message stringlengths 2 44.5k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 276 109k | cluster float64 23 23 | __index_level_0__ int64 552 217k |
|---|---|---|---|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
Lee is going to fashionably decorate his house for a party, using some regular convex polygons...
Lee thinks a regular n-sided (convex) polygon is beautiful if and only if he can rotate it in such a way that at least one of its edges is parallel to the OX-axis and at least one of its edges is parallel to the OY-axis at the same time.
Recall that a regular n-sided polygon is a convex polygon with n vertices such that all the edges and angles are equal.
Now he is shopping: the market has t regular polygons. For each of them print YES if it is beautiful and NO otherwise.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of polygons in the market.
Each of the next t lines contains a single integer n_i (3 ≤ n_i ≤ 10^9): it means that the i-th polygon is a regular n_i-sided polygon.
Output
For each polygon, print YES if it's beautiful or NO otherwise (case insensitive).
Example
Input
4
3
4
12
1000000000
Output
NO
YES
YES
YES
Note
In the example, there are 4 polygons in the market. It's easy to see that an equilateral triangle (a regular 3-sided polygon) is not beautiful, a square (a regular 4-sided polygon) is beautiful and a regular 12-sided polygon (is shown below) is beautiful as well.
<image> | instruction | 0 | 10,942 | 23 | 21,884 |
Tags: geometry, math
Correct Solution:
```
text = int(input())
for i in range(text):
a = int(input())
if a % 4==0:
print('YES')
else:
print('NO')
``` | output | 1 | 10,942 | 23 | 21,885 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Lee is going to fashionably decorate his house for a party, using some regular convex polygons...
Lee thinks a regular n-sided (convex) polygon is beautiful if and only if he can rotate it in such a way that at least one of its edges is parallel to the OX-axis and at least one of its edges is parallel to the OY-axis at the same time.
Recall that a regular n-sided polygon is a convex polygon with n vertices such that all the edges and angles are equal.
Now he is shopping: the market has t regular polygons. For each of them print YES if it is beautiful and NO otherwise.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of polygons in the market.
Each of the next t lines contains a single integer n_i (3 ≤ n_i ≤ 10^9): it means that the i-th polygon is a regular n_i-sided polygon.
Output
For each polygon, print YES if it's beautