha can move a pointer in one string one character to the left or to the right. Strings are cyclic, it means that when we move the pointer which is on the character with index 1 to the left, it moves to the character with the index *m*, and when we move it to the right from the position *m* it moves to the position 1.
You need to determine the minimum number of operations necessary to make the string displayed on the screen a valid password.
Input Specification:
The first line contains two integers *n*, *m* (3<=≤<=*n*<=≤<=50,<=1<=≤<=*m*<=≤<=50) — the length of the password and the length of strings which are assigned to password symbols.
Each of the next *n* lines contains the string which is assigned to the *i*-th symbol of the password string. Its length is *m*, it consists of digits, lowercase English letters, and characters '#', '*' or '&'.
You have such input data that you can always get a valid password.
Output Specification:
Print one integer — the minimum number of operations which is necessary to make the string, which is displayed on the screen, a valid password.
Demo Input:
['3 4\n1**2\na3*0\nc4**\n', '5 5\n#*&#*\n*a1c&\n&q2w*\n#a3c#\n*&#*&\n']
Demo Output:
['1\n', '3\n']
Note:
In the first test it is necessary to move the pointer of the third string to one left to get the optimal answer.
In the second test one of possible algorithms will be:
- to move the pointer of the second symbol once to the right. - to move the pointer of the third symbol twice to the right. | ```python
n,m = list(map(int, input().split(" ")))
strings = []
for i in range(n):
strings.append(input())
spec = set('#*&')
dig = set('0123456789')
alpha = set('abcdefghijklmnopqrstuvwxyz')
def num(checkset, checkstring):
m = len(checkstring)+1
for i, c in enumerate(checkstring):
if c in checkset:
m = min(m, i)
m = min(m, len(checkstring)-i)
if m < len(checkstring)+1:
return m
return -1
ret = float('inf')
for i in range(n):
for j in range(n):
for k in range(n):
if i==j or i==k or j==k:
continue
else:
s1 = num(spec, strings[i])
s2 = num(dig, strings[j])
s3 = num(alpha, strings[k])
if s1 >= 0 and s2 >= 0 and s3 >= 0:
ret = min(ret, s1+s2+s3)
print(str(int(ret)))
``` | 3 | |
940 | B | Our Tanya is Crying Out Loud | PROGRAMMING | 1,400 | [
"dp",
"greedy"
] | null | null | Right now she actually isn't. But she will be, if you don't solve this problem.
You are given integers *n*, *k*, *A* and *B*. There is a number *x*, which is initially equal to *n*. You are allowed to perform two types of operations:
1. Subtract 1 from *x*. This operation costs you *A* coins. 1. Divide *x* by *k*. Can be performed only if *x* is divisible by *k*. This operation costs you *B* coins. | The first line contains a single integer *n* (1<=≤<=*n*<=≤<=2·109).
The second line contains a single integer *k* (1<=≤<=*k*<=≤<=2·109).
The third line contains a single integer *A* (1<=≤<=*A*<=≤<=2·109).
The fourth line contains a single integer *B* (1<=≤<=*B*<=≤<=2·109). | Output a single integer — the minimum amount of coins you have to pay to make *x* equal to 1. | [
"9\n2\n3\n1\n",
"5\n5\n2\n20\n",
"19\n3\n4\n2\n"
] | [
"6\n",
"8\n",
"12\n"
] | In the first testcase, the optimal strategy is as follows:
- Subtract 1 from *x* (9 → 8) paying 3 coins. - Divide *x* by 2 (8 → 4) paying 1 coin. - Divide *x* by 2 (4 → 2) paying 1 coin. - Divide *x* by 2 (2 → 1) paying 1 coin.
The total cost is 6 coins.
In the second test case the optimal strategy is to subtract 1 from *x* 4 times paying 8 coins in total. | 1,250 | [
{
"input": "9\n2\n3\n1",
"output": "6"
},
{
"input": "5\n5\n2\n20",
"output": "8"
},
{
"input": "19\n3\n4\n2",
"output": "12"
},
{
"input": "1845999546\n999435865\n1234234\n2323423",
"output": "1044857680578777"
},
{
"input": "1604353664\n1604353665\n9993432\n1",
... | 1,520,600,893 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 5 | 1,000 | 5,632,000 | n,k,a,b=(int(input()) for i in range(4))
x,s=n,0
while x!=1:
if k>x:
s,x=s+(x-1)*a,1
elif x%k!=0:
s,x=s+(x%k)*a,x-x%k
elif x*(k-1)//k*a>b:
s,x=s+b,x//k
else:
s,x=s+x*(k-1)//k*a,x//k
print(s)
| Title: Our Tanya is Crying Out Loud
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Right now she actually isn't. But she will be, if you don't solve this problem.
You are given integers *n*, *k*, *A* and *B*. There is a number *x*, which is initially equal to *n*. You are allowed to perform two types of operations:
1. Subtract 1 from *x*. This operation costs you *A* coins. 1. Divide *x* by *k*. Can be performed only if *x* is divisible by *k*. This operation costs you *B* coins.
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=2·109).
The second line contains a single integer *k* (1<=≤<=*k*<=≤<=2·109).
The third line contains a single integer *A* (1<=≤<=*A*<=≤<=2·109).
The fourth line contains a single integer *B* (1<=≤<=*B*<=≤<=2·109).
Output Specification:
Output a single integer — the minimum amount of coins you have to pay to make *x* equal to 1.
Demo Input:
['9\n2\n3\n1\n', '5\n5\n2\n20\n', '19\n3\n4\n2\n']
Demo Output:
['6\n', '8\n', '12\n']
Note:
In the first testcase, the optimal strategy is as follows:
- Subtract 1 from *x* (9 → 8) paying 3 coins. - Divide *x* by 2 (8 → 4) paying 1 coin. - Divide *x* by 2 (4 → 2) paying 1 coin. - Divide *x* by 2 (2 → 1) paying 1 coin.
The total cost is 6 coins.
In the second test case the optimal strategy is to subtract 1 from *x* 4 times paying 8 coins in total. | ```python
n,k,a,b=(int(input()) for i in range(4))
x,s=n,0
while x!=1:
if k>x:
s,x=s+(x-1)*a,1
elif x%k!=0:
s,x=s+(x%k)*a,x-x%k
elif x*(k-1)//k*a>b:
s,x=s+b,x//k
else:
s,x=s+x*(k-1)//k*a,x//k
print(s)
``` | 0 | |
445 | A | DZY Loves Chessboard | PROGRAMMING | 1,200 | [
"dfs and similar",
"implementation"
] | null | null | DZY loves chessboard, and he enjoys playing with it.
He has a chessboard of *n* rows and *m* columns. Some cells of the chessboard are bad, others are good. For every good cell, DZY wants to put a chessman on it. Each chessman is either white or black. After putting all chessmen, DZY wants that no two chessmen with the same color are on two adjacent cells. Two cells are adjacent if and only if they share a common edge.
You task is to find any suitable placement of chessmen on the given chessboard. | The first line contains two space-separated integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100).
Each of the next *n* lines contains a string of *m* characters: the *j*-th character of the *i*-th string is either "." or "-". A "." means that the corresponding cell (in the *i*-th row and the *j*-th column) is good, while a "-" means it is bad. | Output must contain *n* lines, each line must contain a string of *m* characters. The *j*-th character of the *i*-th string should be either "W", "B" or "-". Character "W" means the chessman on the cell is white, "B" means it is black, "-" means the cell is a bad cell.
If multiple answers exist, print any of them. It is guaranteed that at least one answer exists. | [
"1 1\n.\n",
"2 2\n..\n..\n",
"3 3\n.-.\n---\n--."
] | [
"B\n",
"BW\nWB\n",
"B-B\n---\n--B"
] | In the first sample, DZY puts a single black chessman. Of course putting a white one is also OK.
In the second sample, all 4 cells are good. No two same chessmen share an edge in the sample output.
In the third sample, no good cells are adjacent. So you can just put 3 chessmen, no matter what their colors are. | 500 | [
{
"input": "1 1\n.",
"output": "B"
},
{
"input": "2 2\n..\n..",
"output": "BW\nWB"
},
{
"input": "3 3\n.-.\n---\n--.",
"output": "B-B\n---\n--B"
},
{
"input": "3 1\n-\n.\n.",
"output": "-\nW\nB"
},
{
"input": "11 11\n-----------\n-----------\n-----------\n--------... | 1,626,249,223 | 2,147,483,647 | Python 3 | OK | TESTS | 37 | 78 | 7,065,600 | def spliter():
d = input()
a = d.split()
r = []
for i in a:
k = int(i)
r.append(k)
return r
n = spliter()
lst = []
final = []
for i in range(n[0]):
lst.append([])
k = input()
for j in range(n[1]):
lst[i].append(k[j])
for i in range(n[0]):
final.append([])
for j in range(n[1]):
if lst[i][j]=='-':
final[i].append('-')
print(final[i][j],end="")
else:
if (i+j)%2==0:
final[i].append('B')
print(final[i][j],end="")
else:
final[i].append('W')
print(final[i][j],end="")
print()
| Title: DZY Loves Chessboard
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
DZY loves chessboard, and he enjoys playing with it.
He has a chessboard of *n* rows and *m* columns. Some cells of the chessboard are bad, others are good. For every good cell, DZY wants to put a chessman on it. Each chessman is either white or black. After putting all chessmen, DZY wants that no two chessmen with the same color are on two adjacent cells. Two cells are adjacent if and only if they share a common edge.
You task is to find any suitable placement of chessmen on the given chessboard.
Input Specification:
The first line contains two space-separated integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100).
Each of the next *n* lines contains a string of *m* characters: the *j*-th character of the *i*-th string is either "." or "-". A "." means that the corresponding cell (in the *i*-th row and the *j*-th column) is good, while a "-" means it is bad.
Output Specification:
Output must contain *n* lines, each line must contain a string of *m* characters. The *j*-th character of the *i*-th string should be either "W", "B" or "-". Character "W" means the chessman on the cell is white, "B" means it is black, "-" means the cell is a bad cell.
If multiple answers exist, print any of them. It is guaranteed that at least one answer exists.
Demo Input:
['1 1\n.\n', '2 2\n..\n..\n', '3 3\n.-.\n---\n--.']
Demo Output:
['B\n', 'BW\nWB\n', 'B-B\n---\n--B']
Note:
In the first sample, DZY puts a single black chessman. Of course putting a white one is also OK.
In the second sample, all 4 cells are good. No two same chessmen share an edge in the sample output.
In the third sample, no good cells are adjacent. So you can just put 3 chessmen, no matter what their colors are. | ```python
def spliter():
d = input()
a = d.split()
r = []
for i in a:
k = int(i)
r.append(k)
return r
n = spliter()
lst = []
final = []
for i in range(n[0]):
lst.append([])
k = input()
for j in range(n[1]):
lst[i].append(k[j])
for i in range(n[0]):
final.append([])
for j in range(n[1]):
if lst[i][j]=='-':
final[i].append('-')
print(final[i][j],end="")
else:
if (i+j)%2==0:
final[i].append('B')
print(final[i][j],end="")
else:
final[i].append('W')
print(final[i][j],end="")
print()
``` | 3 | |
262 | B | Roma and Changing Signs | PROGRAMMING | 1,200 | [
"greedy"
] | null | null | Roma works in a company that sells TVs. Now he has to prepare a report for the last year.
Roma has got a list of the company's incomes. The list is a sequence that consists of *n* integers. The total income of the company is the sum of all integers in sequence. Roma decided to perform exactly *k* changes of signs of several numbers in the sequence. He can also change the sign of a number one, two or more times.
The operation of changing a number's sign is the operation of multiplying this number by -1.
Help Roma perform the changes so as to make the total income of the company (the sum of numbers in the resulting sequence) maximum. Note that Roma should perform exactly *k* changes. | The first line contains two integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=105), showing, how many numbers are in the sequence and how many swaps are to be made.
The second line contains a non-decreasing sequence, consisting of *n* integers *a**i* (|*a**i*|<=≤<=104).
The numbers in the lines are separated by single spaces. Please note that the given sequence is sorted in non-decreasing order. | In the single line print the answer to the problem — the maximum total income that we can obtain after exactly *k* changes. | [
"3 2\n-1 -1 1\n",
"3 1\n-1 -1 1\n"
] | [
"3\n",
"1\n"
] | In the first sample we can get sequence [1, 1, 1], thus the total income equals 3.
In the second test, the optimal strategy is to get sequence [-1, 1, 1], thus the total income equals 1. | 1,000 | [
{
"input": "3 2\n-1 -1 1",
"output": "3"
},
{
"input": "3 1\n-1 -1 1",
"output": "1"
},
{
"input": "17 27\n257 320 676 1136 2068 2505 2639 4225 4951 5786 7677 7697 7851 8337 8429 8469 9343",
"output": "81852"
},
{
"input": "69 28\n-9822 -9264 -9253 -9221 -9139 -9126 -9096 -89... | 1,662,409,940 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 4 | 92 | 0 | n, changes = map(int, input().split())
incomes = sorted(list(map(int, input().split())))
diff = 0
i = 0
while incomes[i] <= 0 and changes > 0:
diff += (incomes[i] * -2)
changes -= 1
i += 1
if changes % 2 == 1:
diff -= abs(sorted(incomes, key=lambda l: abs(l))[0]) * 2
print(sum(incomes) + diff)
| Title: Roma and Changing Signs
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Roma works in a company that sells TVs. Now he has to prepare a report for the last year.
Roma has got a list of the company's incomes. The list is a sequence that consists of *n* integers. The total income of the company is the sum of all integers in sequence. Roma decided to perform exactly *k* changes of signs of several numbers in the sequence. He can also change the sign of a number one, two or more times.
The operation of changing a number's sign is the operation of multiplying this number by -1.
Help Roma perform the changes so as to make the total income of the company (the sum of numbers in the resulting sequence) maximum. Note that Roma should perform exactly *k* changes.
Input Specification:
The first line contains two integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=105), showing, how many numbers are in the sequence and how many swaps are to be made.
The second line contains a non-decreasing sequence, consisting of *n* integers *a**i* (|*a**i*|<=≤<=104).
The numbers in the lines are separated by single spaces. Please note that the given sequence is sorted in non-decreasing order.
Output Specification:
In the single line print the answer to the problem — the maximum total income that we can obtain after exactly *k* changes.
Demo Input:
['3 2\n-1 -1 1\n', '3 1\n-1 -1 1\n']
Demo Output:
['3\n', '1\n']
Note:
In the first sample we can get sequence [1, 1, 1], thus the total income equals 3.
In the second test, the optimal strategy is to get sequence [-1, 1, 1], thus the total income equals 1. | ```python
n, changes = map(int, input().split())
incomes = sorted(list(map(int, input().split())))
diff = 0
i = 0
while incomes[i] <= 0 and changes > 0:
diff += (incomes[i] * -2)
changes -= 1
i += 1
if changes % 2 == 1:
diff -= abs(sorted(incomes, key=lambda l: abs(l))[0]) * 2
print(sum(incomes) + diff)
``` | -1 | |
960 | A | Check the string | PROGRAMMING | 1,200 | [
"implementation"
] | null | null | A has a string consisting of some number of lowercase English letters 'a'. He gives it to his friend B who appends some number of letters 'b' to the end of this string. Since both A and B like the characters 'a' and 'b', they have made sure that at this point, at least one 'a' and one 'b' exist in the string.
B now gives this string to C and he appends some number of letters 'c' to the end of the string. However, since C is a good friend of A and B, the number of letters 'c' he appends is equal to the number of 'a' or to the number of 'b' in the string. It is also possible that the number of letters 'c' equals both to the number of letters 'a' and to the number of letters 'b' at the same time.
You have a string in your hands, and you want to check if it is possible to obtain the string in this way or not. If it is possible to obtain the string, print "YES", otherwise print "NO" (without the quotes). | The first and only line consists of a string $S$ ($ 1 \le |S| \le 5\,000 $). It is guaranteed that the string will only consist of the lowercase English letters 'a', 'b', 'c'. | Print "YES" or "NO", according to the condition. | [
"aaabccc\n",
"bbacc\n",
"aabc\n"
] | [
"YES\n",
"NO\n",
"YES\n"
] | Consider first example: the number of 'c' is equal to the number of 'a'.
Consider second example: although the number of 'c' is equal to the number of the 'b', the order is not correct.
Consider third example: the number of 'c' is equal to the number of 'b'. | 500 | [
{
"input": "aaabccc",
"output": "YES"
},
{
"input": "bbacc",
"output": "NO"
},
{
"input": "aabc",
"output": "YES"
},
{
"input": "aabbcc",
"output": "YES"
},
{
"input": "aaacccbb",
"output": "NO"
},
{
"input": "abc",
"output": "YES"
},
{
"in... | 1,597,862,118 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 32 | 108 | 307,200 | # This is a sample Python script.
# Press Shift+F10 to execute it or replace it with your code.
# Press Double Shift to search everywhere for classes, files, tool windows, actions, and settings.
def print_hi(name):
# Use a breakpoint in the code line below to debug your script.
print(f'Hi, {name}') # Press Ctrl+F8 to toggle the breakpoint.
def check_the_string(s):
if (s.find('a') + 1) and s.find('b') and s.find('c') and s.find('a') < s.find('b') < s.find('c'):
if s.count('a') == s.count('c') or s.count('b') == s.count('c'):
print('YES')
else:
print('NO')
else:
print('NO')
# Press the green button in the gutter to run the script.
if __name__ == '__main__':
s = input()
check_the_string(s)
# print_hi('PyCharm')
# See PyCharm help at https://www.jetbrains.com/help/pycharm/
| Title: Check the string
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A has a string consisting of some number of lowercase English letters 'a'. He gives it to his friend B who appends some number of letters 'b' to the end of this string. Since both A and B like the characters 'a' and 'b', they have made sure that at this point, at least one 'a' and one 'b' exist in the string.
B now gives this string to C and he appends some number of letters 'c' to the end of the string. However, since C is a good friend of A and B, the number of letters 'c' he appends is equal to the number of 'a' or to the number of 'b' in the string. It is also possible that the number of letters 'c' equals both to the number of letters 'a' and to the number of letters 'b' at the same time.
You have a string in your hands, and you want to check if it is possible to obtain the string in this way or not. If it is possible to obtain the string, print "YES", otherwise print "NO" (without the quotes).
Input Specification:
The first and only line consists of a string $S$ ($ 1 \le |S| \le 5\,000 $). It is guaranteed that the string will only consist of the lowercase English letters 'a', 'b', 'c'.
Output Specification:
Print "YES" or "NO", according to the condition.
Demo Input:
['aaabccc\n', 'bbacc\n', 'aabc\n']
Demo Output:
['YES\n', 'NO\n', 'YES\n']
Note:
Consider first example: the number of 'c' is equal to the number of 'a'.
Consider second example: although the number of 'c' is equal to the number of the 'b', the order is not correct.
Consider third example: the number of 'c' is equal to the number of 'b'. | ```python
# This is a sample Python script.
# Press Shift+F10 to execute it or replace it with your code.
# Press Double Shift to search everywhere for classes, files, tool windows, actions, and settings.
def print_hi(name):
# Use a breakpoint in the code line below to debug your script.
print(f'Hi, {name}') # Press Ctrl+F8 to toggle the breakpoint.
def check_the_string(s):
if (s.find('a') + 1) and s.find('b') and s.find('c') and s.find('a') < s.find('b') < s.find('c'):
if s.count('a') == s.count('c') or s.count('b') == s.count('c'):
print('YES')
else:
print('NO')
else:
print('NO')
# Press the green button in the gutter to run the script.
if __name__ == '__main__':
s = input()
check_the_string(s)
# print_hi('PyCharm')
# See PyCharm help at https://www.jetbrains.com/help/pycharm/
``` | 0 | |
61 | A | Ultra-Fast Mathematician | PROGRAMMING | 800 | [
"implementation"
] | A. Ultra-Fast Mathematician | 2 | 256 | Shapur was an extremely gifted student. He was great at everything including Combinatorics, Algebra, Number Theory, Geometry, Calculus, etc. He was not only smart but extraordinarily fast! He could manage to sum 1018 numbers in a single second.
One day in 230 AD Shapur was trying to find out if any one can possibly do calculations faster than him. As a result he made a very great contest and asked every one to come and take part.
In his contest he gave the contestants many different pairs of numbers. Each number is made from digits 0 or 1. The contestants should write a new number corresponding to the given pair of numbers. The rule is simple: The *i*-th digit of the answer is 1 if and only if the *i*-th digit of the two given numbers differ. In the other case the *i*-th digit of the answer is 0.
Shapur made many numbers and first tried his own speed. He saw that he can perform these operations on numbers of length ∞ (length of a number is number of digits in it) in a glance! He always gives correct answers so he expects the contestants to give correct answers, too. He is a good fellow so he won't give anyone very big numbers and he always gives one person numbers of same length.
Now you are going to take part in Shapur's contest. See if you are faster and more accurate. | There are two lines in each input. Each of them contains a single number. It is guaranteed that the numbers are made from 0 and 1 only and that their length is same. The numbers may start with 0. The length of each number doesn't exceed 100. | Write one line — the corresponding answer. Do not omit the leading 0s. | [
"1010100\n0100101\n",
"000\n111\n",
"1110\n1010\n",
"01110\n01100\n"
] | [
"1110001\n",
"111\n",
"0100\n",
"00010\n"
] | none | 500 | [
{
"input": "1010100\n0100101",
"output": "1110001"
},
{
"input": "000\n111",
"output": "111"
},
{
"input": "1110\n1010",
"output": "0100"
},
{
"input": "01110\n01100",
"output": "00010"
},
{
"input": "011101\n000001",
"output": "011100"
},
{
"input": "... | 1,594,114,987 | 2,147,483,647 | Python 3 | OK | TESTS | 102 | 109 | 6,656,000 | s1=input()
s2=input()
x=''
for i in range(len(s1)):
if s1[i]==s2[i]:
x=x+str(0)
else:
x=x+str(1)
print(x) | Title: Ultra-Fast Mathematician
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Shapur was an extremely gifted student. He was great at everything including Combinatorics, Algebra, Number Theory, Geometry, Calculus, etc. He was not only smart but extraordinarily fast! He could manage to sum 1018 numbers in a single second.
One day in 230 AD Shapur was trying to find out if any one can possibly do calculations faster than him. As a result he made a very great contest and asked every one to come and take part.
In his contest he gave the contestants many different pairs of numbers. Each number is made from digits 0 or 1. The contestants should write a new number corresponding to the given pair of numbers. The rule is simple: The *i*-th digit of the answer is 1 if and only if the *i*-th digit of the two given numbers differ. In the other case the *i*-th digit of the answer is 0.
Shapur made many numbers and first tried his own speed. He saw that he can perform these operations on numbers of length ∞ (length of a number is number of digits in it) in a glance! He always gives correct answers so he expects the contestants to give correct answers, too. He is a good fellow so he won't give anyone very big numbers and he always gives one person numbers of same length.
Now you are going to take part in Shapur's contest. See if you are faster and more accurate.
Input Specification:
There are two lines in each input. Each of them contains a single number. It is guaranteed that the numbers are made from 0 and 1 only and that their length is same. The numbers may start with 0. The length of each number doesn't exceed 100.
Output Specification:
Write one line — the corresponding answer. Do not omit the leading 0s.
Demo Input:
['1010100\n0100101\n', '000\n111\n', '1110\n1010\n', '01110\n01100\n']
Demo Output:
['1110001\n', '111\n', '0100\n', '00010\n']
Note:
none | ```python
s1=input()
s2=input()
x=''
for i in range(len(s1)):
if s1[i]==s2[i]:
x=x+str(0)
else:
x=x+str(1)
print(x)
``` | 3.960352 |
50 | A | Domino piling | PROGRAMMING | 800 | [
"greedy",
"math"
] | A. Domino piling | 2 | 256 | You are given a rectangular board of *M*<=×<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=×<=1 squares. You are allowed to rotate the pieces. You are asked to place as many dominoes as possible on the board so as to meet the following conditions:
1. Each domino completely covers two squares.
2. No two dominoes overlap.
3. Each domino lies entirely inside the board. It is allowed to touch the edges of the board.
Find the maximum number of dominoes, which can be placed under these restrictions. | In a single line you are given two integers *M* and *N* — board sizes in squares (1<=≤<=*M*<=≤<=*N*<=≤<=16). | Output one number — the maximal number of dominoes, which can be placed. | [
"2 4\n",
"3 3\n"
] | [
"4\n",
"4\n"
] | none | 500 | [
{
"input": "2 4",
"output": "4"
},
{
"input": "3 3",
"output": "4"
},
{
"input": "1 5",
"output": "2"
},
{
"input": "1 6",
"output": "3"
},
{
"input": "1 15",
"output": "7"
},
{
"input": "1 16",
"output": "8"
},
{
"input": "2 5",
"outpu... | 1,681,489,083 | 2,147,483,647 | Python 3 | OK | TESTS | 35 | 124 | 0 | l,m = map(int, input().split())
if l%2 == 0 or m %2 == 0:
print((l//2) * m if l%2 == 0 else (m//2) * l)
else:
print((max(m, l) - 1)//2 * min(m, l) + min(m, l) //2)
| Title: Domino piling
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
You are given a rectangular board of *M*<=×<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=×<=1 squares. You are allowed to rotate the pieces. You are asked to place as many dominoes as possible on the board so as to meet the following conditions:
1. Each domino completely covers two squares.
2. No two dominoes overlap.
3. Each domino lies entirely inside the board. It is allowed to touch the edges of the board.
Find the maximum number of dominoes, which can be placed under these restrictions.
Input Specification:
In a single line you are given two integers *M* and *N* — board sizes in squares (1<=≤<=*M*<=≤<=*N*<=≤<=16).
Output Specification:
Output one number — the maximal number of dominoes, which can be placed.
Demo Input:
['2 4\n', '3 3\n']
Demo Output:
['4\n', '4\n']
Note:
none | ```python
l,m = map(int, input().split())
if l%2 == 0 or m %2 == 0:
print((l//2) * m if l%2 == 0 else (m//2) * l)
else:
print((max(m, l) - 1)//2 * min(m, l) + min(m, l) //2)
``` | 3.969 |
764 | A | Taymyr is calling you | PROGRAMMING | 800 | [
"brute force",
"implementation",
"math"
] | null | null | Comrade Dujikov is busy choosing artists for Timofey's birthday and is recieving calls from Taymyr from Ilia-alpinist.
Ilia-alpinist calls every *n* minutes, i.e. in minutes *n*, 2*n*, 3*n* and so on. Artists come to the comrade every *m* minutes, i.e. in minutes *m*, 2*m*, 3*m* and so on. The day is *z* minutes long, i.e. the day consists of minutes 1,<=2,<=...,<=*z*. How many artists should be killed so that there are no artists in the room when Ilia calls? Consider that a call and a talk with an artist take exactly one minute. | The only string contains three integers — *n*, *m* and *z* (1<=≤<=*n*,<=*m*,<=*z*<=≤<=104). | Print single integer — the minimum number of artists that should be killed so that there are no artists in the room when Ilia calls. | [
"1 1 10\n",
"1 2 5\n",
"2 3 9\n"
] | [
"10\n",
"2\n",
"1\n"
] | Taymyr is a place in the north of Russia.
In the first test the artists come each minute, as well as the calls, so we need to kill all of them.
In the second test we need to kill artists which come on the second and the fourth minutes.
In the third test — only the artist which comes on the sixth minute. | 500 | [
{
"input": "1 1 10",
"output": "10"
},
{
"input": "1 2 5",
"output": "2"
},
{
"input": "2 3 9",
"output": "1"
},
{
"input": "4 8 9",
"output": "1"
},
{
"input": "7 9 2",
"output": "0"
},
{
"input": "10000 10000 10000",
"output": "1"
},
{
"i... | 1,646,691,986 | 2,147,483,647 | Python 3 | OK | TESTS | 32 | 46 | 1,638,400 | n,m,z = map(int, input().split(" "))
a = set()
c = set()
for i in range(1,z+1):
if (i*m <= z):
a.add(i*m)
if (i*n <= z):
c.add(i*n)
commons = len(set(a) - (set(a) - set(c)))
print(commons) | Title: Taymyr is calling you
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Comrade Dujikov is busy choosing artists for Timofey's birthday and is recieving calls from Taymyr from Ilia-alpinist.
Ilia-alpinist calls every *n* minutes, i.e. in minutes *n*, 2*n*, 3*n* and so on. Artists come to the comrade every *m* minutes, i.e. in minutes *m*, 2*m*, 3*m* and so on. The day is *z* minutes long, i.e. the day consists of minutes 1,<=2,<=...,<=*z*. How many artists should be killed so that there are no artists in the room when Ilia calls? Consider that a call and a talk with an artist take exactly one minute.
Input Specification:
The only string contains three integers — *n*, *m* and *z* (1<=≤<=*n*,<=*m*,<=*z*<=≤<=104).
Output Specification:
Print single integer — the minimum number of artists that should be killed so that there are no artists in the room when Ilia calls.
Demo Input:
['1 1 10\n', '1 2 5\n', '2 3 9\n']
Demo Output:
['10\n', '2\n', '1\n']
Note:
Taymyr is a place in the north of Russia.
In the first test the artists come each minute, as well as the calls, so we need to kill all of them.
In the second test we need to kill artists which come on the second and the fourth minutes.
In the third test — only the artist which comes on the sixth minute. | ```python
n,m,z = map(int, input().split(" "))
a = set()
c = set()
for i in range(1,z+1):
if (i*m <= z):
a.add(i*m)
if (i*n <= z):
c.add(i*n)
commons = len(set(a) - (set(a) - set(c)))
print(commons)
``` | 3 | |
670 | A | Holidays | PROGRAMMING | 900 | [
"brute force",
"constructive algorithms",
"greedy",
"math"
] | null | null | On the planet Mars a year lasts exactly *n* days (there are no leap years on Mars). But Martians have the same weeks as earthlings — 5 work days and then 2 days off. Your task is to determine the minimum possible and the maximum possible number of days off per year on Mars. | The first line of the input contains a positive integer *n* (1<=≤<=*n*<=≤<=1<=000<=000) — the number of days in a year on Mars. | Print two integers — the minimum possible and the maximum possible number of days off per year on Mars. | [
"14\n",
"2\n"
] | [
"4 4\n",
"0 2\n"
] | In the first sample there are 14 days in a year on Mars, and therefore independently of the day a year starts with there will be exactly 4 days off .
In the second sample there are only 2 days in a year on Mars, and they can both be either work days or days off. | 500 | [
{
"input": "14",
"output": "4 4"
},
{
"input": "2",
"output": "0 2"
},
{
"input": "1",
"output": "0 1"
},
{
"input": "3",
"output": "0 2"
},
{
"input": "4",
"output": "0 2"
},
{
"input": "5",
"output": "0 2"
},
{
"input": "6",
"output":... | 1,682,537,594 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 14 | 61 | 512,000 | def holidaymax(n):
if n<=0:
return 0
elif 0<n<=2:
return n
else:
return 2+holidaymax(n-7)
def holidaymin(n):
if n<=5:
return 0
elif 5<n<=7:
return n-5
else:
return 2+holidaymin(n-7)
n=int(input())
print(holidaymin(n),holidaymax(n)) | Title: Holidays
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
On the planet Mars a year lasts exactly *n* days (there are no leap years on Mars). But Martians have the same weeks as earthlings — 5 work days and then 2 days off. Your task is to determine the minimum possible and the maximum possible number of days off per year on Mars.
Input Specification:
The first line of the input contains a positive integer *n* (1<=≤<=*n*<=≤<=1<=000<=000) — the number of days in a year on Mars.
Output Specification:
Print two integers — the minimum possible and the maximum possible number of days off per year on Mars.
Demo Input:
['14\n', '2\n']
Demo Output:
['4 4\n', '0 2\n']
Note:
In the first sample there are 14 days in a year on Mars, and therefore independently of the day a year starts with there will be exactly 4 days off .
In the second sample there are only 2 days in a year on Mars, and they can both be either work days or days off. | ```python
def holidaymax(n):
if n<=0:
return 0
elif 0<n<=2:
return n
else:
return 2+holidaymax(n-7)
def holidaymin(n):
if n<=5:
return 0
elif 5<n<=7:
return n-5
else:
return 2+holidaymin(n-7)
n=int(input())
print(holidaymin(n),holidaymax(n))
``` | -1 | |
11 | A | Increasing Sequence | PROGRAMMING | 900 | [
"constructive algorithms",
"implementation",
"math"
] | A. Increasing Sequence | 1 | 64 | A sequence *a*0,<=*a*1,<=...,<=*a**t*<=-<=1 is called increasing if *a**i*<=-<=1<=<<=*a**i* for each *i*:<=0<=<<=*i*<=<<=*t*.
You are given a sequence *b*0,<=*b*1,<=...,<=*b**n*<=-<=1 and a positive integer *d*. In each move you may choose one element of the given sequence and add *d* to it. What is the least number of moves required to make the given sequence increasing? | The first line of the input contains two integer numbers *n* and *d* (2<=≤<=*n*<=≤<=2000,<=1<=≤<=*d*<=≤<=106). The second line contains space separated sequence *b*0,<=*b*1,<=...,<=*b**n*<=-<=1 (1<=≤<=*b**i*<=≤<=106). | Output the minimal number of moves needed to make the sequence increasing. | [
"4 2\n1 3 3 2\n"
] | [
"3\n"
] | none | 0 | [
{
"input": "4 2\n1 3 3 2",
"output": "3"
},
{
"input": "2 1\n1 1",
"output": "1"
},
{
"input": "2 1\n2 5",
"output": "0"
},
{
"input": "2 1\n1 2",
"output": "0"
},
{
"input": "2 1\n1 1",
"output": "1"
},
{
"input": "2 7\n10 20",
"output": "0"
},
... | 1,592,317,688 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 15 | 1,000 | 0 | n, d = map(int, input().split())
nums = list(map(int, input().split()))
res = 0
for i in range(1, n):
while nums[i] <= nums[i-1]:
nums[i] += d
res += 1
print(res) | Title: Increasing Sequence
Time Limit: 1 seconds
Memory Limit: 64 megabytes
Problem Description:
A sequence *a*0,<=*a*1,<=...,<=*a**t*<=-<=1 is called increasing if *a**i*<=-<=1<=<<=*a**i* for each *i*:<=0<=<<=*i*<=<<=*t*.
You are given a sequence *b*0,<=*b*1,<=...,<=*b**n*<=-<=1 and a positive integer *d*. In each move you may choose one element of the given sequence and add *d* to it. What is the least number of moves required to make the given sequence increasing?
Input Specification:
The first line of the input contains two integer numbers *n* and *d* (2<=≤<=*n*<=≤<=2000,<=1<=≤<=*d*<=≤<=106). The second line contains space separated sequence *b*0,<=*b*1,<=...,<=*b**n*<=-<=1 (1<=≤<=*b**i*<=≤<=106).
Output Specification:
Output the minimal number of moves needed to make the sequence increasing.
Demo Input:
['4 2\n1 3 3 2\n']
Demo Output:
['3\n']
Note:
none | ```python
n, d = map(int, input().split())
nums = list(map(int, input().split()))
res = 0
for i in range(1, n):
while nums[i] <= nums[i-1]:
nums[i] += d
res += 1
print(res)
``` | 0 |
897 | A | Scarborough Fair | PROGRAMMING | 800 | [
"implementation"
] | null | null | Parsley, sage, rosemary and thyme.
Remember me to one who lives there.
He once was the true love of mine.
Willem is taking the girl to the highest building in island No.28, however, neither of them knows how to get there.
Willem asks his friend, Grick for directions, Grick helped them, and gave them a task.
Although the girl wants to help, Willem insists on doing it by himself.
Grick gave Willem a string of length *n*.
Willem needs to do *m* operations, each operation has four parameters *l*,<=*r*,<=*c*1,<=*c*2, which means that all symbols *c*1 in range [*l*,<=*r*] (from *l*-th to *r*-th, including *l* and *r*) are changed into *c*2. String is 1-indexed.
Grick wants to know the final string after all the *m* operations. | The first line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100).
The second line contains a string *s* of length *n*, consisting of lowercase English letters.
Each of the next *m* lines contains four parameters *l*,<=*r*,<=*c*1,<=*c*2 (1<=≤<=*l*<=≤<=*r*<=≤<=*n*, *c*1,<=*c*2 are lowercase English letters), separated by space. | Output string *s* after performing *m* operations described above. | [
"3 1\nioi\n1 1 i n\n",
"5 3\nwxhak\n3 3 h x\n1 5 x a\n1 3 w g\n"
] | [
"noi",
"gaaak"
] | For the second example:
After the first operation, the string is wxxak.
After the second operation, the string is waaak.
After the third operation, the string is gaaak. | 500 | [
{
"input": "3 1\nioi\n1 1 i n",
"output": "noi"
},
{
"input": "5 3\nwxhak\n3 3 h x\n1 5 x a\n1 3 w g",
"output": "gaaak"
},
{
"input": "9 51\nbhfbdcgff\n2 3 b b\n2 8 e f\n3 8 g f\n5 7 d a\n1 5 e b\n3 4 g b\n6 7 c d\n3 6 e g\n3 6 e h\n5 6 a e\n7 9 a c\n4 9 a h\n3 7 c b\n6 9 b g\n1 7 h b\n... | 1,664,970,084 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 47 | 62 | 3,276,800 | def main():
n, m = map(int, input().split())
st = list(input())
for _ in range(m):
l, r, c1, c2 = input().split()
for i in range(int(l)-1, int(r)):
if st[i] == c1:
st[i] = c2
print(''.join(st))
if __name__ == '__main__':
main()
| Title: Scarborough Fair
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Parsley, sage, rosemary and thyme.
Remember me to one who lives there.
He once was the true love of mine.
Willem is taking the girl to the highest building in island No.28, however, neither of them knows how to get there.
Willem asks his friend, Grick for directions, Grick helped them, and gave them a task.
Although the girl wants to help, Willem insists on doing it by himself.
Grick gave Willem a string of length *n*.
Willem needs to do *m* operations, each operation has four parameters *l*,<=*r*,<=*c*1,<=*c*2, which means that all symbols *c*1 in range [*l*,<=*r*] (from *l*-th to *r*-th, including *l* and *r*) are changed into *c*2. String is 1-indexed.
Grick wants to know the final string after all the *m* operations.
Input Specification:
The first line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100).
The second line contains a string *s* of length *n*, consisting of lowercase English letters.
Each of the next *m* lines contains four parameters *l*,<=*r*,<=*c*1,<=*c*2 (1<=≤<=*l*<=≤<=*r*<=≤<=*n*, *c*1,<=*c*2 are lowercase English letters), separated by space.
Output Specification:
Output string *s* after performing *m* operations described above.
Demo Input:
['3 1\nioi\n1 1 i n\n', '5 3\nwxhak\n3 3 h x\n1 5 x a\n1 3 w g\n']
Demo Output:
['noi', 'gaaak']
Note:
For the second example:
After the first operation, the string is wxxak.
After the second operation, the string is waaak.
After the third operation, the string is gaaak. | ```python
def main():
n, m = map(int, input().split())
st = list(input())
for _ in range(m):
l, r, c1, c2 = input().split()
for i in range(int(l)-1, int(r)):
if st[i] == c1:
st[i] = c2
print(''.join(st))
if __name__ == '__main__':
main()
``` | 3 | |
637 | B | Chat Order | PROGRAMMING | 1,200 | [
"*special",
"binary search",
"constructive algorithms",
"data structures",
"sortings"
] | null | null | Polycarp is a big lover of killing time in social networks. A page with a chatlist in his favourite network is made so that when a message is sent to some friend, his friend's chat rises to the very top of the page. The relative order of the other chats doesn't change. If there was no chat with this friend before, then a new chat is simply inserted to the top of the list.
Assuming that the chat list is initially empty, given the sequence of Polycaprus' messages make a list of chats after all of his messages are processed. Assume that no friend wrote any message to Polycarpus. | The first line contains integer *n* (1<=≤<=*n*<=≤<=200<=000) — the number of Polycarpus' messages. Next *n* lines enlist the message recipients in the order in which the messages were sent. The name of each participant is a non-empty sequence of lowercase English letters of length at most 10. | Print all the recipients to who Polycarp talked to in the order of chats with them, from top to bottom. | [
"4\nalex\nivan\nroman\nivan\n",
"8\nalina\nmaria\nekaterina\ndarya\ndarya\nekaterina\nmaria\nalina\n"
] | [
"ivan\nroman\nalex\n",
"alina\nmaria\nekaterina\ndarya\n"
] | In the first test case Polycarpus first writes to friend by name "alex", and the list looks as follows:
1. alex
Then Polycarpus writes to friend by name "ivan" and the list looks as follows:
1. ivan 1. alex
Polycarpus writes the third message to friend by name "roman" and the list looks as follows:
1. roman 1. ivan 1. alex
Polycarpus writes the fourth message to friend by name "ivan", to who he has already sent a message, so the list of chats changes as follows:
1. ivan 1. roman 1. alex | 1,000 | [
{
"input": "4\nalex\nivan\nroman\nivan",
"output": "ivan\nroman\nalex"
},
{
"input": "8\nalina\nmaria\nekaterina\ndarya\ndarya\nekaterina\nmaria\nalina",
"output": "alina\nmaria\nekaterina\ndarya"
},
{
"input": "1\nwdi",
"output": "wdi"
},
{
"input": "2\nypg\nypg",
"outpu... | 1,700,268,753 | 2,147,483,647 | Python 3 | OK | TESTS | 95 | 607 | 47,820,800 | n = int(input())
names = {}
i = 0
for _ in range(n):
names[input()] = i
i += 1
for name, key in sorted(names.items(), key=lambda item: item[1], reverse=True):
print(name) | Title: Chat Order
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarp is a big lover of killing time in social networks. A page with a chatlist in his favourite network is made so that when a message is sent to some friend, his friend's chat rises to the very top of the page. The relative order of the other chats doesn't change. If there was no chat with this friend before, then a new chat is simply inserted to the top of the list.
Assuming that the chat list is initially empty, given the sequence of Polycaprus' messages make a list of chats after all of his messages are processed. Assume that no friend wrote any message to Polycarpus.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=200<=000) — the number of Polycarpus' messages. Next *n* lines enlist the message recipients in the order in which the messages were sent. The name of each participant is a non-empty sequence of lowercase English letters of length at most 10.
Output Specification:
Print all the recipients to who Polycarp talked to in the order of chats with them, from top to bottom.
Demo Input:
['4\nalex\nivan\nroman\nivan\n', '8\nalina\nmaria\nekaterina\ndarya\ndarya\nekaterina\nmaria\nalina\n']
Demo Output:
['ivan\nroman\nalex\n', 'alina\nmaria\nekaterina\ndarya\n']
Note:
In the first test case Polycarpus first writes to friend by name "alex", and the list looks as follows:
1. alex
Then Polycarpus writes to friend by name "ivan" and the list looks as follows:
1. ivan 1. alex
Polycarpus writes the third message to friend by name "roman" and the list looks as follows:
1. roman 1. ivan 1. alex
Polycarpus writes the fourth message to friend by name "ivan", to who he has already sent a message, so the list of chats changes as follows:
1. ivan 1. roman 1. alex | ```python
n = int(input())
names = {}
i = 0
for _ in range(n):
names[input()] = i
i += 1
for name, key in sorted(names.items(), key=lambda item: item[1], reverse=True):
print(name)
``` | 3 | |
19 | B | Checkout Assistant | PROGRAMMING | 1,900 | [
"dp"
] | B. Checkout Assistant | 1 | 256 | Bob came to a cash & carry store, put *n* items into his trolley, and went to the checkout counter to pay. Each item is described by its price *c**i* and time *t**i* in seconds that a checkout assistant spends on this item. While the checkout assistant is occupied with some item, Bob can steal some other items from his trolley. To steal one item Bob needs exactly 1 second. What is the minimum amount of money that Bob will have to pay to the checkout assistant? Remember, please, that it is Bob, who determines the order of items for the checkout assistant. | The first input line contains number *n* (1<=≤<=*n*<=≤<=2000). In each of the following *n* lines each item is described by a pair of numbers *t**i*, *c**i* (0<=≤<=*t**i*<=≤<=2000,<=1<=≤<=*c**i*<=≤<=109). If *t**i* is 0, Bob won't be able to steal anything, while the checkout assistant is occupied with item *i*. | Output one number — answer to the problem: what is the minimum amount of money that Bob will have to pay. | [
"4\n2 10\n0 20\n1 5\n1 3\n",
"3\n0 1\n0 10\n0 100\n"
] | [
"8\n",
"111\n"
] | none | 0 | [
{
"input": "4\n2 10\n0 20\n1 5\n1 3",
"output": "8"
},
{
"input": "3\n0 1\n0 10\n0 100",
"output": "111"
},
{
"input": "2\n0 635254032\n0 75159864",
"output": "710413896"
},
{
"input": "2\n0 861438648\n1 469893784",
"output": "469893784"
},
{
"input": "2\n2 876232... | 1,664,130,761 | 2,147,483,647 | PyPy 3 | RUNTIME_ERROR | TESTS | 0 | 77 | 2,150,400 | from sys import stdin, stdout
def read_list(): # read list of variables
return [int(x) for x in stdin.readline().split()]
def read_list_str(): # read list of variables
return [x for x in stdin.readline().split()]
def readmv(): # read_multiple_variable
return map(int, stdin.readline().split())
def readv(): # read variable
return int(stdin.readline())
def printSol(test_case, x):
ans_to_print = ''
if type(x) == list:
for el in x:
ans_to_print += f'{el} '
if type(x) == int or type(x) == float:
ans_to_print = str(x)
stdout.write(f'Case #{test_case + 1}: ' + ans_to_print + f'\n')
def read():
return stdin.readline().strip()
from heapq import heapify, heappop, heappush
from math import inf
def dijkstra(graph, s):
n = len(graph)
visited = [False]*n
dist = [inf]*n
q = [(s, 0)]
heapify(q)
visited[s] = True
dist[s] = 0
while len(q):
_, u = heappop(q)
for (v, w) in graph[u]:
if not visited[v] and dist[v] > dist[u] + w:
dist[v] = dist[u] + w
heappush(q, (dist[v], v))
return dist
def main():
n, m = readmv()
graph = [[] for _ in range(2*n)]
for _ in range(m):
x, y, w = readmv()
x -= 1
y -= 1
graph[x].append((y, w))
graph[y+n].append((x+n, w))
for x in range(n):
graph[x].append((x+n, 0))
dist = dijkstra(graph, 0)
for i in range(1, n):
if dist[n+i] == inf:
stdout.write('-1 ')
continue
stdout.write(f'{dist[n+i]} ')
stdout.write('\n')
main() | Title: Checkout Assistant
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Bob came to a cash & carry store, put *n* items into his trolley, and went to the checkout counter to pay. Each item is described by its price *c**i* and time *t**i* in seconds that a checkout assistant spends on this item. While the checkout assistant is occupied with some item, Bob can steal some other items from his trolley. To steal one item Bob needs exactly 1 second. What is the minimum amount of money that Bob will have to pay to the checkout assistant? Remember, please, that it is Bob, who determines the order of items for the checkout assistant.
Input Specification:
The first input line contains number *n* (1<=≤<=*n*<=≤<=2000). In each of the following *n* lines each item is described by a pair of numbers *t**i*, *c**i* (0<=≤<=*t**i*<=≤<=2000,<=1<=≤<=*c**i*<=≤<=109). If *t**i* is 0, Bob won't be able to steal anything, while the checkout assistant is occupied with item *i*.
Output Specification:
Output one number — answer to the problem: what is the minimum amount of money that Bob will have to pay.
Demo Input:
['4\n2 10\n0 20\n1 5\n1 3\n', '3\n0 1\n0 10\n0 100\n']
Demo Output:
['8\n', '111\n']
Note:
none | ```python
from sys import stdin, stdout
def read_list(): # read list of variables
return [int(x) for x in stdin.readline().split()]
def read_list_str(): # read list of variables
return [x for x in stdin.readline().split()]
def readmv(): # read_multiple_variable
return map(int, stdin.readline().split())
def readv(): # read variable
return int(stdin.readline())
def printSol(test_case, x):
ans_to_print = ''
if type(x) == list:
for el in x:
ans_to_print += f'{el} '
if type(x) == int or type(x) == float:
ans_to_print = str(x)
stdout.write(f'Case #{test_case + 1}: ' + ans_to_print + f'\n')
def read():
return stdin.readline().strip()
from heapq import heapify, heappop, heappush
from math import inf
def dijkstra(graph, s):
n = len(graph)
visited = [False]*n
dist = [inf]*n
q = [(s, 0)]
heapify(q)
visited[s] = True
dist[s] = 0
while len(q):
_, u = heappop(q)
for (v, w) in graph[u]:
if not visited[v] and dist[v] > dist[u] + w:
dist[v] = dist[u] + w
heappush(q, (dist[v], v))
return dist
def main():
n, m = readmv()
graph = [[] for _ in range(2*n)]
for _ in range(m):
x, y, w = readmv()
x -= 1
y -= 1
graph[x].append((y, w))
graph[y+n].append((x+n, w))
for x in range(n):
graph[x].append((x+n, 0))
dist = dijkstra(graph, 0)
for i in range(1, n):
if dist[n+i] == inf:
stdout.write('-1 ')
continue
stdout.write(f'{dist[n+i]} ')
stdout.write('\n')
main()
``` | -1 |
57 | A | Square Earth? | PROGRAMMING | 1,300 | [
"dfs and similar",
"greedy",
"implementation"
] | A. Square Earth? | 2 | 256 | Meg the Rabbit decided to do something nice, specifically — to determine the shortest distance between two points on the surface of our planet. But Meg... what can you say, she wants everything simple. So, she already regards our planet as a two-dimensional circle. No, wait, it's even worse — as a square of side *n*. Thus, the task has been reduced to finding the shortest path between two dots on a square (the path should go through the square sides). To simplify the task let us consider the vertices of the square to lie at points whose coordinates are: (0,<=0), (*n*,<=0), (0,<=*n*) and (*n*,<=*n*). | The single line contains 5 space-separated integers: *n*,<=*x*1,<=*y*1,<=*x*2,<=*y*2 (1<=≤<=*n*<=≤<=1000,<=0<=≤<=*x*1,<=*y*1,<=*x*2,<=*y*2<=≤<=*n*) which correspondingly represent a side of the square, the coordinates of the first point and the coordinates of the second point. It is guaranteed that the points lie on the sides of the square. | You must print on a single line the shortest distance between the points. | [
"2 0 0 1 0\n",
"2 0 1 2 1\n",
"100 0 0 100 100\n"
] | [
"1\n",
"4\n",
"200\n"
] | none | 500 | [
{
"input": "2 0 0 1 0",
"output": "1"
},
{
"input": "2 0 1 2 1",
"output": "4"
},
{
"input": "100 0 0 100 100",
"output": "200"
},
{
"input": "4 0 3 1 4",
"output": "2"
},
{
"input": "10 8 10 10 0",
"output": "12"
},
{
"input": "26 21 0 26 14",
"ou... | 1,601,799,134 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 25 | 218 | 3,379,200 | n,a,b,c,d=map(int,input().split())
dx=[0,0,-1,1]
dy=[1,-1,0,0]
vis=[[0]*(n+1) for _ in " "*(n+1)]
ok=[[1]*(n+1)]
for i in range(n-1):
ok.append([1]+[0]*(n-1)+[1])
ok.append([1]*(n+1))
def dfs(x,y,k):
if x<0 or x>n or y<0 or y>n:return
if vis[x][y]:return
if ok[x][y]==0:return
vis[x][y]=1
if x==c and y==d:exit(print(min(k,4*n-k)))
for i in range(4):dfs(x+dx[i],y+dy[i],k+1)
dfs(a,b,0) | Title: Square Earth?
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Meg the Rabbit decided to do something nice, specifically — to determine the shortest distance between two points on the surface of our planet. But Meg... what can you say, she wants everything simple. So, she already regards our planet as a two-dimensional circle. No, wait, it's even worse — as a square of side *n*. Thus, the task has been reduced to finding the shortest path between two dots on a square (the path should go through the square sides). To simplify the task let us consider the vertices of the square to lie at points whose coordinates are: (0,<=0), (*n*,<=0), (0,<=*n*) and (*n*,<=*n*).
Input Specification:
The single line contains 5 space-separated integers: *n*,<=*x*1,<=*y*1,<=*x*2,<=*y*2 (1<=≤<=*n*<=≤<=1000,<=0<=≤<=*x*1,<=*y*1,<=*x*2,<=*y*2<=≤<=*n*) which correspondingly represent a side of the square, the coordinates of the first point and the coordinates of the second point. It is guaranteed that the points lie on the sides of the square.
Output Specification:
You must print on a single line the shortest distance between the points.
Demo Input:
['2 0 0 1 0\n', '2 0 1 2 1\n', '100 0 0 100 100\n']
Demo Output:
['1\n', '4\n', '200\n']
Note:
none | ```python
n,a,b,c,d=map(int,input().split())
dx=[0,0,-1,1]
dy=[1,-1,0,0]
vis=[[0]*(n+1) for _ in " "*(n+1)]
ok=[[1]*(n+1)]
for i in range(n-1):
ok.append([1]+[0]*(n-1)+[1])
ok.append([1]*(n+1))
def dfs(x,y,k):
if x<0 or x>n or y<0 or y>n:return
if vis[x][y]:return
if ok[x][y]==0:return
vis[x][y]=1
if x==c and y==d:exit(print(min(k,4*n-k)))
for i in range(4):dfs(x+dx[i],y+dy[i],k+1)
dfs(a,b,0)
``` | -1 |
289 | B | Polo the Penguin and Matrix | PROGRAMMING | 1,400 | [
"brute force",
"dp",
"implementation",
"sortings",
"ternary search"
] | null | null | Little penguin Polo has an *n*<=×<=*m* matrix, consisting of integers. Let's index the matrix rows from 1 to *n* from top to bottom and let's index the columns from 1 to *m* from left to right. Let's represent the matrix element on the intersection of row *i* and column *j* as *a**ij*.
In one move the penguin can add or subtract number *d* from some matrix element. Find the minimum number of moves needed to make all matrix elements equal. If the described plan is impossible to carry out, say so. | The first line contains three integers *n*, *m* and *d* (1<=≤<=*n*,<=*m*<=≤<=100,<=1<=≤<=*d*<=≤<=104) — the matrix sizes and the *d* parameter. Next *n* lines contain the matrix: the *j*-th integer in the *i*-th row is the matrix element *a**ij* (1<=≤<=*a**ij*<=≤<=104). | In a single line print a single integer — the minimum number of moves the penguin needs to make all matrix elements equal. If that is impossible, print "-1" (without the quotes). | [
"2 2 2\n2 4\n6 8\n",
"1 2 7\n6 7\n"
] | [
"4\n",
"-1\n"
] | none | 1,000 | [
{
"input": "2 2 2\n2 4\n6 8",
"output": "4"
},
{
"input": "1 2 7\n6 7",
"output": "-1"
},
{
"input": "3 2 1\n5 7\n1 2\n5 100",
"output": "104"
},
{
"input": "3 3 3\n5 8 5\n11 11 17\n14 5 2",
"output": "12"
},
{
"input": "3 3 3\n5 8 5\n11 11 17\n14 5 3",
"outpu... | 1,610,295,430 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 14 | 2,000 | 614,400 | import collections,math,itertools,bisect,heapq,copy
class Solution:
def solve(self):
n,m,d = list(map(int, input().rstrip().split()))
arr = []
for i in range(n):
a = list(map(int, input().rstrip().split()))
arr.append(a)
res = float('inf')
for i in range(n):
for j in range(m):
num = arr[i][j]
total = 0
for _i in range(n):
for _j in range(m):
if (num-arr[_i][_j]) % d != 0:
print(-1)
return
total += (abs(num-arr[_i][_j])//d)
res = min(res,total)
print(res)
if __name__ == "__main__":
solution = Solution()
solution.solve() | Title: Polo the Penguin and Matrix
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Little penguin Polo has an *n*<=×<=*m* matrix, consisting of integers. Let's index the matrix rows from 1 to *n* from top to bottom and let's index the columns from 1 to *m* from left to right. Let's represent the matrix element on the intersection of row *i* and column *j* as *a**ij*.
In one move the penguin can add or subtract number *d* from some matrix element. Find the minimum number of moves needed to make all matrix elements equal. If the described plan is impossible to carry out, say so.
Input Specification:
The first line contains three integers *n*, *m* and *d* (1<=≤<=*n*,<=*m*<=≤<=100,<=1<=≤<=*d*<=≤<=104) — the matrix sizes and the *d* parameter. Next *n* lines contain the matrix: the *j*-th integer in the *i*-th row is the matrix element *a**ij* (1<=≤<=*a**ij*<=≤<=104).
Output Specification:
In a single line print a single integer — the minimum number of moves the penguin needs to make all matrix elements equal. If that is impossible, print "-1" (without the quotes).
Demo Input:
['2 2 2\n2 4\n6 8\n', '1 2 7\n6 7\n']
Demo Output:
['4\n', '-1\n']
Note:
none | ```python
import collections,math,itertools,bisect,heapq,copy
class Solution:
def solve(self):
n,m,d = list(map(int, input().rstrip().split()))
arr = []
for i in range(n):
a = list(map(int, input().rstrip().split()))
arr.append(a)
res = float('inf')
for i in range(n):
for j in range(m):
num = arr[i][j]
total = 0
for _i in range(n):
for _j in range(m):
if (num-arr[_i][_j]) % d != 0:
print(-1)
return
total += (abs(num-arr[_i][_j])//d)
res = min(res,total)
print(res)
if __name__ == "__main__":
solution = Solution()
solution.solve()
``` | 0 | |
119 | A | Epic Game | PROGRAMMING | 800 | [
"implementation"
] | null | null | Simon and Antisimon play a game. Initially each player receives one fixed positive integer that doesn't change throughout the game. Simon receives number *a* and Antisimon receives number *b*. They also have a heap of *n* stones. The players take turns to make a move and Simon starts. During a move a player should take from the heap the number of stones equal to the greatest common divisor of the fixed number he has received and the number of stones left in the heap. A player loses when he cannot take the required number of stones (i. e. the heap has strictly less stones left than one needs to take).
Your task is to determine by the given *a*, *b* and *n* who wins the game. | The only string contains space-separated integers *a*, *b* and *n* (1<=≤<=*a*,<=*b*,<=*n*<=≤<=100) — the fixed numbers Simon and Antisimon have received correspondingly and the initial number of stones in the pile. | If Simon wins, print "0" (without the quotes), otherwise print "1" (without the quotes). | [
"3 5 9\n",
"1 1 100\n"
] | [
"0",
"1"
] | The greatest common divisor of two non-negative integers *a* and *b* is such maximum positive integer *k*, that *a* is divisible by *k* without remainder and similarly, *b* is divisible by *k* without remainder. Let *gcd*(*a*, *b*) represent the operation of calculating the greatest common divisor of numbers *a* and *b*. Specifically, *gcd*(*x*, 0) = *gcd*(0, *x*) = *x*.
In the first sample the game will go like that:
- Simon should take *gcd*(3, 9) = 3 stones from the heap. After his move the heap has 6 stones left.- Antisimon should take *gcd*(5, 6) = 1 stone from the heap. After his move the heap has 5 stones left.- Simon should take *gcd*(3, 5) = 1 stone from the heap. After his move the heap has 4 stones left.- Antisimon should take *gcd*(5, 4) = 1 stone from the heap. After his move the heap has 3 stones left.- Simon should take *gcd*(3, 3) = 3 stones from the heap. After his move the heap has 0 stones left.- Antisimon should take *gcd*(5, 0) = 5 stones from the heap. As 0 < 5, it is impossible and Antisimon loses.
In the second sample each player during each move takes one stone from the heap. As *n* is even, Antisimon takes the last stone and Simon can't make a move after that. | 500 | [
{
"input": "3 5 9",
"output": "0"
},
{
"input": "1 1 100",
"output": "1"
},
{
"input": "23 12 16",
"output": "1"
},
{
"input": "95 26 29",
"output": "1"
},
{
"input": "73 32 99",
"output": "1"
},
{
"input": "1 1 1",
"output": "0"
},
{
"inpu... | 1,603,086,651 | 2,147,483,647 | Python 3 | OK | TESTS | 45 | 216 | 0 | from math import *
a,b,n=map(int,input().split())
i=0
while n>0:
if i==0:
re=gcd(a,n)
n-=re
i=1
elif i==1:
re=gcd(b,n)
n-=re
i=0
if i==1:
print(0)
else:
print(1) | Title: Epic Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Simon and Antisimon play a game. Initially each player receives one fixed positive integer that doesn't change throughout the game. Simon receives number *a* and Antisimon receives number *b*. They also have a heap of *n* stones. The players take turns to make a move and Simon starts. During a move a player should take from the heap the number of stones equal to the greatest common divisor of the fixed number he has received and the number of stones left in the heap. A player loses when he cannot take the required number of stones (i. e. the heap has strictly less stones left than one needs to take).
Your task is to determine by the given *a*, *b* and *n* who wins the game.
Input Specification:
The only string contains space-separated integers *a*, *b* and *n* (1<=≤<=*a*,<=*b*,<=*n*<=≤<=100) — the fixed numbers Simon and Antisimon have received correspondingly and the initial number of stones in the pile.
Output Specification:
If Simon wins, print "0" (without the quotes), otherwise print "1" (without the quotes).
Demo Input:
['3 5 9\n', '1 1 100\n']
Demo Output:
['0', '1']
Note:
The greatest common divisor of two non-negative integers *a* and *b* is such maximum positive integer *k*, that *a* is divisible by *k* without remainder and similarly, *b* is divisible by *k* without remainder. Let *gcd*(*a*, *b*) represent the operation of calculating the greatest common divisor of numbers *a* and *b*. Specifically, *gcd*(*x*, 0) = *gcd*(0, *x*) = *x*.
In the first sample the game will go like that:
- Simon should take *gcd*(3, 9) = 3 stones from the heap. After his move the heap has 6 stones left.- Antisimon should take *gcd*(5, 6) = 1 stone from the heap. After his move the heap has 5 stones left.- Simon should take *gcd*(3, 5) = 1 stone from the heap. After his move the heap has 4 stones left.- Antisimon should take *gcd*(5, 4) = 1 stone from the heap. After his move the heap has 3 stones left.- Simon should take *gcd*(3, 3) = 3 stones from the heap. After his move the heap has 0 stones left.- Antisimon should take *gcd*(5, 0) = 5 stones from the heap. As 0 < 5, it is impossible and Antisimon loses.
In the second sample each player during each move takes one stone from the heap. As *n* is even, Antisimon takes the last stone and Simon can't make a move after that. | ```python
from math import *
a,b,n=map(int,input().split())
i=0
while n>0:
if i==0:
re=gcd(a,n)
n-=re
i=1
elif i==1:
re=gcd(b,n)
n-=re
i=0
if i==1:
print(0)
else:
print(1)
``` | 3 | |
777 | B | Game of Credit Cards | PROGRAMMING | 1,300 | [
"data structures",
"dp",
"greedy",
"sortings"
] | null | null | After the fourth season Sherlock and Moriary have realized the whole foolishness of the battle between them and decided to continue their competitions in peaceful game of Credit Cards.
Rules of this game are simple: each player bring his favourite *n*-digit credit card. Then both players name the digits written on their cards one by one. If two digits are not equal, then the player, whose digit is smaller gets a flick (knock in the forehead usually made with a forefinger) from the other player. For example, if *n*<==<=3, Sherlock's card is 123 and Moriarty's card has number 321, first Sherlock names 1 and Moriarty names 3 so Sherlock gets a flick. Then they both digit 2 so no one gets a flick. Finally, Sherlock names 3, while Moriarty names 1 and gets a flick.
Of course, Sherlock will play honestly naming digits one by one in the order they are given, while Moriary, as a true villain, plans to cheat. He is going to name his digits in some other order (however, he is not going to change the overall number of occurences of each digit). For example, in case above Moriarty could name 1, 2, 3 and get no flicks at all, or he can name 2, 3 and 1 to give Sherlock two flicks.
Your goal is to find out the minimum possible number of flicks Moriarty will get (no one likes flicks) and the maximum possible number of flicks Sherlock can get from Moriarty. Note, that these two goals are different and the optimal result may be obtained by using different strategies. | The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of digits in the cards Sherlock and Moriarty are going to use.
The second line contains *n* digits — Sherlock's credit card number.
The third line contains *n* digits — Moriarty's credit card number. | First print the minimum possible number of flicks Moriarty will get. Then print the maximum possible number of flicks that Sherlock can get from Moriarty. | [
"3\n123\n321\n",
"2\n88\n00\n"
] | [
"0\n2\n",
"2\n0\n"
] | First sample is elaborated in the problem statement. In the second sample, there is no way Moriarty can avoid getting two flicks. | 1,000 | [
{
"input": "3\n123\n321",
"output": "0\n2"
},
{
"input": "2\n88\n00",
"output": "2\n0"
},
{
"input": "1\n4\n5",
"output": "0\n1"
},
{
"input": "1\n8\n7",
"output": "1\n0"
},
{
"input": "2\n55\n55",
"output": "0\n0"
},
{
"input": "3\n534\n432",
"out... | 1,685,741,149 | 2,147,483,647 | PyPy 3 | OK | TESTS | 55 | 77 | 0 | import sys
input = sys.stdin.readline
n = int(input())
s = sorted(input().strip())
m = sorted(input().strip())
lose, win = 0,0
for i in m:
if i >= s[lose]:
lose += 1
if i > s[win]:
win += 1
print(n - lose, win, sep = '\n') | Title: Game of Credit Cards
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
After the fourth season Sherlock and Moriary have realized the whole foolishness of the battle between them and decided to continue their competitions in peaceful game of Credit Cards.
Rules of this game are simple: each player bring his favourite *n*-digit credit card. Then both players name the digits written on their cards one by one. If two digits are not equal, then the player, whose digit is smaller gets a flick (knock in the forehead usually made with a forefinger) from the other player. For example, if *n*<==<=3, Sherlock's card is 123 and Moriarty's card has number 321, first Sherlock names 1 and Moriarty names 3 so Sherlock gets a flick. Then they both digit 2 so no one gets a flick. Finally, Sherlock names 3, while Moriarty names 1 and gets a flick.
Of course, Sherlock will play honestly naming digits one by one in the order they are given, while Moriary, as a true villain, plans to cheat. He is going to name his digits in some other order (however, he is not going to change the overall number of occurences of each digit). For example, in case above Moriarty could name 1, 2, 3 and get no flicks at all, or he can name 2, 3 and 1 to give Sherlock two flicks.
Your goal is to find out the minimum possible number of flicks Moriarty will get (no one likes flicks) and the maximum possible number of flicks Sherlock can get from Moriarty. Note, that these two goals are different and the optimal result may be obtained by using different strategies.
Input Specification:
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of digits in the cards Sherlock and Moriarty are going to use.
The second line contains *n* digits — Sherlock's credit card number.
The third line contains *n* digits — Moriarty's credit card number.
Output Specification:
First print the minimum possible number of flicks Moriarty will get. Then print the maximum possible number of flicks that Sherlock can get from Moriarty.
Demo Input:
['3\n123\n321\n', '2\n88\n00\n']
Demo Output:
['0\n2\n', '2\n0\n']
Note:
First sample is elaborated in the problem statement. In the second sample, there is no way Moriarty can avoid getting two flicks. | ```python
import sys
input = sys.stdin.readline
n = int(input())
s = sorted(input().strip())
m = sorted(input().strip())
lose, win = 0,0
for i in m:
if i >= s[lose]:
lose += 1
if i > s[win]:
win += 1
print(n - lose, win, sep = '\n')
``` | 3 | |
50 | A | Domino piling | PROGRAMMING | 800 | [
"greedy",
"math"
] | A. Domino piling | 2 | 256 | You are given a rectangular board of *M*<=×<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=×<=1 squares. You are allowed to rotate the pieces. You are asked to place as many dominoes as possible on the board so as to meet the following conditions:
1. Each domino completely covers two squares.
2. No two dominoes overlap.
3. Each domino lies entirely inside the board. It is allowed to touch the edges of the board.
Find the maximum number of dominoes, which can be placed under these restrictions. | In a single line you are given two integers *M* and *N* — board sizes in squares (1<=≤<=*M*<=≤<=*N*<=≤<=16). | Output one number — the maximal number of dominoes, which can be placed. | [
"2 4\n",
"3 3\n"
] | [
"4\n",
"4\n"
] | none | 500 | [
{
"input": "2 4",
"output": "4"
},
{
"input": "3 3",
"output": "4"
},
{
"input": "1 5",
"output": "2"
},
{
"input": "1 6",
"output": "3"
},
{
"input": "1 15",
"output": "7"
},
{
"input": "1 16",
"output": "8"
},
{
"input": "2 5",
"outpu... | 1,669,191,618 | 2,147,483,647 | Python 3 | OK | TESTS | 35 | 92 | 0 | m,n=map(int,input().split())
k=n*m//2
print(k) | Title: Domino piling
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
You are given a rectangular board of *M*<=×<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=×<=1 squares. You are allowed to rotate the pieces. You are asked to place as many dominoes as possible on the board so as to meet the following conditions:
1. Each domino completely covers two squares.
2. No two dominoes overlap.
3. Each domino lies entirely inside the board. It is allowed to touch the edges of the board.
Find the maximum number of dominoes, which can be placed under these restrictions.
Input Specification:
In a single line you are given two integers *M* and *N* — board sizes in squares (1<=≤<=*M*<=≤<=*N*<=≤<=16).
Output Specification:
Output one number — the maximal number of dominoes, which can be placed.
Demo Input:
['2 4\n', '3 3\n']
Demo Output:
['4\n', '4\n']
Note:
none | ```python
m,n=map(int,input().split())
k=n*m//2
print(k)
``` | 3.977 |
847 | G | University Classes | PROGRAMMING | 900 | [
"implementation"
] | null | null | There are *n* student groups at the university. During the study day, each group can take no more than 7 classes. Seven time slots numbered from 1 to 7 are allocated for the classes.
The schedule on Monday is known for each group, i. e. time slots when group will have classes are known.
Your task is to determine the minimum number of rooms needed to hold classes for all groups on Monday. Note that one room can hold at most one group class in a single time slot. | The first line contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of groups.
Each of the following *n* lines contains a sequence consisting of 7 zeroes and ones — the schedule of classes on Monday for a group. If the symbol in a position equals to 1 then the group has class in the corresponding time slot. In the other case, the group has no class in the corresponding time slot. | Print minimum number of rooms needed to hold all groups classes on Monday. | [
"2\n0101010\n1010101\n",
"3\n0101011\n0011001\n0110111\n"
] | [
"1\n",
"3\n"
] | In the first example one room is enough. It will be occupied in each of the seven time slot by the first group or by the second group.
In the second example three rooms is enough, because in the seventh time slot all three groups have classes. | 0 | [
{
"input": "2\n0101010\n1010101",
"output": "1"
},
{
"input": "3\n0101011\n0011001\n0110111",
"output": "3"
},
{
"input": "1\n0111000",
"output": "1"
},
{
"input": "1\n0000000",
"output": "0"
},
{
"input": "1\n1111111",
"output": "1"
},
{
"input": "2\n... | 1,568,546,710 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 12 | 124 | 0 | a = int(input())
i = 0
s = 0
while i < a :
i+=1
x = int(input())
s += x
##print(s)
c1 = s // 1000000
c2 = s // 100000 -10*c1
c3 = s // 10000 -100*c1 -10*c2
c4 = s // 1000 -1000*c1 -100*c2 - 10*c3
c7 = s % 10
c6 = (s % 100 - c7)//10
c5 = (s % 1000 - c6*10 -c7)//100
##print(c1,c2,c3,c4,c5,c6,c7)
if c1 < c2 :
c1 = c2
if c3 < c4 :
c3 = c4
if c5 < c6 :
c5 = c6
if c1 < c3 :
c1 = c3
if c5 < c7 :
c5 = c7
if c1 < c5 :
c1 = c5
print(c1)
# Sun Sep 15 2019 14:25:09 GMT+0300 (MSK)
| Title: University Classes
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are *n* student groups at the university. During the study day, each group can take no more than 7 classes. Seven time slots numbered from 1 to 7 are allocated for the classes.
The schedule on Monday is known for each group, i. e. time slots when group will have classes are known.
Your task is to determine the minimum number of rooms needed to hold classes for all groups on Monday. Note that one room can hold at most one group class in a single time slot.
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of groups.
Each of the following *n* lines contains a sequence consisting of 7 zeroes and ones — the schedule of classes on Monday for a group. If the symbol in a position equals to 1 then the group has class in the corresponding time slot. In the other case, the group has no class in the corresponding time slot.
Output Specification:
Print minimum number of rooms needed to hold all groups classes on Monday.
Demo Input:
['2\n0101010\n1010101\n', '3\n0101011\n0011001\n0110111\n']
Demo Output:
['1\n', '3\n']
Note:
In the first example one room is enough. It will be occupied in each of the seven time slot by the first group or by the second group.
In the second example three rooms is enough, because in the seventh time slot all three groups have classes. | ```python
a = int(input())
i = 0
s = 0
while i < a :
i+=1
x = int(input())
s += x
##print(s)
c1 = s // 1000000
c2 = s // 100000 -10*c1
c3 = s // 10000 -100*c1 -10*c2
c4 = s // 1000 -1000*c1 -100*c2 - 10*c3
c7 = s % 10
c6 = (s % 100 - c7)//10
c5 = (s % 1000 - c6*10 -c7)//100
##print(c1,c2,c3,c4,c5,c6,c7)
if c1 < c2 :
c1 = c2
if c3 < c4 :
c3 = c4
if c5 < c6 :
c5 = c6
if c1 < c3 :
c1 = c3
if c5 < c7 :
c5 = c7
if c1 < c5 :
c1 = c5
print(c1)
# Sun Sep 15 2019 14:25:09 GMT+0300 (MSK)
``` | 0 | |
340 | A | The Wall | PROGRAMMING | 1,200 | [
"math"
] | null | null | Iahub and his friend Floyd have started painting a wall. Iahub is painting the wall red and Floyd is painting it pink. You can consider the wall being made of a very large number of bricks, numbered 1, 2, 3 and so on.
Iahub has the following scheme of painting: he skips *x*<=-<=1 consecutive bricks, then he paints the *x*-th one. That is, he'll paint bricks *x*, 2·*x*, 3·*x* and so on red. Similarly, Floyd skips *y*<=-<=1 consecutive bricks, then he paints the *y*-th one. Hence he'll paint bricks *y*, 2·*y*, 3·*y* and so on pink.
After painting the wall all day, the boys observed that some bricks are painted both red and pink. Iahub has a lucky number *a* and Floyd has a lucky number *b*. Boys wonder how many bricks numbered no less than *a* and no greater than *b* are painted both red and pink. This is exactly your task: compute and print the answer to the question. | The input will have a single line containing four integers in this order: *x*, *y*, *a*, *b*. (1<=≤<=*x*,<=*y*<=≤<=1000, 1<=≤<=*a*,<=*b*<=≤<=2·109, *a*<=≤<=*b*). | Output a single integer — the number of bricks numbered no less than *a* and no greater than *b* that are painted both red and pink. | [
"2 3 6 18\n"
] | [
"3"
] | Let's look at the bricks from *a* to *b* (*a* = 6, *b* = 18). The bricks colored in red are numbered 6, 8, 10, 12, 14, 16, 18. The bricks colored in pink are numbered 6, 9, 12, 15, 18. The bricks colored in both red and pink are numbered with 6, 12 and 18. | 500 | [
{
"input": "2 3 6 18",
"output": "3"
},
{
"input": "4 6 20 201",
"output": "15"
},
{
"input": "15 27 100 10000",
"output": "74"
},
{
"input": "105 60 3456 78910",
"output": "179"
},
{
"input": "1 1 1000 100000",
"output": "99001"
},
{
"input": "3 2 5 5... | 1,418,080,288 | 2,147,483,647 | Python 3 | OK | TESTS | 35 | 124 | 0 | def main():
x, y, a, b = map(int, input().split())
u, v = x, y
while v:
u, v = v, u % v
v = x * y // u
print(b // v - (a - 1) // v)
if __name__ == '__main__':
main()
| Title: The Wall
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Iahub and his friend Floyd have started painting a wall. Iahub is painting the wall red and Floyd is painting it pink. You can consider the wall being made of a very large number of bricks, numbered 1, 2, 3 and so on.
Iahub has the following scheme of painting: he skips *x*<=-<=1 consecutive bricks, then he paints the *x*-th one. That is, he'll paint bricks *x*, 2·*x*, 3·*x* and so on red. Similarly, Floyd skips *y*<=-<=1 consecutive bricks, then he paints the *y*-th one. Hence he'll paint bricks *y*, 2·*y*, 3·*y* and so on pink.
After painting the wall all day, the boys observed that some bricks are painted both red and pink. Iahub has a lucky number *a* and Floyd has a lucky number *b*. Boys wonder how many bricks numbered no less than *a* and no greater than *b* are painted both red and pink. This is exactly your task: compute and print the answer to the question.
Input Specification:
The input will have a single line containing four integers in this order: *x*, *y*, *a*, *b*. (1<=≤<=*x*,<=*y*<=≤<=1000, 1<=≤<=*a*,<=*b*<=≤<=2·109, *a*<=≤<=*b*).
Output Specification:
Output a single integer — the number of bricks numbered no less than *a* and no greater than *b* that are painted both red and pink.
Demo Input:
['2 3 6 18\n']
Demo Output:
['3']
Note:
Let's look at the bricks from *a* to *b* (*a* = 6, *b* = 18). The bricks colored in red are numbered 6, 8, 10, 12, 14, 16, 18. The bricks colored in pink are numbered 6, 9, 12, 15, 18. The bricks colored in both red and pink are numbered with 6, 12 and 18. | ```python
def main():
x, y, a, b = map(int, input().split())
u, v = x, y
while v:
u, v = v, u % v
v = x * y // u
print(b // v - (a - 1) // v)
if __name__ == '__main__':
main()
``` | 3 | |
712 | A | Memory and Crow | PROGRAMMING | 800 | [
"implementation",
"math"
] | null | null | There are *n* integers *b*1,<=*b*2,<=...,<=*b**n* written in a row. For all *i* from 1 to *n*, values *a**i* are defined by the crows performing the following procedure:
- The crow sets *a**i* initially 0. - The crow then adds *b**i* to *a**i*, subtracts *b**i*<=+<=1, adds the *b**i*<=+<=2 number, and so on until the *n*'th number. Thus, *a**i*<==<=*b**i*<=-<=*b**i*<=+<=1<=+<=*b**i*<=+<=2<=-<=*b**i*<=+<=3....
Memory gives you the values *a*1,<=*a*2,<=...,<=*a**n*, and he now wants you to find the initial numbers *b*1,<=*b*2,<=...,<=*b**n* written in the row? Can you do it? | The first line of the input contains a single integer *n* (2<=≤<=*n*<=≤<=100<=000) — the number of integers written in the row.
The next line contains *n*, the *i*'th of which is *a**i* (<=-<=109<=≤<=*a**i*<=≤<=109) — the value of the *i*'th number. | Print *n* integers corresponding to the sequence *b*1,<=*b*2,<=...,<=*b**n*. It's guaranteed that the answer is unique and fits in 32-bit integer type. | [
"5\n6 -4 8 -2 3\n",
"5\n3 -2 -1 5 6\n"
] | [
"2 4 6 1 3 \n",
"1 -3 4 11 6 \n"
] | In the first sample test, the crows report the numbers 6, - 4, 8, - 2, and 3 when he starts at indices 1, 2, 3, 4 and 5 respectively. It is easy to check that the sequence 2 4 6 1 3 satisfies the reports. For example, 6 = 2 - 4 + 6 - 1 + 3, and - 4 = 4 - 6 + 1 - 3.
In the second sample test, the sequence 1, - 3, 4, 11, 6 satisfies the reports. For example, 5 = 11 - 6 and 6 = 6. | 500 | [
{
"input": "5\n6 -4 8 -2 3",
"output": "2 4 6 1 3 "
},
{
"input": "5\n3 -2 -1 5 6",
"output": "1 -3 4 11 6 "
},
{
"input": "10\n13 -2 532 -63 -23 -63 -64 -23 12 10",
"output": "11 530 469 -86 -86 -127 -87 -11 22 10 "
},
{
"input": "10\n0 0 0 0 0 0 0 0 0 0",
"output": "0 0... | 1,615,842,845 | 2,147,483,647 | Python 3 | OK | TESTS | 49 | 265 | 8,499,200 | num = int(input())
arr = list(map(int, input().split())) + [0]
ans = []
for i in range(num):
ans.append(arr[i]+arr[i+1])
print(*ans)
| Title: Memory and Crow
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are *n* integers *b*1,<=*b*2,<=...,<=*b**n* written in a row. For all *i* from 1 to *n*, values *a**i* are defined by the crows performing the following procedure:
- The crow sets *a**i* initially 0. - The crow then adds *b**i* to *a**i*, subtracts *b**i*<=+<=1, adds the *b**i*<=+<=2 number, and so on until the *n*'th number. Thus, *a**i*<==<=*b**i*<=-<=*b**i*<=+<=1<=+<=*b**i*<=+<=2<=-<=*b**i*<=+<=3....
Memory gives you the values *a*1,<=*a*2,<=...,<=*a**n*, and he now wants you to find the initial numbers *b*1,<=*b*2,<=...,<=*b**n* written in the row? Can you do it?
Input Specification:
The first line of the input contains a single integer *n* (2<=≤<=*n*<=≤<=100<=000) — the number of integers written in the row.
The next line contains *n*, the *i*'th of which is *a**i* (<=-<=109<=≤<=*a**i*<=≤<=109) — the value of the *i*'th number.
Output Specification:
Print *n* integers corresponding to the sequence *b*1,<=*b*2,<=...,<=*b**n*. It's guaranteed that the answer is unique and fits in 32-bit integer type.
Demo Input:
['5\n6 -4 8 -2 3\n', '5\n3 -2 -1 5 6\n']
Demo Output:
['2 4 6 1 3 \n', '1 -3 4 11 6 \n']
Note:
In the first sample test, the crows report the numbers 6, - 4, 8, - 2, and 3 when he starts at indices 1, 2, 3, 4 and 5 respectively. It is easy to check that the sequence 2 4 6 1 3 satisfies the reports. For example, 6 = 2 - 4 + 6 - 1 + 3, and - 4 = 4 - 6 + 1 - 3.
In the second sample test, the sequence 1, - 3, 4, 11, 6 satisfies the reports. For example, 5 = 11 - 6 and 6 = 6. | ```python
num = int(input())
arr = list(map(int, input().split())) + [0]
ans = []
for i in range(num):
ans.append(arr[i]+arr[i+1])
print(*ans)
``` | 3 | |
146 | A | Lucky Ticket | PROGRAMMING | 800 | [
"implementation"
] | null | null | Petya loves lucky numbers very much. Everybody knows that lucky numbers are positive integers whose decimal record contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Petya loves tickets very much. As we know, each ticket has a number that is a positive integer. Its length equals *n* (*n* is always even). Petya calls a ticket lucky if the ticket's number is a lucky number and the sum of digits in the first half (the sum of the first *n*<=/<=2 digits) equals the sum of digits in the second half (the sum of the last *n*<=/<=2 digits). Check if the given ticket is lucky. | The first line contains an even integer *n* (2<=≤<=*n*<=≤<=50) — the length of the ticket number that needs to be checked. The second line contains an integer whose length equals exactly *n* — the ticket number. The number may contain leading zeros. | On the first line print "YES" if the given ticket number is lucky. Otherwise, print "NO" (without the quotes). | [
"2\n47\n",
"4\n4738\n",
"4\n4774\n"
] | [
"NO\n",
"NO\n",
"YES\n"
] | In the first sample the sum of digits in the first half does not equal the sum of digits in the second half (4 ≠ 7).
In the second sample the ticket number is not the lucky number. | 500 | [
{
"input": "2\n47",
"output": "NO"
},
{
"input": "4\n4738",
"output": "NO"
},
{
"input": "4\n4774",
"output": "YES"
},
{
"input": "4\n4570",
"output": "NO"
},
{
"input": "6\n477477",
"output": "YES"
},
{
"input": "6\n777777",
"output": "YES"
},
... | 1,646,652,413 | 2,147,483,647 | Python 3 | OK | TESTS | 46 | 92 | 0 | n =int(input())
m = input()
lis1 = []
lis2 = []
c = True
c2 = True
for i in range(int(n/2)):
lis1.append(m[i])
for j in range(int(n/2) , n):
lis2.append(m[j])
for k in range(int(n/2)):
lis1[k] = int(lis1[k])
if int(lis1[k]) == 4 or lis1[k] == 7:
c = True
else:
c= False
break
for o in range(int(n/2)):
lis2[o] = int(lis2[o])
if int(lis2[o]) == 4 or int(lis2[o]) == 7:
c2 = True
else:
c2= False
break
if c and c2 :
if sum(lis1) == sum(lis2):
print("YES")
else:
print("NO")
else:
print("NO") | Title: Lucky Ticket
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Petya loves lucky numbers very much. Everybody knows that lucky numbers are positive integers whose decimal record contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Petya loves tickets very much. As we know, each ticket has a number that is a positive integer. Its length equals *n* (*n* is always even). Petya calls a ticket lucky if the ticket's number is a lucky number and the sum of digits in the first half (the sum of the first *n*<=/<=2 digits) equals the sum of digits in the second half (the sum of the last *n*<=/<=2 digits). Check if the given ticket is lucky.
Input Specification:
The first line contains an even integer *n* (2<=≤<=*n*<=≤<=50) — the length of the ticket number that needs to be checked. The second line contains an integer whose length equals exactly *n* — the ticket number. The number may contain leading zeros.
Output Specification:
On the first line print "YES" if the given ticket number is lucky. Otherwise, print "NO" (without the quotes).
Demo Input:
['2\n47\n', '4\n4738\n', '4\n4774\n']
Demo Output:
['NO\n', 'NO\n', 'YES\n']
Note:
In the first sample the sum of digits in the first half does not equal the sum of digits in the second half (4 ≠ 7).
In the second sample the ticket number is not the lucky number. | ```python
n =int(input())
m = input()
lis1 = []
lis2 = []
c = True
c2 = True
for i in range(int(n/2)):
lis1.append(m[i])
for j in range(int(n/2) , n):
lis2.append(m[j])
for k in range(int(n/2)):
lis1[k] = int(lis1[k])
if int(lis1[k]) == 4 or lis1[k] == 7:
c = True
else:
c= False
break
for o in range(int(n/2)):
lis2[o] = int(lis2[o])
if int(lis2[o]) == 4 or int(lis2[o]) == 7:
c2 = True
else:
c2= False
break
if c and c2 :
if sum(lis1) == sum(lis2):
print("YES")
else:
print("NO")
else:
print("NO")
``` | 3 | |
979 | A | Pizza, Pizza, Pizza!!! | PROGRAMMING | 1,000 | [
"math"
] | null | null | Katie, Kuro and Shiro are best friends. They have known each other since kindergarten. That's why they often share everything with each other and work together on some very hard problems.
Today is Shiro's birthday. She really loves pizza so she wants to invite her friends to the pizza restaurant near her house to celebrate her birthday, including her best friends Katie and Kuro.
She has ordered a very big round pizza, in order to serve her many friends. Exactly $n$ of Shiro's friends are here. That's why she has to divide the pizza into $n + 1$ slices (Shiro also needs to eat). She wants the slices to be exactly the same size and shape. If not, some of her friends will get mad and go home early, and the party will be over.
Shiro is now hungry. She wants to cut the pizza with minimum of straight cuts. A cut is a straight segment, it might have ends inside or outside the pizza. But she is too lazy to pick up the calculator.
As usual, she will ask Katie and Kuro for help. But they haven't come yet. Could you help Shiro with this problem? | A single line contains one non-negative integer $n$ ($0 \le n \leq 10^{18}$) — the number of Shiro's friends. The circular pizza has to be sliced into $n + 1$ pieces. | A single integer — the number of straight cuts Shiro needs. | [
"3\n",
"4\n"
] | [
"2",
"5"
] | To cut the round pizza into quarters one has to make two cuts through the center with angle $90^{\circ}$ between them.
To cut the round pizza into five equal parts one has to make five cuts. | 500 | [
{
"input": "3",
"output": "2"
},
{
"input": "4",
"output": "5"
},
{
"input": "10",
"output": "11"
},
{
"input": "10000000000",
"output": "10000000001"
},
{
"input": "1234567891",
"output": "617283946"
},
{
"input": "7509213957",
"output": "37546069... | 1,564,558,575 | 2,147,483,647 | PyPy 3 | OK | TESTS | 47 | 155 | 0 | n=int(input())
if n==0:
print(0)
else:
if (n+1)%2==0:
print((n+1)//2)
else:
print((n + 1))
| Title: Pizza, Pizza, Pizza!!!
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Katie, Kuro and Shiro are best friends. They have known each other since kindergarten. That's why they often share everything with each other and work together on some very hard problems.
Today is Shiro's birthday. She really loves pizza so she wants to invite her friends to the pizza restaurant near her house to celebrate her birthday, including her best friends Katie and Kuro.
She has ordered a very big round pizza, in order to serve her many friends. Exactly $n$ of Shiro's friends are here. That's why she has to divide the pizza into $n + 1$ slices (Shiro also needs to eat). She wants the slices to be exactly the same size and shape. If not, some of her friends will get mad and go home early, and the party will be over.
Shiro is now hungry. She wants to cut the pizza with minimum of straight cuts. A cut is a straight segment, it might have ends inside or outside the pizza. But she is too lazy to pick up the calculator.
As usual, she will ask Katie and Kuro for help. But they haven't come yet. Could you help Shiro with this problem?
Input Specification:
A single line contains one non-negative integer $n$ ($0 \le n \leq 10^{18}$) — the number of Shiro's friends. The circular pizza has to be sliced into $n + 1$ pieces.
Output Specification:
A single integer — the number of straight cuts Shiro needs.
Demo Input:
['3\n', '4\n']
Demo Output:
['2', '5']
Note:
To cut the round pizza into quarters one has to make two cuts through the center with angle $90^{\circ}$ between them.
To cut the round pizza into five equal parts one has to make five cuts. | ```python
n=int(input())
if n==0:
print(0)
else:
if (n+1)%2==0:
print((n+1)//2)
else:
print((n + 1))
``` | 3 | |
248 | A | Cupboards | PROGRAMMING | 800 | [
"implementation"
] | null | null | One foggy Stockholm morning, Karlsson decided to snack on some jam in his friend Lillebror Svantenson's house. Fortunately for Karlsson, there wasn't anybody in his friend's house. Karlsson was not going to be hungry any longer, so he decided to get some food in the house.
Karlsson's gaze immediately fell on *n* wooden cupboards, standing in the kitchen. He immediately realized that these cupboards have hidden jam stocks. Karlsson began to fly greedily around the kitchen, opening and closing the cupboards' doors, grab and empty all the jars of jam that he could find.
And now all jars of jam are empty, Karlsson has had enough and does not want to leave traces of his stay, so as not to let down his friend. Each of the cupboards has two doors: the left one and the right one. Karlsson remembers that when he rushed to the kitchen, all the cupboards' left doors were in the same position (open or closed), similarly, all the cupboards' right doors were in the same position (open or closed). Karlsson wants the doors to meet this condition as well by the time the family returns. Karlsson does not remember the position of all the left doors, also, he cannot remember the position of all the right doors. Therefore, it does not matter to him in what position will be all left or right doors. It is important to leave all the left doors in the same position, and all the right doors in the same position. For example, all the left doors may be closed, and all the right ones may be open.
Karlsson needs one second to open or close a door of a cupboard. He understands that he has very little time before the family returns, so he wants to know the minimum number of seconds *t*, in which he is able to bring all the cupboard doors in the required position.
Your task is to write a program that will determine the required number of seconds *t*. | The first input line contains a single integer *n* — the number of cupboards in the kitchen (2<=≤<=*n*<=≤<=104). Then follow *n* lines, each containing two integers *l**i* and *r**i* (0<=≤<=*l**i*,<=*r**i*<=≤<=1). Number *l**i* equals one, if the left door of the *i*-th cupboard is opened, otherwise number *l**i* equals zero. Similarly, number *r**i* equals one, if the right door of the *i*-th cupboard is opened, otherwise number *r**i* equals zero.
The numbers in the lines are separated by single spaces. | In the only output line print a single integer *t* — the minimum number of seconds Karlsson needs to change the doors of all cupboards to the position he needs. | [
"5\n0 1\n1 0\n0 1\n1 1\n0 1\n"
] | [
"3\n"
] | none | 500 | [
{
"input": "5\n0 1\n1 0\n0 1\n1 1\n0 1",
"output": "3"
},
{
"input": "2\n0 0\n0 0",
"output": "0"
},
{
"input": "3\n0 1\n1 1\n1 1",
"output": "1"
},
{
"input": "8\n0 1\n1 0\n0 1\n1 1\n0 1\n1 0\n0 1\n1 0",
"output": "7"
},
{
"input": "8\n1 0\n1 0\n1 0\n0 1\n0 1\n1 ... | 1,595,434,283 | 2,147,483,647 | Python 3 | OK | TESTS | 57 | 248 | 6,656,000 | n=int(input())
u=v=0
for i in range(n):
a,b =map(int, input().split())
u+=a
v+=b
x=min(u,n-u)
y=min(v,n-v)
print(x+y)
| Title: Cupboards
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One foggy Stockholm morning, Karlsson decided to snack on some jam in his friend Lillebror Svantenson's house. Fortunately for Karlsson, there wasn't anybody in his friend's house. Karlsson was not going to be hungry any longer, so he decided to get some food in the house.
Karlsson's gaze immediately fell on *n* wooden cupboards, standing in the kitchen. He immediately realized that these cupboards have hidden jam stocks. Karlsson began to fly greedily around the kitchen, opening and closing the cupboards' doors, grab and empty all the jars of jam that he could find.
And now all jars of jam are empty, Karlsson has had enough and does not want to leave traces of his stay, so as not to let down his friend. Each of the cupboards has two doors: the left one and the right one. Karlsson remembers that when he rushed to the kitchen, all the cupboards' left doors were in the same position (open or closed), similarly, all the cupboards' right doors were in the same position (open or closed). Karlsson wants the doors to meet this condition as well by the time the family returns. Karlsson does not remember the position of all the left doors, also, he cannot remember the position of all the right doors. Therefore, it does not matter to him in what position will be all left or right doors. It is important to leave all the left doors in the same position, and all the right doors in the same position. For example, all the left doors may be closed, and all the right ones may be open.
Karlsson needs one second to open or close a door of a cupboard. He understands that he has very little time before the family returns, so he wants to know the minimum number of seconds *t*, in which he is able to bring all the cupboard doors in the required position.
Your task is to write a program that will determine the required number of seconds *t*.
Input Specification:
The first input line contains a single integer *n* — the number of cupboards in the kitchen (2<=≤<=*n*<=≤<=104). Then follow *n* lines, each containing two integers *l**i* and *r**i* (0<=≤<=*l**i*,<=*r**i*<=≤<=1). Number *l**i* equals one, if the left door of the *i*-th cupboard is opened, otherwise number *l**i* equals zero. Similarly, number *r**i* equals one, if the right door of the *i*-th cupboard is opened, otherwise number *r**i* equals zero.
The numbers in the lines are separated by single spaces.
Output Specification:
In the only output line print a single integer *t* — the minimum number of seconds Karlsson needs to change the doors of all cupboards to the position he needs.
Demo Input:
['5\n0 1\n1 0\n0 1\n1 1\n0 1\n']
Demo Output:
['3\n']
Note:
none | ```python
n=int(input())
u=v=0
for i in range(n):
a,b =map(int, input().split())
u+=a
v+=b
x=min(u,n-u)
y=min(v,n-v)
print(x+y)
``` | 3 | |
940 | B | Our Tanya is Crying Out Loud | PROGRAMMING | 1,400 | [
"dp",
"greedy"
] | null | null | Right now she actually isn't. But she will be, if you don't solve this problem.
You are given integers *n*, *k*, *A* and *B*. There is a number *x*, which is initially equal to *n*. You are allowed to perform two types of operations:
1. Subtract 1 from *x*. This operation costs you *A* coins. 1. Divide *x* by *k*. Can be performed only if *x* is divisible by *k*. This operation costs you *B* coins. | The first line contains a single integer *n* (1<=≤<=*n*<=≤<=2·109).
The second line contains a single integer *k* (1<=≤<=*k*<=≤<=2·109).
The third line contains a single integer *A* (1<=≤<=*A*<=≤<=2·109).
The fourth line contains a single integer *B* (1<=≤<=*B*<=≤<=2·109). | Output a single integer — the minimum amount of coins you have to pay to make *x* equal to 1. | [
"9\n2\n3\n1\n",
"5\n5\n2\n20\n",
"19\n3\n4\n2\n"
] | [
"6\n",
"8\n",
"12\n"
] | In the first testcase, the optimal strategy is as follows:
- Subtract 1 from *x* (9 → 8) paying 3 coins. - Divide *x* by 2 (8 → 4) paying 1 coin. - Divide *x* by 2 (4 → 2) paying 1 coin. - Divide *x* by 2 (2 → 1) paying 1 coin.
The total cost is 6 coins.
In the second test case the optimal strategy is to subtract 1 from *x* 4 times paying 8 coins in total. | 1,250 | [
{
"input": "9\n2\n3\n1",
"output": "6"
},
{
"input": "5\n5\n2\n20",
"output": "8"
},
{
"input": "19\n3\n4\n2",
"output": "12"
},
{
"input": "1845999546\n999435865\n1234234\n2323423",
"output": "1044857680578777"
},
{
"input": "1604353664\n1604353665\n9993432\n1",
... | 1,590,408,549 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 0 | 109 | 0 |
def solve(n,k,a,b):
ans = 0
while n != 1:
if n < k:
ans += (n-1)*a
n = 1
elif n%k != 0:
n -= n%k
ans += (n%k) *a
else:
if b < (n-n//k)*a:
n /= k
ans += b
else:
n /= k
ans += (n-n//k)*a
print(int(ans))
if __name__ == '__main__':
n = int(input())
k = int(input())
a = int(input())
b = int(input())
solve(n,k,a,b) | Title: Our Tanya is Crying Out Loud
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Right now she actually isn't. But she will be, if you don't solve this problem.
You are given integers *n*, *k*, *A* and *B*. There is a number *x*, which is initially equal to *n*. You are allowed to perform two types of operations:
1. Subtract 1 from *x*. This operation costs you *A* coins. 1. Divide *x* by *k*. Can be performed only if *x* is divisible by *k*. This operation costs you *B* coins.
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=2·109).
The second line contains a single integer *k* (1<=≤<=*k*<=≤<=2·109).
The third line contains a single integer *A* (1<=≤<=*A*<=≤<=2·109).
The fourth line contains a single integer *B* (1<=≤<=*B*<=≤<=2·109).
Output Specification:
Output a single integer — the minimum amount of coins you have to pay to make *x* equal to 1.
Demo Input:
['9\n2\n3\n1\n', '5\n5\n2\n20\n', '19\n3\n4\n2\n']
Demo Output:
['6\n', '8\n', '12\n']
Note:
In the first testcase, the optimal strategy is as follows:
- Subtract 1 from *x* (9 → 8) paying 3 coins. - Divide *x* by 2 (8 → 4) paying 1 coin. - Divide *x* by 2 (4 → 2) paying 1 coin. - Divide *x* by 2 (2 → 1) paying 1 coin.
The total cost is 6 coins.
In the second test case the optimal strategy is to subtract 1 from *x* 4 times paying 8 coins in total. | ```python
def solve(n,k,a,b):
ans = 0
while n != 1:
if n < k:
ans += (n-1)*a
n = 1
elif n%k != 0:
n -= n%k
ans += (n%k) *a
else:
if b < (n-n//k)*a:
n /= k
ans += b
else:
n /= k
ans += (n-n//k)*a
print(int(ans))
if __name__ == '__main__':
n = int(input())
k = int(input())
a = int(input())
b = int(input())
solve(n,k,a,b)
``` | 0 | |
124 | B | Permutations | PROGRAMMING | 1,400 | [
"brute force",
"combinatorics",
"implementation"
] | null | null | You are given *n* *k*-digit integers. You have to rearrange the digits in the integers so that the difference between the largest and the smallest number was minimum. Digits should be rearranged by the same rule in all integers. | The first line contains integers *n* and *k* — the number and digit capacity of numbers correspondingly (1<=≤<=*n*,<=*k*<=≤<=8). Next *n* lines contain *k*-digit positive integers. Leading zeroes are allowed both in the initial integers and the integers resulting from the rearranging of digits. | Print a single number: the minimally possible difference between the largest and the smallest number after the digits are rearranged in all integers by the same rule. | [
"6 4\n5237\n2753\n7523\n5723\n5327\n2537\n",
"3 3\n010\n909\n012\n",
"7 5\n50808\n36603\n37198\n44911\n29994\n42543\n50156\n"
] | [
"2700\n",
"3\n",
"20522\n"
] | In the first sample, if we rearrange the digits in numbers as (3,1,4,2), then the 2-nd and the 4-th numbers will equal 5237 and 2537 correspondingly (they will be maximum and minimum for such order of digits).
In the second sample, if we swap the second digits and the first ones, we get integers 100, 99 and 102. | 1,000 | [
{
"input": "6 4\n5237\n2753\n7523\n5723\n5327\n2537",
"output": "2700"
},
{
"input": "3 3\n010\n909\n012",
"output": "3"
},
{
"input": "7 5\n50808\n36603\n37198\n44911\n29994\n42543\n50156",
"output": "20522"
},
{
"input": "5 5\n61374\n74304\n41924\n46010\n09118",
"output... | 1,692,367,115 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 128 | 342 | 4,812,800 | import sys
input = lambda: sys.stdin.readline().rstrip()
import math
from heapq import heappush , heappop
from collections import defaultdict,deque,Counter
from bisect import *
import itertools
MOD = 10**9+7
N,K = map(int, input().split())
S = []
for _ in range(N):
S.append(input())
def deal(p):
A = [[] for _ in range(N)]
for i in p:
for j in range(N):
A[j].append(S[j][i])
B = []
for a in A:
B.append(int(''.join(a)))
B.sort()
return B[-1]-B[0]
A = [i for i in range(K)]
ans = float('inf')
for p in itertools.permutations(A,K):
ans = min(ans, deal(p))
print(ans)
| Title: Permutations
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given *n* *k*-digit integers. You have to rearrange the digits in the integers so that the difference between the largest and the smallest number was minimum. Digits should be rearranged by the same rule in all integers.
Input Specification:
The first line contains integers *n* and *k* — the number and digit capacity of numbers correspondingly (1<=≤<=*n*,<=*k*<=≤<=8). Next *n* lines contain *k*-digit positive integers. Leading zeroes are allowed both in the initial integers and the integers resulting from the rearranging of digits.
Output Specification:
Print a single number: the minimally possible difference between the largest and the smallest number after the digits are rearranged in all integers by the same rule.
Demo Input:
['6 4\n5237\n2753\n7523\n5723\n5327\n2537\n', '3 3\n010\n909\n012\n', '7 5\n50808\n36603\n37198\n44911\n29994\n42543\n50156\n']
Demo Output:
['2700\n', '3\n', '20522\n']
Note:
In the first sample, if we rearrange the digits in numbers as (3,1,4,2), then the 2-nd and the 4-th numbers will equal 5237 and 2537 correspondingly (they will be maximum and minimum for such order of digits).
In the second sample, if we swap the second digits and the first ones, we get integers 100, 99 and 102. | ```python
import sys
input = lambda: sys.stdin.readline().rstrip()
import math
from heapq import heappush , heappop
from collections import defaultdict,deque,Counter
from bisect import *
import itertools
MOD = 10**9+7
N,K = map(int, input().split())
S = []
for _ in range(N):
S.append(input())
def deal(p):
A = [[] for _ in range(N)]
for i in p:
for j in range(N):
A[j].append(S[j][i])
B = []
for a in A:
B.append(int(''.join(a)))
B.sort()
return B[-1]-B[0]
A = [i for i in range(K)]
ans = float('inf')
for p in itertools.permutations(A,K):
ans = min(ans, deal(p))
print(ans)
``` | 3 | |
677 | A | Vanya and Fence | PROGRAMMING | 800 | [
"implementation"
] | null | null | Vanya and his friends are walking along the fence of height *h* and they do not want the guard to notice them. In order to achieve this the height of each of the friends should not exceed *h*. If the height of some person is greater than *h* he can bend down and then he surely won't be noticed by the guard. The height of the *i*-th person is equal to *a**i*.
Consider the width of the person walking as usual to be equal to 1, while the width of the bent person is equal to 2. Friends want to talk to each other while walking, so they would like to walk in a single row. What is the minimum width of the road, such that friends can walk in a row and remain unattended by the guard? | The first line of the input contains two integers *n* and *h* (1<=≤<=*n*<=≤<=1000, 1<=≤<=*h*<=≤<=1000) — the number of friends and the height of the fence, respectively.
The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=2*h*), the *i*-th of them is equal to the height of the *i*-th person. | Print a single integer — the minimum possible valid width of the road. | [
"3 7\n4 5 14\n",
"6 1\n1 1 1 1 1 1\n",
"6 5\n7 6 8 9 10 5\n"
] | [
"4\n",
"6\n",
"11\n"
] | In the first sample, only person number 3 must bend down, so the required width is equal to 1 + 1 + 2 = 4.
In the second sample, all friends are short enough and no one has to bend, so the width 1 + 1 + 1 + 1 + 1 + 1 = 6 is enough.
In the third sample, all the persons have to bend, except the last one. The required minimum width of the road is equal to 2 + 2 + 2 + 2 + 2 + 1 = 11. | 500 | [
{
"input": "3 7\n4 5 14",
"output": "4"
},
{
"input": "6 1\n1 1 1 1 1 1",
"output": "6"
},
{
"input": "6 5\n7 6 8 9 10 5",
"output": "11"
},
{
"input": "10 420\n214 614 297 675 82 740 174 23 255 15",
"output": "13"
},
{
"input": "10 561\n657 23 1096 487 785 66 481... | 1,695,919,930 | 2,147,483,647 | Python 3 | OK | TESTS | 29 | 46 | 0 | def rng(s,k,lst):
#s is the total number of people like minimum widith and k is the max height possible
count=s
for f in lst:
if f>k:
count+=1
return count
y=input()
alist=[int(x) for x in y.split()]
d=input()
blist=[int(x) for x in d.split()]
print(rng(alist[0],alist[1],blist)) | Title: Vanya and Fence
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vanya and his friends are walking along the fence of height *h* and they do not want the guard to notice them. In order to achieve this the height of each of the friends should not exceed *h*. If the height of some person is greater than *h* he can bend down and then he surely won't be noticed by the guard. The height of the *i*-th person is equal to *a**i*.
Consider the width of the person walking as usual to be equal to 1, while the width of the bent person is equal to 2. Friends want to talk to each other while walking, so they would like to walk in a single row. What is the minimum width of the road, such that friends can walk in a row and remain unattended by the guard?
Input Specification:
The first line of the input contains two integers *n* and *h* (1<=≤<=*n*<=≤<=1000, 1<=≤<=*h*<=≤<=1000) — the number of friends and the height of the fence, respectively.
The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=2*h*), the *i*-th of them is equal to the height of the *i*-th person.
Output Specification:
Print a single integer — the minimum possible valid width of the road.
Demo Input:
['3 7\n4 5 14\n', '6 1\n1 1 1 1 1 1\n', '6 5\n7 6 8 9 10 5\n']
Demo Output:
['4\n', '6\n', '11\n']
Note:
In the first sample, only person number 3 must bend down, so the required width is equal to 1 + 1 + 2 = 4.
In the second sample, all friends are short enough and no one has to bend, so the width 1 + 1 + 1 + 1 + 1 + 1 = 6 is enough.
In the third sample, all the persons have to bend, except the last one. The required minimum width of the road is equal to 2 + 2 + 2 + 2 + 2 + 1 = 11. | ```python
def rng(s,k,lst):
#s is the total number of people like minimum widith and k is the max height possible
count=s
for f in lst:
if f>k:
count+=1
return count
y=input()
alist=[int(x) for x in y.split()]
d=input()
blist=[int(x) for x in d.split()]
print(rng(alist[0],alist[1],blist))
``` | 3 | |
1 | A | Theatre Square | PROGRAMMING | 1,000 | [
"math"
] | A. Theatre Square | 1 | 256 | Theatre Square in the capital city of Berland has a rectangular shape with the size *n*<=×<=*m* meters. On the occasion of the city's anniversary, a decision was taken to pave the Square with square granite flagstones. Each flagstone is of the size *a*<=×<=*a*.
What is the least number of flagstones needed to pave the Square? It's allowed to cover the surface larger than the Theatre Square, but the Square has to be covered. It's not allowed to break the flagstones. The sides of flagstones should be parallel to the sides of the Square. | The input contains three positive integer numbers in the first line: *n*,<=<=*m* and *a* (1<=≤<=<=*n*,<=*m*,<=*a*<=≤<=109). | Write the needed number of flagstones. | [
"6 6 4\n"
] | [
"4\n"
] | none | 0 | [
{
"input": "6 6 4",
"output": "4"
},
{
"input": "1 1 1",
"output": "1"
},
{
"input": "2 1 1",
"output": "2"
},
{
"input": "1 2 1",
"output": "2"
},
{
"input": "2 2 1",
"output": "4"
},
{
"input": "2 1 2",
"output": "1"
},
{
"input": "1 1 3"... | 1,692,902,411 | 2,147,483,647 | Python 3 | OK | TESTS | 20 | 46 | 0 | n, m, a = map(int, input().split())
w = (n + a - 1) // a
h = (m + a - 1) // a
total = w * h
print(total) | Title: Theatre Square
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Theatre Square in the capital city of Berland has a rectangular shape with the size *n*<=×<=*m* meters. On the occasion of the city's anniversary, a decision was taken to pave the Square with square granite flagstones. Each flagstone is of the size *a*<=×<=*a*.
What is the least number of flagstones needed to pave the Square? It's allowed to cover the surface larger than the Theatre Square, but the Square has to be covered. It's not allowed to break the flagstones. The sides of flagstones should be parallel to the sides of the Square.
Input Specification:
The input contains three positive integer numbers in the first line: *n*,<=<=*m* and *a* (1<=≤<=<=*n*,<=*m*,<=*a*<=≤<=109).
Output Specification:
Write the needed number of flagstones.
Demo Input:
['6 6 4\n']
Demo Output:
['4\n']
Note:
none | ```python
n, m, a = map(int, input().split())
w = (n + a - 1) // a
h = (m + a - 1) // a
total = w * h
print(total)
``` | 3.977 |
61 | A | Ultra-Fast Mathematician | PROGRAMMING | 800 | [
"implementation"
] | A. Ultra-Fast Mathematician | 2 | 256 | Shapur was an extremely gifted student. He was great at everything including Combinatorics, Algebra, Number Theory, Geometry, Calculus, etc. He was not only smart but extraordinarily fast! He could manage to sum 1018 numbers in a single second.
One day in 230 AD Shapur was trying to find out if any one can possibly do calculations faster than him. As a result he made a very great contest and asked every one to come and take part.
In his contest he gave the contestants many different pairs of numbers. Each number is made from digits 0 or 1. The contestants should write a new number corresponding to the given pair of numbers. The rule is simple: The *i*-th digit of the answer is 1 if and only if the *i*-th digit of the two given numbers differ. In the other case the *i*-th digit of the answer is 0.
Shapur made many numbers and first tried his own speed. He saw that he can perform these operations on numbers of length ∞ (length of a number is number of digits in it) in a glance! He always gives correct answers so he expects the contestants to give correct answers, too. He is a good fellow so he won't give anyone very big numbers and he always gives one person numbers of same length.
Now you are going to take part in Shapur's contest. See if you are faster and more accurate. | There are two lines in each input. Each of them contains a single number. It is guaranteed that the numbers are made from 0 and 1 only and that their length is same. The numbers may start with 0. The length of each number doesn't exceed 100. | Write one line — the corresponding answer. Do not omit the leading 0s. | [
"1010100\n0100101\n",
"000\n111\n",
"1110\n1010\n",
"01110\n01100\n"
] | [
"1110001\n",
"111\n",
"0100\n",
"00010\n"
] | none | 500 | [
{
"input": "1010100\n0100101",
"output": "1110001"
},
{
"input": "000\n111",
"output": "111"
},
{
"input": "1110\n1010",
"output": "0100"
},
{
"input": "01110\n01100",
"output": "00010"
},
{
"input": "011101\n000001",
"output": "011100"
},
{
"input": "... | 1,607,777,572 | 2,147,483,647 | Python 3 | OK | TESTS | 102 | 109 | 0 | x=input()
y=input()
for i in range(len(x)):
if(x[i]==y[i]):
print('0',end="")
else:
print('1',end="") | Title: Ultra-Fast Mathematician
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Shapur was an extremely gifted student. He was great at everything including Combinatorics, Algebra, Number Theory, Geometry, Calculus, etc. He was not only smart but extraordinarily fast! He could manage to sum 1018 numbers in a single second.
One day in 230 AD Shapur was trying to find out if any one can possibly do calculations faster than him. As a result he made a very great contest and asked every one to come and take part.
In his contest he gave the contestants many different pairs of numbers. Each number is made from digits 0 or 1. The contestants should write a new number corresponding to the given pair of numbers. The rule is simple: The *i*-th digit of the answer is 1 if and only if the *i*-th digit of the two given numbers differ. In the other case the *i*-th digit of the answer is 0.
Shapur made many numbers and first tried his own speed. He saw that he can perform these operations on numbers of length ∞ (length of a number is number of digits in it) in a glance! He always gives correct answers so he expects the contestants to give correct answers, too. He is a good fellow so he won't give anyone very big numbers and he always gives one person numbers of same length.
Now you are going to take part in Shapur's contest. See if you are faster and more accurate.
Input Specification:
There are two lines in each input. Each of them contains a single number. It is guaranteed that the numbers are made from 0 and 1 only and that their length is same. The numbers may start with 0. The length of each number doesn't exceed 100.
Output Specification:
Write one line — the corresponding answer. Do not omit the leading 0s.
Demo Input:
['1010100\n0100101\n', '000\n111\n', '1110\n1010\n', '01110\n01100\n']
Demo Output:
['1110001\n', '111\n', '0100\n', '00010\n']
Note:
none | ```python
x=input()
y=input()
for i in range(len(x)):
if(x[i]==y[i]):
print('0',end="")
else:
print('1',end="")
``` | 3.97275 |
898 | A | Rounding | PROGRAMMING | 800 | [
"implementation",
"math"
] | null | null | Vasya has a non-negative integer *n*. He wants to round it to nearest integer, which ends up with 0. If *n* already ends up with 0, Vasya considers it already rounded.
For example, if *n*<==<=4722 answer is 4720. If *n*<==<=5 Vasya can round it to 0 or to 10. Both ways are correct.
For given *n* find out to which integer will Vasya round it. | The first line contains single integer *n* (0<=≤<=*n*<=≤<=109) — number that Vasya has. | Print result of rounding *n*. Pay attention that in some cases answer isn't unique. In that case print any correct answer. | [
"5\n",
"113\n",
"1000000000\n",
"5432359\n"
] | [
"0\n",
"110\n",
"1000000000\n",
"5432360\n"
] | In the first example *n* = 5. Nearest integers, that ends up with zero are 0 and 10. Any of these answers is correct, so you can print 0 or 10. | 500 | [
{
"input": "5",
"output": "0"
},
{
"input": "113",
"output": "110"
},
{
"input": "1000000000",
"output": "1000000000"
},
{
"input": "5432359",
"output": "5432360"
},
{
"input": "999999994",
"output": "999999990"
},
{
"input": "10",
"output": "10"
... | 1,584,635,240 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 93 | 0 | n=int(input())
if n%10>5:
print(n+10-(n%5))
else:
print(n-(n%5)) | Title: Rounding
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya has a non-negative integer *n*. He wants to round it to nearest integer, which ends up with 0. If *n* already ends up with 0, Vasya considers it already rounded.
For example, if *n*<==<=4722 answer is 4720. If *n*<==<=5 Vasya can round it to 0 or to 10. Both ways are correct.
For given *n* find out to which integer will Vasya round it.
Input Specification:
The first line contains single integer *n* (0<=≤<=*n*<=≤<=109) — number that Vasya has.
Output Specification:
Print result of rounding *n*. Pay attention that in some cases answer isn't unique. In that case print any correct answer.
Demo Input:
['5\n', '113\n', '1000000000\n', '5432359\n']
Demo Output:
['0\n', '110\n', '1000000000\n', '5432360\n']
Note:
In the first example *n* = 5. Nearest integers, that ends up with zero are 0 and 10. Any of these answers is correct, so you can print 0 or 10. | ```python
n=int(input())
if n%10>5:
print(n+10-(n%5))
else:
print(n-(n%5))
``` | 0 | |
784 | E | Twisted Circuit | PROGRAMMING | 2,000 | [
"*special",
"brute force",
"implementation"
] | null | null | The input consists of four lines, each line containing a single digit 0 or 1. | Output a single digit, 0 or 1. | [
"0\n1\n1\n0\n"
] | [
"0\n"
] | none | 0 | [
{
"input": "0\n1\n1\n0",
"output": "0"
},
{
"input": "0\n0\n0\n0",
"output": "0"
},
{
"input": "1\n0\n0\n0",
"output": "1"
},
{
"input": "0\n1\n0\n0",
"output": "0"
},
{
"input": "1\n1\n0\n0",
"output": "1"
},
{
"input": "0\n0\n1\n0",
"output": "0"... | 1,692,431,434 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 16 | 61 | 0 | from sys import stdin,stdout
# from os import _exit
# from bisect import bisect_left,bisect
# from heapq import heapify,heappop,heappush
# from sys import setrecursionlimit
# from collections import defaultdict,Counter
# from itertools import permutations
# from math import gcd,ceil,sqrt,factorial
# setrecursionlimit(int(1e5))
input,print = stdin.readline,stdout.write
def xor(x,y):
return (not x and y) or (not y and x)
a = int(input())
b = int(input())
c = int(input())
d = int(input())
x = xor(a,b)
y = c or d
z = b and c
w = xor(a,d)
p = x and y
q = z or w
ans = xor(p,q)
print(str(int(ans))+"\n")
| Title: Twisted Circuit
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Input Specification:
The input consists of four lines, each line containing a single digit 0 or 1.
Output Specification:
Output a single digit, 0 or 1.
Demo Input:
['0\n1\n1\n0\n']
Demo Output:
['0\n']
Note:
none | ```python
from sys import stdin,stdout
# from os import _exit
# from bisect import bisect_left,bisect
# from heapq import heapify,heappop,heappush
# from sys import setrecursionlimit
# from collections import defaultdict,Counter
# from itertools import permutations
# from math import gcd,ceil,sqrt,factorial
# setrecursionlimit(int(1e5))
input,print = stdin.readline,stdout.write
def xor(x,y):
return (not x and y) or (not y and x)
a = int(input())
b = int(input())
c = int(input())
d = int(input())
x = xor(a,b)
y = c or d
z = b and c
w = xor(a,d)
p = x and y
q = z or w
ans = xor(p,q)
print(str(int(ans))+"\n")
``` | 3 | ||
938 | D | Buy a Ticket | PROGRAMMING | 2,000 | [
"data structures",
"graphs",
"shortest paths"
] | null | null | Musicians of a popular band "Flayer" have announced that they are going to "make their exit" with a world tour. Of course, they will visit Berland as well.
There are *n* cities in Berland. People can travel between cities using two-directional train routes; there are exactly *m* routes, *i*-th route can be used to go from city *v**i* to city *u**i* (and from *u**i* to *v**i*), and it costs *w**i* coins to use this route.
Each city will be visited by "Flayer", and the cost of the concert ticket in *i*-th city is *a**i* coins.
You have friends in every city of Berland, and they, knowing about your programming skills, asked you to calculate the minimum possible number of coins they have to pay to visit the concert. For every city *i* you have to compute the minimum number of coins a person from city *i* has to spend to travel to some city *j* (or possibly stay in city *i*), attend a concert there, and return to city *i* (if *j*<=≠<=*i*).
Formally, for every you have to calculate , where *d*(*i*,<=*j*) is the minimum number of coins you have to spend to travel from city *i* to city *j*. If there is no way to reach city *j* from city *i*, then we consider *d*(*i*,<=*j*) to be infinitely large. | The first line contains two integers *n* and *m* (2<=≤<=*n*<=≤<=2·105, 1<=≤<=*m*<=≤<=2·105).
Then *m* lines follow, *i*-th contains three integers *v**i*, *u**i* and *w**i* (1<=≤<=*v**i*,<=*u**i*<=≤<=*n*,<=*v**i*<=≠<=*u**i*, 1<=≤<=*w**i*<=≤<=1012) denoting *i*-th train route. There are no multiple train routes connecting the same pair of cities, that is, for each (*v*,<=*u*) neither extra (*v*,<=*u*) nor (*u*,<=*v*) present in input.
The next line contains *n* integers *a*1,<=*a*2,<=... *a**k* (1<=≤<=*a**i*<=≤<=1012) — price to attend the concert in *i*-th city. | Print *n* integers. *i*-th of them must be equal to the minimum number of coins a person from city *i* has to spend to travel to some city *j* (or possibly stay in city *i*), attend a concert there, and return to city *i* (if *j*<=≠<=*i*). | [
"4 2\n1 2 4\n2 3 7\n6 20 1 25\n",
"3 3\n1 2 1\n2 3 1\n1 3 1\n30 10 20\n"
] | [
"6 14 1 25 \n",
"12 10 12 \n"
] | none | 0 | [
{
"input": "4 2\n1 2 4\n2 3 7\n6 20 1 25",
"output": "6 14 1 25 "
},
{
"input": "3 3\n1 2 1\n2 3 1\n1 3 1\n30 10 20",
"output": "12 10 12 "
},
{
"input": "7 7\n1 6 745325\n2 3 3581176\n2 4 19\n3 6 71263060078\n5 4 141198\n7 4 163953\n5 6 15994\n1 297404206755 82096176217 14663411 1873897... | 1,601,356,328 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 2 | 2,000 | 70,860,800 | class Edge:
def __init__(self, s, d, w):
self.source = s
self.destination = d
self.weight = w
self.next = None
class Solution:
def find_min_vertex(self):
min_index = -1
for i in range(self.len):
if not self.visited[i]:
if min_index == -1 or self.result[min_index] > self.result[i]:
min_index = i
return min_index
def solve(self, A, B):
n = len(A)
self.len = n
self.graph = [0 for index in range(n)]
self.result = A
self.visited = [False for index in range(n)]
for edge in B:
if self.graph[edge[0]-1] == 0:
self.graph[edge[0]-1] = Edge(edge[0]-1, edge[1]-1, edge[2])
else:
t = self.graph[edge[0]-1]
while t:
if t.next is None:
t.next = Edge(edge[0]-1, edge[1]-1, edge[2])
t = None
else:
t = t.next
if self.graph[edge[1] - 1] == 0:
self.graph[edge[1] - 1] = Edge(edge[1] - 1, edge[0] - 1, edge[2])
else:
t = self.graph[edge[1] - 1]
while t:
if t.next is None:
t.next = Edge(edge[0]-1, edge[1]-1, edge[2])
t = None
else:
t = t.next
for index in range(n):
minVertex = self.find_min_vertex()
self.visited[minVertex] = True
t = self.graph[minVertex]
while t:
ans = (2*t.weight) + A[t.destination]
if ans < self.result[minVertex]:
self.result[minVertex] = ans
# self.traverse_min(minVertex,t.destination,A)
t = t.next
print(*self.result)
if __name__ == "__main__":
num = input().split(" ")
B = []
for i in range(int(num[1])):
r= []
r = list(map(int, input().split(" ")))
B.append(r)
A = list(map(int,input().split(" ")))
s = Solution()
s.solve(A,B) | Title: Buy a Ticket
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Musicians of a popular band "Flayer" have announced that they are going to "make their exit" with a world tour. Of course, they will visit Berland as well.
There are *n* cities in Berland. People can travel between cities using two-directional train routes; there are exactly *m* routes, *i*-th route can be used to go from city *v**i* to city *u**i* (and from *u**i* to *v**i*), and it costs *w**i* coins to use this route.
Each city will be visited by "Flayer", and the cost of the concert ticket in *i*-th city is *a**i* coins.
You have friends in every city of Berland, and they, knowing about your programming skills, asked you to calculate the minimum possible number of coins they have to pay to visit the concert. For every city *i* you have to compute the minimum number of coins a person from city *i* has to spend to travel to some city *j* (or possibly stay in city *i*), attend a concert there, and return to city *i* (if *j*<=≠<=*i*).
Formally, for every you have to calculate , where *d*(*i*,<=*j*) is the minimum number of coins you have to spend to travel from city *i* to city *j*. If there is no way to reach city *j* from city *i*, then we consider *d*(*i*,<=*j*) to be infinitely large.
Input Specification:
The first line contains two integers *n* and *m* (2<=≤<=*n*<=≤<=2·105, 1<=≤<=*m*<=≤<=2·105).
Then *m* lines follow, *i*-th contains three integers *v**i*, *u**i* and *w**i* (1<=≤<=*v**i*,<=*u**i*<=≤<=*n*,<=*v**i*<=≠<=*u**i*, 1<=≤<=*w**i*<=≤<=1012) denoting *i*-th train route. There are no multiple train routes connecting the same pair of cities, that is, for each (*v*,<=*u*) neither extra (*v*,<=*u*) nor (*u*,<=*v*) present in input.
The next line contains *n* integers *a*1,<=*a*2,<=... *a**k* (1<=≤<=*a**i*<=≤<=1012) — price to attend the concert in *i*-th city.
Output Specification:
Print *n* integers. *i*-th of them must be equal to the minimum number of coins a person from city *i* has to spend to travel to some city *j* (or possibly stay in city *i*), attend a concert there, and return to city *i* (if *j*<=≠<=*i*).
Demo Input:
['4 2\n1 2 4\n2 3 7\n6 20 1 25\n', '3 3\n1 2 1\n2 3 1\n1 3 1\n30 10 20\n']
Demo Output:
['6 14 1 25 \n', '12 10 12 \n']
Note:
none | ```python
class Edge:
def __init__(self, s, d, w):
self.source = s
self.destination = d
self.weight = w
self.next = None
class Solution:
def find_min_vertex(self):
min_index = -1
for i in range(self.len):
if not self.visited[i]:
if min_index == -1 or self.result[min_index] > self.result[i]:
min_index = i
return min_index
def solve(self, A, B):
n = len(A)
self.len = n
self.graph = [0 for index in range(n)]
self.result = A
self.visited = [False for index in range(n)]
for edge in B:
if self.graph[edge[0]-1] == 0:
self.graph[edge[0]-1] = Edge(edge[0]-1, edge[1]-1, edge[2])
else:
t = self.graph[edge[0]-1]
while t:
if t.next is None:
t.next = Edge(edge[0]-1, edge[1]-1, edge[2])
t = None
else:
t = t.next
if self.graph[edge[1] - 1] == 0:
self.graph[edge[1] - 1] = Edge(edge[1] - 1, edge[0] - 1, edge[2])
else:
t = self.graph[edge[1] - 1]
while t:
if t.next is None:
t.next = Edge(edge[0]-1, edge[1]-1, edge[2])
t = None
else:
t = t.next
for index in range(n):
minVertex = self.find_min_vertex()
self.visited[minVertex] = True
t = self.graph[minVertex]
while t:
ans = (2*t.weight) + A[t.destination]
if ans < self.result[minVertex]:
self.result[minVertex] = ans
# self.traverse_min(minVertex,t.destination,A)
t = t.next
print(*self.result)
if __name__ == "__main__":
num = input().split(" ")
B = []
for i in range(int(num[1])):
r= []
r = list(map(int, input().split(" ")))
B.append(r)
A = list(map(int,input().split(" ")))
s = Solution()
s.solve(A,B)
``` | 0 | |
545 | C | Woodcutters | PROGRAMMING | 1,500 | [
"dp",
"greedy"
] | null | null | Little Susie listens to fairy tales before bed every day. Today's fairy tale was about wood cutters and the little girl immediately started imagining the choppers cutting wood. She imagined the situation that is described below.
There are *n* trees located along the road at points with coordinates *x*1,<=*x*2,<=...,<=*x**n*. Each tree has its height *h**i*. Woodcutters can cut down a tree and fell it to the left or to the right. After that it occupies one of the segments [*x**i*<=-<=*h**i*,<=*x**i*] or [*x**i*;*x**i*<=+<=*h**i*]. The tree that is not cut down occupies a single point with coordinate *x**i*. Woodcutters can fell a tree if the segment to be occupied by the fallen tree doesn't contain any occupied point. The woodcutters want to process as many trees as possible, so Susie wonders, what is the maximum number of trees to fell. | The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of trees.
Next *n* lines contain pairs of integers *x**i*,<=*h**i* (1<=≤<=*x**i*,<=*h**i*<=≤<=109) — the coordinate and the height of the *і*-th tree.
The pairs are given in the order of ascending *x**i*. No two trees are located at the point with the same coordinate. | Print a single number — the maximum number of trees that you can cut down by the given rules. | [
"5\n1 2\n2 1\n5 10\n10 9\n19 1\n",
"5\n1 2\n2 1\n5 10\n10 9\n20 1\n"
] | [
"3\n",
"4\n"
] | In the first sample you can fell the trees like that:
- fell the 1-st tree to the left — now it occupies segment [ - 1;1] - fell the 2-nd tree to the right — now it occupies segment [2;3] - leave the 3-rd tree — it occupies point 5 - leave the 4-th tree — it occupies point 10 - fell the 5-th tree to the right — now it occupies segment [19;20]
In the second sample you can also fell 4-th tree to the right, after that it will occupy segment [10;19]. | 1,750 | [
{
"input": "5\n1 2\n2 1\n5 10\n10 9\n19 1",
"output": "3"
},
{
"input": "5\n1 2\n2 1\n5 10\n10 9\n20 1",
"output": "4"
},
{
"input": "4\n10 4\n15 1\n19 3\n20 1",
"output": "4"
},
{
"input": "35\n1 7\n3 11\n6 12\n7 6\n8 5\n9 11\n15 3\n16 10\n22 2\n23 3\n25 7\n27 3\n34 5\n35 10... | 1,666,861,330 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 67 | 655 | 13,107,200 | n=int(input())
jk=0
jl=0
x=[] #pos of trees 0_n-1
h=[] #height of trees 0_n-1
for i in range(n):#input
jk,jl=[int(x) for x in input().split()]
x.append(jk)
h.append(jl)
x=[-10000000000000000]+x+[1000000000000000]#pos of trees 1_n
h=[0]+h+[0]#height of trees 1_n
s=0
for i in range(1,n+1):#1_n
if x[i]-x[i-1]>h[i]:
s+=1
elif x[i+1]-x[i]>h[i]:
s+=1
x[i]+=h[i]
print(s)
| Title: Woodcutters
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Little Susie listens to fairy tales before bed every day. Today's fairy tale was about wood cutters and the little girl immediately started imagining the choppers cutting wood. She imagined the situation that is described below.
There are *n* trees located along the road at points with coordinates *x*1,<=*x*2,<=...,<=*x**n*. Each tree has its height *h**i*. Woodcutters can cut down a tree and fell it to the left or to the right. After that it occupies one of the segments [*x**i*<=-<=*h**i*,<=*x**i*] or [*x**i*;*x**i*<=+<=*h**i*]. The tree that is not cut down occupies a single point with coordinate *x**i*. Woodcutters can fell a tree if the segment to be occupied by the fallen tree doesn't contain any occupied point. The woodcutters want to process as many trees as possible, so Susie wonders, what is the maximum number of trees to fell.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of trees.
Next *n* lines contain pairs of integers *x**i*,<=*h**i* (1<=≤<=*x**i*,<=*h**i*<=≤<=109) — the coordinate and the height of the *і*-th tree.
The pairs are given in the order of ascending *x**i*. No two trees are located at the point with the same coordinate.
Output Specification:
Print a single number — the maximum number of trees that you can cut down by the given rules.
Demo Input:
['5\n1 2\n2 1\n5 10\n10 9\n19 1\n', '5\n1 2\n2 1\n5 10\n10 9\n20 1\n']
Demo Output:
['3\n', '4\n']
Note:
In the first sample you can fell the trees like that:
- fell the 1-st tree to the left — now it occupies segment [ - 1;1] - fell the 2-nd tree to the right — now it occupies segment [2;3] - leave the 3-rd tree — it occupies point 5 - leave the 4-th tree — it occupies point 10 - fell the 5-th tree to the right — now it occupies segment [19;20]
In the second sample you can also fell 4-th tree to the right, after that it will occupy segment [10;19]. | ```python
n=int(input())
jk=0
jl=0
x=[] #pos of trees 0_n-1
h=[] #height of trees 0_n-1
for i in range(n):#input
jk,jl=[int(x) for x in input().split()]
x.append(jk)
h.append(jl)
x=[-10000000000000000]+x+[1000000000000000]#pos of trees 1_n
h=[0]+h+[0]#height of trees 1_n
s=0
for i in range(1,n+1):#1_n
if x[i]-x[i-1]>h[i]:
s+=1
elif x[i+1]-x[i]>h[i]:
s+=1
x[i]+=h[i]
print(s)
``` | 3 | |
331 | C1 | The Great Julya Calendar | PROGRAMMING | 1,100 | [
"dp"
] | null | null | Yet another Armageddon is coming! This time the culprit is the Julya tribe calendar.
The beavers in this tribe knew math very well. Smart Beaver, an archaeologist, got a sacred plate with a magic integer on it. The translation from Old Beaverish is as follows:
"May the Great Beaver bless you! May your chacres open and may your third eye never turn blind from beholding the Truth! Take the magic number, subtract a digit from it (the digit must occur in the number) and get a new magic number. Repeat this operation until a magic number equals zero. The Earth will stand on Three Beavers for the time, equal to the number of subtractions you perform!"
Distinct subtraction sequences can obviously get you different number of operations. But the Smart Beaver is ready to face the worst and is asking you to count the minimum number of operations he needs to reduce the magic number to zero. | The single line contains the magic integer *n*, 0<=≤<=*n*.
- to get 20 points, you need to solve the problem with constraints: *n*<=≤<=106 (subproblem C1); - to get 40 points, you need to solve the problem with constraints: *n*<=≤<=1012 (subproblems C1+C2); - to get 100 points, you need to solve the problem with constraints: *n*<=≤<=1018 (subproblems C1+C2+C3). | Print a single integer — the minimum number of subtractions that turns the magic number to a zero. | [
"24\n"
] | [
"5"
] | In the first test sample the minimum number of operations can be reached by the following sequence of subtractions: | 20 | [
{
"input": "24",
"output": "5"
},
{
"input": "0",
"output": "0"
},
{
"input": "3",
"output": "1"
},
{
"input": "8",
"output": "1"
},
{
"input": "9",
"output": "1"
},
{
"input": "10",
"output": "2"
},
{
"input": "31",
"output": "6"
},
... | 1,599,544,514 | 2,147,483,647 | Python 3 | COMPILATION_ERROR | TESTS1 | 0 | 0 | 0 | memoization_table = {}
def compute_count(previous_max, rest):
global memoization_table
original_rest = rest
if (previous_max, rest) in memoization_table:
return memoization_table[(previous_max, rest)]
num_digits = len(str(rest))
if num_digits == 1:
memoization_table[(previous_max, original_rest)] = 1, min(0, rest-previous_max)
return 1, min(0, rest-previous_max)
sum_count = 0
while rest > 0:
s = str(rest).zfill(num_digits)
new_max = max(previous_max, int(s[0]))
new_rest = int(s[1:])
count, leftover = compute_count(new_max, new_rest)
sum_count += count
rest -= new_rest
rest += leftover
memoization_table[(previous_max, original_rest)] = sum_count, rest
return sum_count, rest
print(compute_count(0, int(input())[0])
| Title: The Great Julya Calendar
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Yet another Armageddon is coming! This time the culprit is the Julya tribe calendar.
The beavers in this tribe knew math very well. Smart Beaver, an archaeologist, got a sacred plate with a magic integer on it. The translation from Old Beaverish is as follows:
"May the Great Beaver bless you! May your chacres open and may your third eye never turn blind from beholding the Truth! Take the magic number, subtract a digit from it (the digit must occur in the number) and get a new magic number. Repeat this operation until a magic number equals zero. The Earth will stand on Three Beavers for the time, equal to the number of subtractions you perform!"
Distinct subtraction sequences can obviously get you different number of operations. But the Smart Beaver is ready to face the worst and is asking you to count the minimum number of operations he needs to reduce the magic number to zero.
Input Specification:
The single line contains the magic integer *n*, 0<=≤<=*n*.
- to get 20 points, you need to solve the problem with constraints: *n*<=≤<=106 (subproblem C1); - to get 40 points, you need to solve the problem with constraints: *n*<=≤<=1012 (subproblems C1+C2); - to get 100 points, you need to solve the problem with constraints: *n*<=≤<=1018 (subproblems C1+C2+C3).
Output Specification:
Print a single integer — the minimum number of subtractions that turns the magic number to a zero.
Demo Input:
['24\n']
Demo Output:
['5']
Note:
In the first test sample the minimum number of operations can be reached by the following sequence of subtractions: | ```python
memoization_table = {}
def compute_count(previous_max, rest):
global memoization_table
original_rest = rest
if (previous_max, rest) in memoization_table:
return memoization_table[(previous_max, rest)]
num_digits = len(str(rest))
if num_digits == 1:
memoization_table[(previous_max, original_rest)] = 1, min(0, rest-previous_max)
return 1, min(0, rest-previous_max)
sum_count = 0
while rest > 0:
s = str(rest).zfill(num_digits)
new_max = max(previous_max, int(s[0]))
new_rest = int(s[1:])
count, leftover = compute_count(new_max, new_rest)
sum_count += count
rest -= new_rest
rest += leftover
memoization_table[(previous_max, original_rest)] = sum_count, rest
return sum_count, rest
print(compute_count(0, int(input())[0])
``` | -1 | |
706 | B | Interesting drink | PROGRAMMING | 1,100 | [
"binary search",
"dp",
"implementation"
] | null | null | Vasiliy likes to rest after a hard work, so you may often meet him in some bar nearby. As all programmers do, he loves the famous drink "Beecola", which can be bought in *n* different shops in the city. It's known that the price of one bottle in the shop *i* is equal to *x**i* coins.
Vasiliy plans to buy his favorite drink for *q* consecutive days. He knows, that on the *i*-th day he will be able to spent *m**i* coins. Now, for each of the days he want to know in how many different shops he can buy a bottle of "Beecola". | The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=100<=000) — the number of shops in the city that sell Vasiliy's favourite drink.
The second line contains *n* integers *x**i* (1<=≤<=*x**i*<=≤<=100<=000) — prices of the bottles of the drink in the *i*-th shop.
The third line contains a single integer *q* (1<=≤<=*q*<=≤<=100<=000) — the number of days Vasiliy plans to buy the drink.
Then follow *q* lines each containing one integer *m**i* (1<=≤<=*m**i*<=≤<=109) — the number of coins Vasiliy can spent on the *i*-th day. | Print *q* integers. The *i*-th of them should be equal to the number of shops where Vasiliy will be able to buy a bottle of the drink on the *i*-th day. | [
"5\n3 10 8 6 11\n4\n1\n10\n3\n11\n"
] | [
"0\n4\n1\n5\n"
] | On the first day, Vasiliy won't be able to buy a drink in any of the shops.
On the second day, Vasiliy can buy a drink in the shops 1, 2, 3 and 4.
On the third day, Vasiliy can buy a drink only in the shop number 1.
Finally, on the last day Vasiliy can buy a drink in any shop. | 1,000 | [
{
"input": "5\n3 10 8 6 11\n4\n1\n10\n3\n11",
"output": "0\n4\n1\n5"
},
{
"input": "5\n868 987 714 168 123\n10\n424\n192\n795\n873\n117\n914\n735\n158\n631\n471",
"output": "2\n2\n3\n4\n0\n4\n3\n1\n2\n2"
},
{
"input": "3\n435 482 309\n7\n245\n241\n909\n745\n980\n29\n521",
"output": "... | 1,696,483,281 | 2,147,483,647 | PyPy 3 | TIME_LIMIT_EXCEEDED | TESTS | 10 | 2,000 | 13,312,000 | import bisect
n = int(input())
prices = list(map(int, input().split()))
q = int(input())
sprices = sorted(prices)
for _ in range(q):
stat = 0
coins = int(input())
i = bisect.bisect_right(sprices, coins)
print(len(sprices[:i])) | Title: Interesting drink
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasiliy likes to rest after a hard work, so you may often meet him in some bar nearby. As all programmers do, he loves the famous drink "Beecola", which can be bought in *n* different shops in the city. It's known that the price of one bottle in the shop *i* is equal to *x**i* coins.
Vasiliy plans to buy his favorite drink for *q* consecutive days. He knows, that on the *i*-th day he will be able to spent *m**i* coins. Now, for each of the days he want to know in how many different shops he can buy a bottle of "Beecola".
Input Specification:
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=100<=000) — the number of shops in the city that sell Vasiliy's favourite drink.
The second line contains *n* integers *x**i* (1<=≤<=*x**i*<=≤<=100<=000) — prices of the bottles of the drink in the *i*-th shop.
The third line contains a single integer *q* (1<=≤<=*q*<=≤<=100<=000) — the number of days Vasiliy plans to buy the drink.
Then follow *q* lines each containing one integer *m**i* (1<=≤<=*m**i*<=≤<=109) — the number of coins Vasiliy can spent on the *i*-th day.
Output Specification:
Print *q* integers. The *i*-th of them should be equal to the number of shops where Vasiliy will be able to buy a bottle of the drink on the *i*-th day.
Demo Input:
['5\n3 10 8 6 11\n4\n1\n10\n3\n11\n']
Demo Output:
['0\n4\n1\n5\n']
Note:
On the first day, Vasiliy won't be able to buy a drink in any of the shops.
On the second day, Vasiliy can buy a drink in the shops 1, 2, 3 and 4.
On the third day, Vasiliy can buy a drink only in the shop number 1.
Finally, on the last day Vasiliy can buy a drink in any shop. | ```python
import bisect
n = int(input())
prices = list(map(int, input().split()))
q = int(input())
sprices = sorted(prices)
for _ in range(q):
stat = 0
coins = int(input())
i = bisect.bisect_right(sprices, coins)
print(len(sprices[:i]))
``` | 0 | |
625 | B | War of the Corporations | PROGRAMMING | 1,200 | [
"constructive algorithms",
"greedy",
"strings"
] | null | null | A long time ago, in a galaxy far far away two giant IT-corporations Pineapple and Gogol continue their fierce competition. Crucial moment is just around the corner: Gogol is ready to release it's new tablet Lastus 3000.
This new device is equipped with specially designed artificial intelligence (AI). Employees of Pineapple did their best to postpone the release of Lastus 3000 as long as possible. Finally, they found out, that the name of the new artificial intelligence is similar to the name of the phone, that Pineapple released 200 years ago. As all rights on its name belong to Pineapple, they stand on changing the name of Gogol's artificial intelligence.
Pineapple insists, that the name of their phone occurs in the name of AI as a substring. Because the name of technology was already printed on all devices, the Gogol's director decided to replace some characters in AI name with "#". As this operation is pretty expensive, you should find the minimum number of characters to replace with "#", such that the name of AI doesn't contain the name of the phone as a substring.
Substring is a continuous subsequence of a string. | The first line of the input contains the name of AI designed by Gogol, its length doesn't exceed 100<=000 characters. Second line contains the name of the phone released by Pineapple 200 years ago, its length doesn't exceed 30. Both string are non-empty and consist of only small English letters. | Print the minimum number of characters that must be replaced with "#" in order to obtain that the name of the phone doesn't occur in the name of AI as a substring. | [
"intellect\ntell\n",
"google\napple\n",
"sirisiri\nsir\n"
] | [
"1",
"0",
"2"
] | In the first sample AI's name may be replaced with "int#llect".
In the second sample Gogol can just keep things as they are.
In the third sample one of the new possible names of AI may be "s#ris#ri". | 750 | [
{
"input": "intellect\ntell",
"output": "1"
},
{
"input": "google\napple",
"output": "0"
},
{
"input": "sirisiri\nsir",
"output": "2"
},
{
"input": "sirisiri\nsiri",
"output": "2"
},
{
"input": "aaaaaaa\naaaa",
"output": "1"
},
{
"input": "bbbbbb\nbb",... | 1,479,351,909 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 3 | 46 | 0 |
def main():
name = input()
substring = input()
count = 0
i = 0
while(i + len(substring) < len(name)):
if name[i] == substring[0]:
differ = False
for j in range(len(substring)):
if substring[j] != name[j+i]:
differ = True
break
if not differ:
count += 1
i += len(substring) - 1
i+=1
print(count)
return 0
if __name__ == '__main__':
main()
| Title: War of the Corporations
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A long time ago, in a galaxy far far away two giant IT-corporations Pineapple and Gogol continue their fierce competition. Crucial moment is just around the corner: Gogol is ready to release it's new tablet Lastus 3000.
This new device is equipped with specially designed artificial intelligence (AI). Employees of Pineapple did their best to postpone the release of Lastus 3000 as long as possible. Finally, they found out, that the name of the new artificial intelligence is similar to the name of the phone, that Pineapple released 200 years ago. As all rights on its name belong to Pineapple, they stand on changing the name of Gogol's artificial intelligence.
Pineapple insists, that the name of their phone occurs in the name of AI as a substring. Because the name of technology was already printed on all devices, the Gogol's director decided to replace some characters in AI name with "#". As this operation is pretty expensive, you should find the minimum number of characters to replace with "#", such that the name of AI doesn't contain the name of the phone as a substring.
Substring is a continuous subsequence of a string.
Input Specification:
The first line of the input contains the name of AI designed by Gogol, its length doesn't exceed 100<=000 characters. Second line contains the name of the phone released by Pineapple 200 years ago, its length doesn't exceed 30. Both string are non-empty and consist of only small English letters.
Output Specification:
Print the minimum number of characters that must be replaced with "#" in order to obtain that the name of the phone doesn't occur in the name of AI as a substring.
Demo Input:
['intellect\ntell\n', 'google\napple\n', 'sirisiri\nsir\n']
Demo Output:
['1', '0', '2']
Note:
In the first sample AI's name may be replaced with "int#llect".
In the second sample Gogol can just keep things as they are.
In the third sample one of the new possible names of AI may be "s#ris#ri". | ```python
def main():
name = input()
substring = input()
count = 0
i = 0
while(i + len(substring) < len(name)):
if name[i] == substring[0]:
differ = False
for j in range(len(substring)):
if substring[j] != name[j+i]:
differ = True
break
if not differ:
count += 1
i += len(substring) - 1
i+=1
print(count)
return 0
if __name__ == '__main__':
main()
``` | 0 | |
124 | A | The number of positions | PROGRAMMING | 1,000 | [
"math"
] | null | null | Petr stands in line of *n* people, but he doesn't know exactly which position he occupies. He can say that there are no less than *a* people standing in front of him and no more than *b* people standing behind him. Find the number of different positions Petr can occupy. | The only line contains three integers *n*, *a* and *b* (0<=≤<=*a*,<=*b*<=<<=*n*<=≤<=100). | Print the single number — the number of the sought positions. | [
"3 1 1\n",
"5 2 3\n"
] | [
"2\n",
"3\n"
] | The possible positions in the first sample are: 2 and 3 (if we number the positions starting with 1).
In the second sample they are 3, 4 and 5. | 500 | [
{
"input": "3 1 1",
"output": "2"
},
{
"input": "5 2 3",
"output": "3"
},
{
"input": "5 4 0",
"output": "1"
},
{
"input": "6 5 5",
"output": "1"
},
{
"input": "9 4 3",
"output": "4"
},
{
"input": "11 4 6",
"output": "7"
},
{
"input": "13 8 ... | 1,670,745,872 | 2,147,483,647 | Python 3 | OK | TESTS | 50 | 92 | 0 | x=list(map(int,input().split()))
g=x[0]-x[1]
h=x[2]+1
print(min(g,h)) | Title: The number of positions
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Petr stands in line of *n* people, but he doesn't know exactly which position he occupies. He can say that there are no less than *a* people standing in front of him and no more than *b* people standing behind him. Find the number of different positions Petr can occupy.
Input Specification:
The only line contains three integers *n*, *a* and *b* (0<=≤<=*a*,<=*b*<=<<=*n*<=≤<=100).
Output Specification:
Print the single number — the number of the sought positions.
Demo Input:
['3 1 1\n', '5 2 3\n']
Demo Output:
['2\n', '3\n']
Note:
The possible positions in the first sample are: 2 and 3 (if we number the positions starting with 1).
In the second sample they are 3, 4 and 5. | ```python
x=list(map(int,input().split()))
g=x[0]-x[1]
h=x[2]+1
print(min(g,h))
``` | 3 | |
41 | A | Translation | PROGRAMMING | 800 | [
"implementation",
"strings"
] | A. Translation | 2 | 256 | The translation from the Berland language into the Birland language is not an easy task. Those languages are very similar: a berlandish word differs from a birlandish word with the same meaning a little: it is spelled (and pronounced) reversely. For example, a Berlandish word code corresponds to a Birlandish word edoc. However, it's easy to make a mistake during the «translation». Vasya translated word *s* from Berlandish into Birlandish as *t*. Help him: find out if he translated the word correctly. | The first line contains word *s*, the second line contains word *t*. The words consist of lowercase Latin letters. The input data do not consist unnecessary spaces. The words are not empty and their lengths do not exceed 100 symbols. | If the word *t* is a word *s*, written reversely, print YES, otherwise print NO. | [
"code\nedoc\n",
"abb\naba\n",
"code\ncode\n"
] | [
"YES\n",
"NO\n",
"NO\n"
] | none | 500 | [
{
"input": "code\nedoc",
"output": "YES"
},
{
"input": "abb\naba",
"output": "NO"
},
{
"input": "code\ncode",
"output": "NO"
},
{
"input": "abacaba\nabacaba",
"output": "YES"
},
{
"input": "q\nq",
"output": "YES"
},
{
"input": "asrgdfngfnmfgnhweratgjkk... | 1,671,968,936 | 2,147,483,647 | Python 3 | OK | TESTS | 40 | 92 | 0 | s = input() # read the first line as string s
t = input() # read the second line as string t
# check if t is equal to the reverse of s
if t == s[::-1]:
print("YES")
else:
print("NO")
| Title: Translation
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
The translation from the Berland language into the Birland language is not an easy task. Those languages are very similar: a berlandish word differs from a birlandish word with the same meaning a little: it is spelled (and pronounced) reversely. For example, a Berlandish word code corresponds to a Birlandish word edoc. However, it's easy to make a mistake during the «translation». Vasya translated word *s* from Berlandish into Birlandish as *t*. Help him: find out if he translated the word correctly.
Input Specification:
The first line contains word *s*, the second line contains word *t*. The words consist of lowercase Latin letters. The input data do not consist unnecessary spaces. The words are not empty and their lengths do not exceed 100 symbols.
Output Specification:
If the word *t* is a word *s*, written reversely, print YES, otherwise print NO.
Demo Input:
['code\nedoc\n', 'abb\naba\n', 'code\ncode\n']
Demo Output:
['YES\n', 'NO\n', 'NO\n']
Note:
none | ```python
s = input() # read the first line as string s
t = input() # read the second line as string t
# check if t is equal to the reverse of s
if t == s[::-1]:
print("YES")
else:
print("NO")
``` | 3.977 |
611 | A | New Year and Days | PROGRAMMING | 900 | [
"implementation"
] | null | null | Today is Wednesday, the third day of the week. What's more interesting is that tomorrow is the last day of the year 2015.
Limak is a little polar bear. He enjoyed this year a lot. Now, he is so eager to the coming year 2016.
Limak wants to prove how responsible a bear he is. He is going to regularly save candies for the entire year 2016! He considers various saving plans. He can save one candy either on some fixed day of the week or on some fixed day of the month.
Limak chose one particular plan. He isn't sure how many candies he will save in the 2016 with his plan. Please, calculate it and tell him. | The only line of the input is in one of the following two formats:
- "*x* of week" where *x* (1<=≤<=*x*<=≤<=7) denotes the day of the week. The 1-st day is Monday and the 7-th one is Sunday. - "*x* of month" where *x* (1<=≤<=*x*<=≤<=31) denotes the day of the month. | Print one integer — the number of candies Limak will save in the year 2016. | [
"4 of week\n",
"30 of month\n"
] | [
"52\n",
"11\n"
] | Polar bears use the Gregorian calendar. It is the most common calendar and you likely use it too. You can read about it on Wikipedia if you want to – [https://en.wikipedia.org/wiki/Gregorian_calendar](https://en.wikipedia.org/wiki/Gregorian_calendar). The week starts with Monday.
In the first sample Limak wants to save one candy on each Thursday (the 4-th day of the week). There are 52 Thursdays in the 2016. Thus, he will save 52 candies in total.
In the second sample Limak wants to save one candy on the 30-th day of each month. There is the 30-th day in exactly 11 months in the 2016 — all months but February. It means that Limak will save 11 candies in total. | 500 | [
{
"input": "4 of week",
"output": "52"
},
{
"input": "30 of month",
"output": "11"
},
{
"input": "17 of month",
"output": "12"
},
{
"input": "31 of month",
"output": "7"
},
{
"input": "6 of week",
"output": "53"
},
{
"input": "1 of week",
"output":... | 1,452,911,783 | 2,147,483,647 | Python 3 | OK | TESTS | 38 | 62 | 0 | def main():
x, _, w = input().split()
x = int(x)
if w == "week":
print((366 + x) // 7 - x // 7)
else:
print(7 + 4 * (x < 31) + (x < 30))
if __name__ == '__main__':
main()
| Title: New Year and Days
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Today is Wednesday, the third day of the week. What's more interesting is that tomorrow is the last day of the year 2015.
Limak is a little polar bear. He enjoyed this year a lot. Now, he is so eager to the coming year 2016.
Limak wants to prove how responsible a bear he is. He is going to regularly save candies for the entire year 2016! He considers various saving plans. He can save one candy either on some fixed day of the week or on some fixed day of the month.
Limak chose one particular plan. He isn't sure how many candies he will save in the 2016 with his plan. Please, calculate it and tell him.
Input Specification:
The only line of the input is in one of the following two formats:
- "*x* of week" where *x* (1<=≤<=*x*<=≤<=7) denotes the day of the week. The 1-st day is Monday and the 7-th one is Sunday. - "*x* of month" where *x* (1<=≤<=*x*<=≤<=31) denotes the day of the month.
Output Specification:
Print one integer — the number of candies Limak will save in the year 2016.
Demo Input:
['4 of week\n', '30 of month\n']
Demo Output:
['52\n', '11\n']
Note:
Polar bears use the Gregorian calendar. It is the most common calendar and you likely use it too. You can read about it on Wikipedia if you want to – [https://en.wikipedia.org/wiki/Gregorian_calendar](https://en.wikipedia.org/wiki/Gregorian_calendar). The week starts with Monday.
In the first sample Limak wants to save one candy on each Thursday (the 4-th day of the week). There are 52 Thursdays in the 2016. Thus, he will save 52 candies in total.
In the second sample Limak wants to save one candy on the 30-th day of each month. There is the 30-th day in exactly 11 months in the 2016 — all months but February. It means that Limak will save 11 candies in total. | ```python
def main():
x, _, w = input().split()
x = int(x)
if w == "week":
print((366 + x) // 7 - x // 7)
else:
print(7 + 4 * (x < 31) + (x < 30))
if __name__ == '__main__':
main()
``` | 3 | |
1 | A | Theatre Square | PROGRAMMING | 1,000 | [
"math"
] | A. Theatre Square | 1 | 256 | Theatre Square in the capital city of Berland has a rectangular shape with the size *n*<=×<=*m* meters. On the occasion of the city's anniversary, a decision was taken to pave the Square with square granite flagstones. Each flagstone is of the size *a*<=×<=*a*.
What is the least number of flagstones needed to pave the Square? It's allowed to cover the surface larger than the Theatre Square, but the Square has to be covered. It's not allowed to break the flagstones. The sides of flagstones should be parallel to the sides of the Square. | The input contains three positive integer numbers in the first line: *n*,<=<=*m* and *a* (1<=≤<=<=*n*,<=*m*,<=*a*<=≤<=109). | Write the needed number of flagstones. | [
"6 6 4\n"
] | [
"4\n"
] | none | 0 | [
{
"input": "6 6 4",
"output": "4"
},
{
"input": "1 1 1",
"output": "1"
},
{
"input": "2 1 1",
"output": "2"
},
{
"input": "1 2 1",
"output": "2"
},
{
"input": "2 2 1",
"output": "4"
},
{
"input": "2 1 2",
"output": "1"
},
{
"input": "1 1 3"... | 1,696,261,189 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 20 | 61 | 0 | #initializes variables to be used
variables = input().split(' ')
n = int(variables[0])
m = int(variables[1])
a = int(variables[2])
#floor divides dimensions n and m with a to get number of flagstones are needed to cover, assuming it can go beyond the dimension
flagstones = ((-n // a) * (-m // a))
print(flagstones) | Title: Theatre Square
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Theatre Square in the capital city of Berland has a rectangular shape with the size *n*<=×<=*m* meters. On the occasion of the city's anniversary, a decision was taken to pave the Square with square granite flagstones. Each flagstone is of the size *a*<=×<=*a*.
What is the least number of flagstones needed to pave the Square? It's allowed to cover the surface larger than the Theatre Square, but the Square has to be covered. It's not allowed to break the flagstones. The sides of flagstones should be parallel to the sides of the Square.
Input Specification:
The input contains three positive integer numbers in the first line: *n*,<=<=*m* and *a* (1<=≤<=<=*n*,<=*m*,<=*a*<=≤<=109).
Output Specification:
Write the needed number of flagstones.
Demo Input:
['6 6 4\n']
Demo Output:
['4\n']
Note:
none | ```python
#initializes variables to be used
variables = input().split(' ')
n = int(variables[0])
m = int(variables[1])
a = int(variables[2])
#floor divides dimensions n and m with a to get number of flagstones are needed to cover, assuming it can go beyond the dimension
flagstones = ((-n // a) * (-m // a))
print(flagstones)
``` | 3.9695 |
456 | A | Laptops | PROGRAMMING | 1,100 | [
"sortings"
] | null | null | One day Dima and Alex had an argument about the price and quality of laptops. Dima thinks that the more expensive a laptop is, the better it is. Alex disagrees. Alex thinks that there are two laptops, such that the price of the first laptop is less (strictly smaller) than the price of the second laptop but the quality of the first laptop is higher (strictly greater) than the quality of the second laptop.
Please, check the guess of Alex. You are given descriptions of *n* laptops. Determine whether two described above laptops exist. | The first line contains an integer *n* (1<=≤<=*n*<=≤<=105) — the number of laptops.
Next *n* lines contain two integers each, *a**i* and *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*), where *a**i* is the price of the *i*-th laptop, and *b**i* is the number that represents the quality of the *i*-th laptop (the larger the number is, the higher is the quality).
All *a**i* are distinct. All *b**i* are distinct. | If Alex is correct, print "Happy Alex", otherwise print "Poor Alex" (without the quotes). | [
"2\n1 2\n2 1\n"
] | [
"Happy Alex\n"
] | none | 500 | [
{
"input": "2\n1 2\n2 1",
"output": "Happy Alex"
},
{
"input": "2\n1 1\n2 2",
"output": "Poor Alex"
},
{
"input": "3\n2 2\n3 3\n1 1",
"output": "Poor Alex"
},
{
"input": "3\n3 3\n1 2\n2 1",
"output": "Happy Alex"
},
{
"input": "1\n1 1",
"output": "Poor Alex"
... | 1,690,011,993 | 2,147,483,647 | Python 3 | OK | TESTS | 46 | 405 | 18,227,200 | n=int(input())
a=[]
b=[]
for i in range(n):
_,__=list(map(int,input().split()))
a.append(_)
b.append(__)
c=[i for j, i in sorted(zip(a,b))]
ans=False
for i in range(1,n):
if not (c[i]>=c[i-1]):
ans=True
break
if ans:
print("Happy Alex")
else:
print("Poor Alex") | Title: Laptops
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day Dima and Alex had an argument about the price and quality of laptops. Dima thinks that the more expensive a laptop is, the better it is. Alex disagrees. Alex thinks that there are two laptops, such that the price of the first laptop is less (strictly smaller) than the price of the second laptop but the quality of the first laptop is higher (strictly greater) than the quality of the second laptop.
Please, check the guess of Alex. You are given descriptions of *n* laptops. Determine whether two described above laptops exist.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=105) — the number of laptops.
Next *n* lines contain two integers each, *a**i* and *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*), where *a**i* is the price of the *i*-th laptop, and *b**i* is the number that represents the quality of the *i*-th laptop (the larger the number is, the higher is the quality).
All *a**i* are distinct. All *b**i* are distinct.
Output Specification:
If Alex is correct, print "Happy Alex", otherwise print "Poor Alex" (without the quotes).
Demo Input:
['2\n1 2\n2 1\n']
Demo Output:
['Happy Alex\n']
Note:
none | ```python
n=int(input())
a=[]
b=[]
for i in range(n):
_,__=list(map(int,input().split()))
a.append(_)
b.append(__)
c=[i for j, i in sorted(zip(a,b))]
ans=False
for i in range(1,n):
if not (c[i]>=c[i-1]):
ans=True
break
if ans:
print("Happy Alex")
else:
print("Poor Alex")
``` | 3 | |
180 | C | Letter | PROGRAMMING | 1,400 | [
"dp"
] | null | null | Patrick has just finished writing a message to his sweetheart Stacey when he noticed that the message didn't look fancy. Patrick was nervous while writing the message, so some of the letters there were lowercase and some of them were uppercase.
Patrick believes that a message is fancy if any uppercase letter stands to the left of any lowercase one. In other words, this rule describes the strings where first go zero or more uppercase letters, and then — zero or more lowercase letters.
To make the message fancy, Patrick can erase some letter and add the same letter in the same place in the opposite case (that is, he can replace an uppercase letter with the lowercase one and vice versa). Patrick got interested in the following question: what minimum number of actions do we need to make a message fancy? Changing a letter's case in the message counts as one action. Patrick cannot perform any other actions. | The only line of the input contains a non-empty string consisting of uppercase and lowercase letters. The string's length does not exceed 105. | Print a single number — the least number of actions needed to make the message fancy. | [
"PRuvetSTAaYA\n",
"OYPROSTIYAOPECHATALSYAPRIVETSTASYA\n",
"helloworld\n"
] | [
"5\n",
"0\n",
"0\n"
] | none | 0 | [
{
"input": "PRuvetSTAaYA",
"output": "5"
},
{
"input": "OYPROSTIYAOPECHATALSYAPRIVETSTASYA",
"output": "0"
},
{
"input": "helloworld",
"output": "0"
},
{
"input": "P",
"output": "0"
},
{
"input": "t",
"output": "0"
},
{
"input": "XdJ",
"output": "1... | 1,594,144,374 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 12 | 218 | 6,656,000 | s = input()
upper,lower = 0,0
for i in s:
if i.isupper():
upper+=1
else:
lower+=1
print(min(upper,lower)) | Title: Letter
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Patrick has just finished writing a message to his sweetheart Stacey when he noticed that the message didn't look fancy. Patrick was nervous while writing the message, so some of the letters there were lowercase and some of them were uppercase.
Patrick believes that a message is fancy if any uppercase letter stands to the left of any lowercase one. In other words, this rule describes the strings where first go zero or more uppercase letters, and then — zero or more lowercase letters.
To make the message fancy, Patrick can erase some letter and add the same letter in the same place in the opposite case (that is, he can replace an uppercase letter with the lowercase one and vice versa). Patrick got interested in the following question: what minimum number of actions do we need to make a message fancy? Changing a letter's case in the message counts as one action. Patrick cannot perform any other actions.
Input Specification:
The only line of the input contains a non-empty string consisting of uppercase and lowercase letters. The string's length does not exceed 105.
Output Specification:
Print a single number — the least number of actions needed to make the message fancy.
Demo Input:
['PRuvetSTAaYA\n', 'OYPROSTIYAOPECHATALSYAPRIVETSTASYA\n', 'helloworld\n']
Demo Output:
['5\n', '0\n', '0\n']
Note:
none | ```python
s = input()
upper,lower = 0,0
for i in s:
if i.isupper():
upper+=1
else:
lower+=1
print(min(upper,lower))
``` | 0 | |
731 | A | Night at the Museum | PROGRAMMING | 800 | [
"implementation",
"strings"
] | null | null | Grigoriy, like the hero of one famous comedy film, found a job as a night security guard at the museum. At first night he received embosser and was to take stock of the whole exposition.
Embosser is a special devise that allows to "print" the text of a plastic tape. Text is printed sequentially, character by character. The device consists of a wheel with a lowercase English letters written in a circle, static pointer to the current letter and a button that print the chosen letter. At one move it's allowed to rotate the alphabetic wheel one step clockwise or counterclockwise. Initially, static pointer points to letter 'a'. Other letters are located as shown on the picture:
After Grigoriy add new item to the base he has to print its name on the plastic tape and attach it to the corresponding exhibit. It's not required to return the wheel to its initial position with pointer on the letter 'a'.
Our hero is afraid that some exhibits may become alive and start to attack him, so he wants to print the names as fast as possible. Help him, for the given string find the minimum number of rotations of the wheel required to print it. | The only line of input contains the name of some exhibit — the non-empty string consisting of no more than 100 characters. It's guaranteed that the string consists of only lowercase English letters. | Print one integer — the minimum number of rotations of the wheel, required to print the name given in the input. | [
"zeus\n",
"map\n",
"ares\n"
] | [
"18\n",
"35\n",
"34\n"
] | To print the string from the first sample it would be optimal to perform the following sequence of rotations:
1. from 'a' to 'z' (1 rotation counterclockwise), 1. from 'z' to 'e' (5 clockwise rotations), 1. from 'e' to 'u' (10 rotations counterclockwise), 1. from 'u' to 's' (2 counterclockwise rotations). | 500 | [
{
"input": "zeus",
"output": "18"
},
{
"input": "map",
"output": "35"
},
{
"input": "ares",
"output": "34"
},
{
"input": "l",
"output": "11"
},
{
"input": "abcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuv",
"... | 1,628,139,100 | 2,147,483,647 | Python 3 | OK | TESTS | 44 | 77 | 6,963,200 | a=input()
k=0
n=0
while True :
if k==len(a):
print(n)
break
elif k==0 :
n=min(abs(ord('a')-ord(a[k])),26-abs(ord('a')-ord(a[k])))
k=1
elif k>0 :
n+=min(abs(ord(a[k-1])-ord(a[k])),26-abs(ord(a[k-1])-ord(a[k])))
k+=1 | Title: Night at the Museum
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Grigoriy, like the hero of one famous comedy film, found a job as a night security guard at the museum. At first night he received embosser and was to take stock of the whole exposition.
Embosser is a special devise that allows to "print" the text of a plastic tape. Text is printed sequentially, character by character. The device consists of a wheel with a lowercase English letters written in a circle, static pointer to the current letter and a button that print the chosen letter. At one move it's allowed to rotate the alphabetic wheel one step clockwise or counterclockwise. Initially, static pointer points to letter 'a'. Other letters are located as shown on the picture:
After Grigoriy add new item to the base he has to print its name on the plastic tape and attach it to the corresponding exhibit. It's not required to return the wheel to its initial position with pointer on the letter 'a'.
Our hero is afraid that some exhibits may become alive and start to attack him, so he wants to print the names as fast as possible. Help him, for the given string find the minimum number of rotations of the wheel required to print it.
Input Specification:
The only line of input contains the name of some exhibit — the non-empty string consisting of no more than 100 characters. It's guaranteed that the string consists of only lowercase English letters.
Output Specification:
Print one integer — the minimum number of rotations of the wheel, required to print the name given in the input.
Demo Input:
['zeus\n', 'map\n', 'ares\n']
Demo Output:
['18\n', '35\n', '34\n']
Note:
To print the string from the first sample it would be optimal to perform the following sequence of rotations:
1. from 'a' to 'z' (1 rotation counterclockwise), 1. from 'z' to 'e' (5 clockwise rotations), 1. from 'e' to 'u' (10 rotations counterclockwise), 1. from 'u' to 's' (2 counterclockwise rotations). | ```python
a=input()
k=0
n=0
while True :
if k==len(a):
print(n)
break
elif k==0 :
n=min(abs(ord('a')-ord(a[k])),26-abs(ord('a')-ord(a[k])))
k=1
elif k>0 :
n+=min(abs(ord(a[k-1])-ord(a[k])),26-abs(ord(a[k-1])-ord(a[k])))
k+=1
``` | 3 | |
69 | A | Young Physicist | PROGRAMMING | 1,000 | [
"implementation",
"math"
] | A. Young Physicist | 2 | 256 | A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" — thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces. | The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100), then follow *n* lines containing three integers each: the *x**i* coordinate, the *y**i* coordinate and the *z**i* coordinate of the force vector, applied to the body (<=-<=100<=≤<=*x**i*,<=*y**i*,<=*z**i*<=≤<=100). | Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not. | [
"3\n4 1 7\n-2 4 -1\n1 -5 -3\n",
"3\n3 -1 7\n-5 2 -4\n2 -1 -3\n"
] | [
"NO",
"YES"
] | none | 500 | [
{
"input": "3\n4 1 7\n-2 4 -1\n1 -5 -3",
"output": "NO"
},
{
"input": "3\n3 -1 7\n-5 2 -4\n2 -1 -3",
"output": "YES"
},
{
"input": "10\n21 32 -46\n43 -35 21\n42 2 -50\n22 40 20\n-27 -9 38\n-4 1 1\n-40 6 -31\n-13 -2 34\n-21 34 -12\n-32 -29 41",
"output": "NO"
},
{
"input": "10... | 1,675,085,852 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 60 | 0 | import sys
sys.stdin, sys.stdout = open("input.txt", "r"), open("output.txt", "w")
def main():
n = int(input())
arr = [[int(a) for a in input().split()]for _ in range(n)]
a,b,c=0,0,0
for i in range(n):
a+=arr[i][0]
b+=arr[i][1]
c+=arr[i][2]
if a==0 and b==0 and c==0 :
print("YES")
else:
print("NO")
if __name__ == "__main__":
main() | Title: Young Physicist
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" — thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces.
Input Specification:
The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100), then follow *n* lines containing three integers each: the *x**i* coordinate, the *y**i* coordinate and the *z**i* coordinate of the force vector, applied to the body (<=-<=100<=≤<=*x**i*,<=*y**i*,<=*z**i*<=≤<=100).
Output Specification:
Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not.
Demo Input:
['3\n4 1 7\n-2 4 -1\n1 -5 -3\n', '3\n3 -1 7\n-5 2 -4\n2 -1 -3\n']
Demo Output:
['NO', 'YES']
Note:
none | ```python
import sys
sys.stdin, sys.stdout = open("input.txt", "r"), open("output.txt", "w")
def main():
n = int(input())
arr = [[int(a) for a in input().split()]for _ in range(n)]
a,b,c=0,0,0
for i in range(n):
a+=arr[i][0]
b+=arr[i][1]
c+=arr[i][2]
if a==0 and b==0 and c==0 :
print("YES")
else:
print("NO")
if __name__ == "__main__":
main()
``` | -1 |
868 | A | Bark to Unlock | PROGRAMMING | 900 | [
"brute force",
"implementation",
"strings"
] | null | null | As technologies develop, manufacturers are making the process of unlocking a phone as user-friendly as possible. To unlock its new phone, Arkady's pet dog Mu-mu has to bark the password once. The phone represents a password as a string of two lowercase English letters.
Mu-mu's enemy Kashtanka wants to unlock Mu-mu's phone to steal some sensible information, but it can only bark *n* distinct words, each of which can be represented as a string of two lowercase English letters. Kashtanka wants to bark several words (not necessarily distinct) one after another to pronounce a string containing the password as a substring. Tell if it's possible to unlock the phone in this way, or not. | The first line contains two lowercase English letters — the password on the phone.
The second line contains single integer *n* (1<=≤<=*n*<=≤<=100) — the number of words Kashtanka knows.
The next *n* lines contain two lowercase English letters each, representing the words Kashtanka knows. The words are guaranteed to be distinct. | Print "YES" if Kashtanka can bark several words in a line forming a string containing the password, and "NO" otherwise.
You can print each letter in arbitrary case (upper or lower). | [
"ya\n4\nah\noy\nto\nha\n",
"hp\n2\nht\ntp\n",
"ah\n1\nha\n"
] | [
"YES\n",
"NO\n",
"YES\n"
] | In the first example the password is "ya", and Kashtanka can bark "oy" and then "ah", and then "ha" to form the string "oyahha" which contains the password. So, the answer is "YES".
In the second example Kashtanka can't produce a string containing password as a substring. Note that it can bark "ht" and then "tp" producing "http", but it doesn't contain the password "hp" as a substring.
In the third example the string "hahahaha" contains "ah" as a substring. | 250 | [
{
"input": "ya\n4\nah\noy\nto\nha",
"output": "YES"
},
{
"input": "hp\n2\nht\ntp",
"output": "NO"
},
{
"input": "ah\n1\nha",
"output": "YES"
},
{
"input": "bb\n4\nba\nab\naa\nbb",
"output": "YES"
},
{
"input": "bc\n4\nca\nba\nbb\ncc",
"output": "YES"
},
{
... | 1,680,971,357 | 2,147,483,647 | Python 3 | MEMORY_LIMIT_EXCEEDED | TESTS | 7 | 592 | 268,390,400 | from itertools import permutations
password = input()
n = int(input())
vals = []
for i in range(n):
s = input()
s *= 3
vals.append(s)
poss = list(permutations(vals))
ans = "NO"
for lst in poss:
lst = "".join(lst)
if password in lst:
ans = "YES"
break
print(ans) | Title: Bark to Unlock
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
As technologies develop, manufacturers are making the process of unlocking a phone as user-friendly as possible. To unlock its new phone, Arkady's pet dog Mu-mu has to bark the password once. The phone represents a password as a string of two lowercase English letters.
Mu-mu's enemy Kashtanka wants to unlock Mu-mu's phone to steal some sensible information, but it can only bark *n* distinct words, each of which can be represented as a string of two lowercase English letters. Kashtanka wants to bark several words (not necessarily distinct) one after another to pronounce a string containing the password as a substring. Tell if it's possible to unlock the phone in this way, or not.
Input Specification:
The first line contains two lowercase English letters — the password on the phone.
The second line contains single integer *n* (1<=≤<=*n*<=≤<=100) — the number of words Kashtanka knows.
The next *n* lines contain two lowercase English letters each, representing the words Kashtanka knows. The words are guaranteed to be distinct.
Output Specification:
Print "YES" if Kashtanka can bark several words in a line forming a string containing the password, and "NO" otherwise.
You can print each letter in arbitrary case (upper or lower).
Demo Input:
['ya\n4\nah\noy\nto\nha\n', 'hp\n2\nht\ntp\n', 'ah\n1\nha\n']
Demo Output:
['YES\n', 'NO\n', 'YES\n']
Note:
In the first example the password is "ya", and Kashtanka can bark "oy" and then "ah", and then "ha" to form the string "oyahha" which contains the password. So, the answer is "YES".
In the second example Kashtanka can't produce a string containing password as a substring. Note that it can bark "ht" and then "tp" producing "http", but it doesn't contain the password "hp" as a substring.
In the third example the string "hahahaha" contains "ah" as a substring. | ```python
from itertools import permutations
password = input()
n = int(input())
vals = []
for i in range(n):
s = input()
s *= 3
vals.append(s)
poss = list(permutations(vals))
ans = "NO"
for lst in poss:
lst = "".join(lst)
if password in lst:
ans = "YES"
break
print(ans)
``` | 0 | |
884 | B | Japanese Crosswords Strike Back | PROGRAMMING | 1,100 | [
"implementation"
] | null | null | A one-dimensional Japanese crossword can be represented as a binary string of length *x*. An encoding of this crossword is an array *a* of size *n*, where *n* is the number of segments formed completely of 1's, and *a**i* is the length of *i*-th segment. No two segments touch or intersect.
For example:
- If *x*<==<=6 and the crossword is 111011, then its encoding is an array {3,<=2}; - If *x*<==<=8 and the crossword is 01101010, then its encoding is an array {2,<=1,<=1}; - If *x*<==<=5 and the crossword is 11111, then its encoding is an array {5}; - If *x*<==<=5 and the crossword is 00000, then its encoding is an empty array.
Mishka wants to create a new one-dimensional Japanese crossword. He has already picked the length and the encoding for this crossword. And now he needs to check if there is exactly one crossword such that its length and encoding are equal to the length and encoding he picked. Help him to check it! | The first line contains two integer numbers *n* and *x* (1<=≤<=*n*<=≤<=100000, 1<=≤<=*x*<=≤<=109) — the number of elements in the encoding and the length of the crossword Mishka picked.
The second line contains *n* integer numbers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=10000) — the encoding. | Print YES if there exists exaclty one crossword with chosen length and encoding. Otherwise, print NO. | [
"2 4\n1 3\n",
"3 10\n3 3 2\n",
"2 10\n1 3\n"
] | [
"NO\n",
"YES\n",
"NO\n"
] | none | 0 | [
{
"input": "2 4\n1 3",
"output": "NO"
},
{
"input": "3 10\n3 3 2",
"output": "YES"
},
{
"input": "2 10\n1 3",
"output": "NO"
},
{
"input": "1 1\n1",
"output": "YES"
},
{
"input": "1 10\n10",
"output": "YES"
},
{
"input": "1 10000\n10000",
"output":... | 1,509,214,523 | 2,147,483,647 | Python 3 | OK | TESTS | 66 | 77 | 7,372,800 | n,x=map(int,input().split())
L=list(map(int,input().split()))
if x-sum(L)==n-1:
print('Yes')
else:
print('No')
| Title: Japanese Crosswords Strike Back
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A one-dimensional Japanese crossword can be represented as a binary string of length *x*. An encoding of this crossword is an array *a* of size *n*, where *n* is the number of segments formed completely of 1's, and *a**i* is the length of *i*-th segment. No two segments touch or intersect.
For example:
- If *x*<==<=6 and the crossword is 111011, then its encoding is an array {3,<=2}; - If *x*<==<=8 and the crossword is 01101010, then its encoding is an array {2,<=1,<=1}; - If *x*<==<=5 and the crossword is 11111, then its encoding is an array {5}; - If *x*<==<=5 and the crossword is 00000, then its encoding is an empty array.
Mishka wants to create a new one-dimensional Japanese crossword. He has already picked the length and the encoding for this crossword. And now he needs to check if there is exactly one crossword such that its length and encoding are equal to the length and encoding he picked. Help him to check it!
Input Specification:
The first line contains two integer numbers *n* and *x* (1<=≤<=*n*<=≤<=100000, 1<=≤<=*x*<=≤<=109) — the number of elements in the encoding and the length of the crossword Mishka picked.
The second line contains *n* integer numbers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=10000) — the encoding.
Output Specification:
Print YES if there exists exaclty one crossword with chosen length and encoding. Otherwise, print NO.
Demo Input:
['2 4\n1 3\n', '3 10\n3 3 2\n', '2 10\n1 3\n']
Demo Output:
['NO\n', 'YES\n', 'NO\n']
Note:
none | ```python
n,x=map(int,input().split())
L=list(map(int,input().split()))
if x-sum(L)==n-1:
print('Yes')
else:
print('No')
``` | 3 | |
69 | A | Young Physicist | PROGRAMMING | 1,000 | [
"implementation",
"math"
] | A. Young Physicist | 2 | 256 | A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" — thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces. | The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100), then follow *n* lines containing three integers each: the *x**i* coordinate, the *y**i* coordinate and the *z**i* coordinate of the force vector, applied to the body (<=-<=100<=≤<=*x**i*,<=*y**i*,<=*z**i*<=≤<=100). | Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not. | [
"3\n4 1 7\n-2 4 -1\n1 -5 -3\n",
"3\n3 -1 7\n-5 2 -4\n2 -1 -3\n"
] | [
"NO",
"YES"
] | none | 500 | [
{
"input": "3\n4 1 7\n-2 4 -1\n1 -5 -3",
"output": "NO"
},
{
"input": "3\n3 -1 7\n-5 2 -4\n2 -1 -3",
"output": "YES"
},
{
"input": "10\n21 32 -46\n43 -35 21\n42 2 -50\n22 40 20\n-27 -9 38\n-4 1 1\n-40 6 -31\n-13 -2 34\n-21 34 -12\n-32 -29 41",
"output": "NO"
},
{
"input": "10... | 1,556,626,125 | 2,147,483,647 | Python 3 | OK | TESTS | 81 | 218 | 0 | n=int(input())
x=y=z=0
for i in range(0,n):
inp = list(map(int,input().split()))
x+=inp[0]
y+=inp[1]
z+=inp[2]
if x**2+y**2+z**2==0:
print("YES")
else:
print("NO") | Title: Young Physicist
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" — thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces.
Input Specification:
The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100), then follow *n* lines containing three integers each: the *x**i* coordinate, the *y**i* coordinate and the *z**i* coordinate of the force vector, applied to the body (<=-<=100<=≤<=*x**i*,<=*y**i*,<=*z**i*<=≤<=100).
Output Specification:
Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not.
Demo Input:
['3\n4 1 7\n-2 4 -1\n1 -5 -3\n', '3\n3 -1 7\n-5 2 -4\n2 -1 -3\n']
Demo Output:
['NO', 'YES']
Note:
none | ```python
n=int(input())
x=y=z=0
for i in range(0,n):
inp = list(map(int,input().split()))
x+=inp[0]
y+=inp[1]
z+=inp[2]
if x**2+y**2+z**2==0:
print("YES")
else:
print("NO")
``` | 3.9455 |
909 | C | Python Indentation | PROGRAMMING | 1,800 | [
"dp"
] | null | null | In Python, code blocks don't have explicit begin/end or curly braces to mark beginning and end of the block. Instead, code blocks are defined by indentation.
We will consider an extremely simplified subset of Python with only two types of statements.
Simple statements are written in a single line, one per line. An example of a simple statement is assignment.
For statements are compound statements: they contain one or several other statements. For statement consists of a header written in a separate line which starts with "for" prefix, and loop body. Loop body is a block of statements indented one level further than the header of the loop. Loop body can contain both types of statements. Loop body can't be empty.
You are given a sequence of statements without indentation. Find the number of ways in which the statements can be indented to form a valid Python program. | The first line contains a single integer *N* (1<=≤<=*N*<=≤<=5000) — the number of commands in the program. *N* lines of the program follow, each line describing a single command. Each command is either "f" (denoting "for statement") or "s" ("simple statement"). It is guaranteed that the last line is a simple statement. | Output one line containing an integer - the number of ways the given sequence of statements can be indented modulo 109<=+<=7. | [
"4\ns\nf\nf\ns\n",
"4\nf\ns\nf\ns\n"
] | [
"1\n",
"2\n"
] | In the first test case, there is only one way to indent the program: the second for statement must be part of the body of the first one.
In the second test case, there are two ways to indent the program: the second for statement can either be part of the first one's body or a separate statement following the first one.
or | 1,500 | [
{
"input": "4\ns\nf\nf\ns",
"output": "1"
},
{
"input": "4\nf\ns\nf\ns",
"output": "2"
},
{
"input": "156\nf\ns\nf\ns\nf\ns\ns\ns\ns\nf\ns\ns\nf\nf\ns\nf\nf\nf\nf\ns\ns\ns\nf\ns\ns\nf\nf\nf\nf\nf\nf\ns\ns\ns\ns\nf\ns\nf\ns\nf\ns\nf\nf\nf\nf\ns\ns\nf\nf\ns\ns\ns\ns\nf\ns\nf\ns\nf\ns\nf\ns... | 1,514,431,177 | 2,147,483,647 | Python 3 | COMPILATION_ERROR | TESTS | 0 | 0 | 0 | #include <bits/stdc++.h>
using namespace std;
const int MAX_N = 5005;
const int MOD = 1000000007;
int dp[MAX_N][MAX_N];
int n;
char a[MAX_N];
int main () {
scanf("%d", &n);
for (int i = 1; i <= n; i++) {
char s[2];
scanf("%s", s);
a[i] = s[0];
}
dp[1][1] = 1;
for (int i = 1; i < n; i++) {
for (int j = 1; j <= i; j++) {
if (dp[i][j] > 0) {
if (a[i] == 'f') {
(dp[i + 1][j + 1] += dp[i][j]) %= MOD;
}
else {
(dp[i + 1][j] += dp[i][j]) %= MOD;
if (i > 1) {
(dp[i + 1][j + 1] += dp[i][j]) %= MOD;
}
}
}
}
}
int res = 0;
for (int j = 1; j <= n; j++) {
(res += dp[n][j]) %= MOD;
}
printf("%d", res);
return 0;
} | Title: Python Indentation
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
In Python, code blocks don't have explicit begin/end or curly braces to mark beginning and end of the block. Instead, code blocks are defined by indentation.
We will consider an extremely simplified subset of Python with only two types of statements.
Simple statements are written in a single line, one per line. An example of a simple statement is assignment.
For statements are compound statements: they contain one or several other statements. For statement consists of a header written in a separate line which starts with "for" prefix, and loop body. Loop body is a block of statements indented one level further than the header of the loop. Loop body can contain both types of statements. Loop body can't be empty.
You are given a sequence of statements without indentation. Find the number of ways in which the statements can be indented to form a valid Python program.
Input Specification:
The first line contains a single integer *N* (1<=≤<=*N*<=≤<=5000) — the number of commands in the program. *N* lines of the program follow, each line describing a single command. Each command is either "f" (denoting "for statement") or "s" ("simple statement"). It is guaranteed that the last line is a simple statement.
Output Specification:
Output one line containing an integer - the number of ways the given sequence of statements can be indented modulo 109<=+<=7.
Demo Input:
['4\ns\nf\nf\ns\n', '4\nf\ns\nf\ns\n']
Demo Output:
['1\n', '2\n']
Note:
In the first test case, there is only one way to indent the program: the second for statement must be part of the body of the first one.
In the second test case, there are two ways to indent the program: the second for statement can either be part of the first one's body or a separate statement following the first one.
or | ```python
#include <bits/stdc++.h>
using namespace std;
const int MAX_N = 5005;
const int MOD = 1000000007;
int dp[MAX_N][MAX_N];
int n;
char a[MAX_N];
int main () {
scanf("%d", &n);
for (int i = 1; i <= n; i++) {
char s[2];
scanf("%s", s);
a[i] = s[0];
}
dp[1][1] = 1;
for (int i = 1; i < n; i++) {
for (int j = 1; j <= i; j++) {
if (dp[i][j] > 0) {
if (a[i] == 'f') {
(dp[i + 1][j + 1] += dp[i][j]) %= MOD;
}
else {
(dp[i + 1][j] += dp[i][j]) %= MOD;
if (i > 1) {
(dp[i + 1][j + 1] += dp[i][j]) %= MOD;
}
}
}
}
}
int res = 0;
for (int j = 1; j <= n; j++) {
(res += dp[n][j]) %= MOD;
}
printf("%d", res);
return 0;
}
``` | -1 | |
266 | A | Stones on the Table | PROGRAMMING | 800 | [
"implementation"
] | null | null | There are *n* stones on the table in a row, each of them can be red, green or blue. Count the minimum number of stones to take from the table so that any two neighboring stones had different colors. Stones in a row are considered neighboring if there are no other stones between them. | The first line contains integer *n* (1<=≤<=*n*<=≤<=50) — the number of stones on the table.
The next line contains string *s*, which represents the colors of the stones. We'll consider the stones in the row numbered from 1 to *n* from left to right. Then the *i*-th character *s* equals "R", if the *i*-th stone is red, "G", if it's green and "B", if it's blue. | Print a single integer — the answer to the problem. | [
"3\nRRG\n",
"5\nRRRRR\n",
"4\nBRBG\n"
] | [
"1\n",
"4\n",
"0\n"
] | none | 500 | [
{
"input": "3\nRRG",
"output": "1"
},
{
"input": "5\nRRRRR",
"output": "4"
},
{
"input": "4\nBRBG",
"output": "0"
},
{
"input": "1\nB",
"output": "0"
},
{
"input": "2\nBG",
"output": "0"
},
{
"input": "3\nBGB",
"output": "0"
},
{
"input": "... | 1,694,251,565 | 2,147,483,647 | Python 3 | OK | TESTS | 34 | 92 | 0 | n = int(input())
s = list(input())
if len(s) >= 1 and len(s) <= 50:
min_stones = 0
for i in range(1, len(s)):
if s[i] == s[i-1]: # إذا Ùا٠ÙدÙÙا Øج��ÙÙ Ù
تتاÙÙÙ٠بÙÙس اÙÙÙÙ
min_stones += 1 # ÙزÙد عدد اÙØ£Øجار اÙÙ
Ø·ÙÙبة بÙاØد
print(min_stones)
| Title: Stones on the Table
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are *n* stones on the table in a row, each of them can be red, green or blue. Count the minimum number of stones to take from the table so that any two neighboring stones had different colors. Stones in a row are considered neighboring if there are no other stones between them.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=50) — the number of stones on the table.
The next line contains string *s*, which represents the colors of the stones. We'll consider the stones in the row numbered from 1 to *n* from left to right. Then the *i*-th character *s* equals "R", if the *i*-th stone is red, "G", if it's green and "B", if it's blue.
Output Specification:
Print a single integer — the answer to the problem.
Demo Input:
['3\nRRG\n', '5\nRRRRR\n', '4\nBRBG\n']
Demo Output:
['1\n', '4\n', '0\n']
Note:
none | ```python
n = int(input())
s = list(input())
if len(s) >= 1 and len(s) <= 50:
min_stones = 0
for i in range(1, len(s)):
if s[i] == s[i-1]: # إذا Ùا٠ÙدÙÙا Øج��ÙÙ Ù
تتاÙÙÙ٠بÙÙس اÙÙÙÙ
min_stones += 1 # ÙزÙد عدد اÙØ£Øجار اÙÙ
Ø·ÙÙبة بÙاØد
print(min_stones)
``` | 3 | |
427 | C | Checkposts | PROGRAMMING | 1,700 | [
"dfs and similar",
"graphs",
"two pointers"
] | null | null | Your city has *n* junctions. There are *m* one-way roads between the junctions. As a mayor of the city, you have to ensure the security of all the junctions.
To ensure the security, you have to build some police checkposts. Checkposts can only be built in a junction. A checkpost at junction *i* can protect junction *j* if either *i*<==<=*j* or the police patrol car can go to *j* from *i* and then come back to *i*.
Building checkposts costs some money. As some areas of the city are more expensive than others, building checkpost at some junctions might cost more money than other junctions.
You have to determine the minimum possible money needed to ensure the security of all the junctions. Also you have to find the number of ways to ensure the security in minimum price and in addition in minimum number of checkposts. Two ways are different if any of the junctions contains a checkpost in one of them and do not contain in the other. | In the first line, you will be given an integer *n*, number of junctions (1<=≤<=*n*<=≤<=105). In the next line, *n* space-separated integers will be given. The *i**th* integer is the cost of building checkpost at the *i**th* junction (costs will be non-negative and will not exceed 109).
The next line will contain an integer *m* (0<=≤<=*m*<=≤<=3·105). And each of the next *m* lines contains two integers *u**i* and *v**i* (1<=≤<=*u**i*,<=*v**i*<=≤<=*n*; *u*<=≠<=*v*). A pair *u**i*,<=*v**i* means, that there is a one-way road which goes from *u**i* to *v**i*. There will not be more than one road between two nodes in the same direction. | Print two integers separated by spaces. The first one is the minimum possible money needed to ensure the security of all the junctions. And the second one is the number of ways you can ensure the security modulo 1000000007 (109<=+<=7). | [
"3\n1 2 3\n3\n1 2\n2 3\n3 2\n",
"5\n2 8 0 6 0\n6\n1 4\n1 3\n2 4\n3 4\n4 5\n5 1\n",
"10\n1 3 2 2 1 3 1 4 10 10\n12\n1 2\n2 3\n3 1\n3 4\n4 5\n5 6\n5 7\n6 4\n7 3\n8 9\n9 10\n10 9\n",
"2\n7 91\n2\n1 2\n2 1\n"
] | [
"3 1\n",
"8 2\n",
"15 6\n",
"7 1\n"
] | none | 1,500 | [
{
"input": "3\n1 2 3\n3\n1 2\n2 3\n3 2",
"output": "3 1"
},
{
"input": "5\n2 8 0 6 0\n6\n1 4\n1 3\n2 4\n3 4\n4 5\n5 1",
"output": "8 2"
},
{
"input": "10\n1 3 2 2 1 3 1 4 10 10\n12\n1 2\n2 3\n3 1\n3 4\n4 5\n5 6\n5 7\n6 4\n7 3\n8 9\n9 10\n10 9",
"output": "15 6"
},
{
"input": ... | 1,680,606,633 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 70 | 810 | 22,630,400 | # import sys
from collections import defaultdict
# sys.setrecursionlimit(1000000)
def add_edge(a, b, c):
edges[a] = c
next_edge[a] = adjacency_list[b]
adjacency_list[b] = a
def depth_first_search(node):
global time, stack_top, component_count
low[node] = visited[node] = time + 1
time += 1
is_visited[node] = True
stack_top += 1
stack[stack_top] = node
stack_position = stack_top
current_edge = adjacency_list[node]
while current_edge != 0:
if is_visited[edges[current_edge]]:
low[node] = min(low[node], visited[edges[current_edge]])
else:
if not visited[edges[current_edge]]:
depth_first_search(edges[current_edge])
low[node] = min(low[node], low[edges[current_edge]])
current_edge = next_edge[current_edge]
if low[node] == visited[node]:
component_count += 1
for i in range(stack_position, stack_top + 1):
components[stack[i]] = component_count
is_visited[stack[i]] = False
stack_top = stack_position - 1
n = int(input().strip())
values = list(map(int, input().strip().split()))
m = int(input().strip())
ans1 = ans2 = 0
stack_top = edge_count = component_count = time = 0
visited = [0] * (n + 5)
low = [0] * (n + 5)
stack = [0] * (n + 5)
edges = [0] * (3 * 10**5 + 5)
next_edge = [0] * (3 * 10**5 + 5)
adjacency_list = [0] * (n + 5)
components = [0] * (n + 5)
is_visited = [False] * (n + 5)
min_value = [1000000007] * (n + 5)
component_count_list = [0] * (n + 5)
for _ in range(m):
A, B = map(int, input().strip().split())
edge_count += 1
add_edge(edge_count, A, B)
for i in range(1, n + 1):
if not visited[i]:
depth_first_search(i)
for i in range(1, n + 1):
if min_value[components[i]] > values[i - 1]:
min_value[components[i]] = values[i - 1]
component_count_list[components[i]] = 1
elif min_value[components[i]] == values[i - 1]:
component_count_list[components[i]] += 1
MOD = 1000000007
ans2 = 1
for i in range(1, component_count + 1):
ans1 += min_value[i]
ans2 = (ans2 * component_count_list[i]) % MOD
print(ans1, ans2)
| Title: Checkposts
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Your city has *n* junctions. There are *m* one-way roads between the junctions. As a mayor of the city, you have to ensure the security of all the junctions.
To ensure the security, you have to build some police checkposts. Checkposts can only be built in a junction. A checkpost at junction *i* can protect junction *j* if either *i*<==<=*j* or the police patrol car can go to *j* from *i* and then come back to *i*.
Building checkposts costs some money. As some areas of the city are more expensive than others, building checkpost at some junctions might cost more money than other junctions.
You have to determine the minimum possible money needed to ensure the security of all the junctions. Also you have to find the number of ways to ensure the security in minimum price and in addition in minimum number of checkposts. Two ways are different if any of the junctions contains a checkpost in one of them and do not contain in the other.
Input Specification:
In the first line, you will be given an integer *n*, number of junctions (1<=≤<=*n*<=≤<=105). In the next line, *n* space-separated integers will be given. The *i**th* integer is the cost of building checkpost at the *i**th* junction (costs will be non-negative and will not exceed 109).
The next line will contain an integer *m* (0<=≤<=*m*<=≤<=3·105). And each of the next *m* lines contains two integers *u**i* and *v**i* (1<=≤<=*u**i*,<=*v**i*<=≤<=*n*; *u*<=≠<=*v*). A pair *u**i*,<=*v**i* means, that there is a one-way road which goes from *u**i* to *v**i*. There will not be more than one road between two nodes in the same direction.
Output Specification:
Print two integers separated by spaces. The first one is the minimum possible money needed to ensure the security of all the junctions. And the second one is the number of ways you can ensure the security modulo 1000000007 (109<=+<=7).
Demo Input:
['3\n1 2 3\n3\n1 2\n2 3\n3 2\n', '5\n2 8 0 6 0\n6\n1 4\n1 3\n2 4\n3 4\n4 5\n5 1\n', '10\n1 3 2 2 1 3 1 4 10 10\n12\n1 2\n2 3\n3 1\n3 4\n4 5\n5 6\n5 7\n6 4\n7 3\n8 9\n9 10\n10 9\n', '2\n7 91\n2\n1 2\n2 1\n']
Demo Output:
['3 1\n', '8 2\n', '15 6\n', '7 1\n']
Note:
none | ```python
# import sys
from collections import defaultdict
# sys.setrecursionlimit(1000000)
def add_edge(a, b, c):
edges[a] = c
next_edge[a] = adjacency_list[b]
adjacency_list[b] = a
def depth_first_search(node):
global time, stack_top, component_count
low[node] = visited[node] = time + 1
time += 1
is_visited[node] = True
stack_top += 1
stack[stack_top] = node
stack_position = stack_top
current_edge = adjacency_list[node]
while current_edge != 0:
if is_visited[edges[current_edge]]:
low[node] = min(low[node], visited[edges[current_edge]])
else:
if not visited[edges[current_edge]]:
depth_first_search(edges[current_edge])
low[node] = min(low[node], low[edges[current_edge]])
current_edge = next_edge[current_edge]
if low[node] == visited[node]:
component_count += 1
for i in range(stack_position, stack_top + 1):
components[stack[i]] = component_count
is_visited[stack[i]] = False
stack_top = stack_position - 1
n = int(input().strip())
values = list(map(int, input().strip().split()))
m = int(input().strip())
ans1 = ans2 = 0
stack_top = edge_count = component_count = time = 0
visited = [0] * (n + 5)
low = [0] * (n + 5)
stack = [0] * (n + 5)
edges = [0] * (3 * 10**5 + 5)
next_edge = [0] * (3 * 10**5 + 5)
adjacency_list = [0] * (n + 5)
components = [0] * (n + 5)
is_visited = [False] * (n + 5)
min_value = [1000000007] * (n + 5)
component_count_list = [0] * (n + 5)
for _ in range(m):
A, B = map(int, input().strip().split())
edge_count += 1
add_edge(edge_count, A, B)
for i in range(1, n + 1):
if not visited[i]:
depth_first_search(i)
for i in range(1, n + 1):
if min_value[components[i]] > values[i - 1]:
min_value[components[i]] = values[i - 1]
component_count_list[components[i]] = 1
elif min_value[components[i]] == values[i - 1]:
component_count_list[components[i]] += 1
MOD = 1000000007
ans2 = 1
for i in range(1, component_count + 1):
ans1 += min_value[i]
ans2 = (ans2 * component_count_list[i]) % MOD
print(ans1, ans2)
``` | -1 | |
56 | A | Bar | PROGRAMMING | 1,000 | [
"implementation"
] | A. Bar | 2 | 256 | According to Berland laws it is only allowed to sell alcohol to people not younger than 18 years. Vasya's job is to monitor the law's enforcement. Tonight he entered a bar and saw *n* people sitting there. For every one of them Vasya happened to determine either the age or the drink the person is having. Vasya can check any person, i.e. learn his age and the drink he is having at the same time. What minimal number of people should Vasya check additionally to make sure that there are no clients under 18 having alcohol drinks?
The list of all alcohol drinks in Berland is: ABSINTH, BEER, BRANDY, CHAMPAGNE, GIN, RUM, SAKE, TEQUILA, VODKA, WHISKEY, WINE | The first line contains an integer *n* (1<=≤<=*n*<=≤<=100) which is the number of the bar's clients. Then follow *n* lines, each describing one visitor. A line either contains his age (an integer from 0 to 1000) or his drink (a string of capital Latin letters from 1 to 100 in length). It is guaranteed that the input data does not contain spaces and other unnecessary separators.
Only the drinks from the list given above should be considered alcohol. | Print a single number which is the number of people Vasya should check to guarantee the law enforcement. | [
"5\n18\nVODKA\nCOKE\n19\n17\n"
] | [
"2\n"
] | In the sample test the second and fifth clients should be checked. | 500 | [
{
"input": "5\n18\nVODKA\nCOKE\n19\n17",
"output": "2"
},
{
"input": "2\n2\nGIN",
"output": "2"
},
{
"input": "3\nWHISKEY\n3\nGIN",
"output": "3"
},
{
"input": "4\n813\nIORBQITQXMPTFAEMEQDQIKFGKGOTNKTOSZCBRPXJLUKVLVHJYNRUJXK\nRUM\nRHVRWGODYWWTYZFLFYKCVUFFRTQDINKNWPKFHZBFWBHWI... | 1,576,350,687 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 1 | 186 | 0 | a = ['ABSINTH', 'BEER', 'BRANDY', 'CHAMPAGNE', 'GIN', 'RUM', 'SAKE', 'TEQUILA', 'VODKA', 'WHISKEY', 'WINE']
a1=0
a2=0
n=int(input())
for i in range(n) :
N = input()
if N.isdigit():
if int(N)>=18:
a1=a1+1
if N==a[0] or N==a[1] or N==a[2] or N==a[3] or N==a[4] or N==a[5] or N==a[6] or N==a[7] or N==a[8] or N==a[9] or N==a[10]:
a2=a2+1
print(a1)
| Title: Bar
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
According to Berland laws it is only allowed to sell alcohol to people not younger than 18 years. Vasya's job is to monitor the law's enforcement. Tonight he entered a bar and saw *n* people sitting there. For every one of them Vasya happened to determine either the age or the drink the person is having. Vasya can check any person, i.e. learn his age and the drink he is having at the same time. What minimal number of people should Vasya check additionally to make sure that there are no clients under 18 having alcohol drinks?
The list of all alcohol drinks in Berland is: ABSINTH, BEER, BRANDY, CHAMPAGNE, GIN, RUM, SAKE, TEQUILA, VODKA, WHISKEY, WINE
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100) which is the number of the bar's clients. Then follow *n* lines, each describing one visitor. A line either contains his age (an integer from 0 to 1000) or his drink (a string of capital Latin letters from 1 to 100 in length). It is guaranteed that the input data does not contain spaces and other unnecessary separators.
Only the drinks from the list given above should be considered alcohol.
Output Specification:
Print a single number which is the number of people Vasya should check to guarantee the law enforcement.
Demo Input:
['5\n18\nVODKA\nCOKE\n19\n17\n']
Demo Output:
['2\n']
Note:
In the sample test the second and fifth clients should be checked. | ```python
a = ['ABSINTH', 'BEER', 'BRANDY', 'CHAMPAGNE', 'GIN', 'RUM', 'SAKE', 'TEQUILA', 'VODKA', 'WHISKEY', 'WINE']
a1=0
a2=0
n=int(input())
for i in range(n) :
N = input()
if N.isdigit():
if int(N)>=18:
a1=a1+1
if N==a[0] or N==a[1] or N==a[2] or N==a[3] or N==a[4] or N==a[5] or N==a[6] or N==a[7] or N==a[8] or N==a[9] or N==a[10]:
a2=a2+1
print(a1)
``` | 0 |
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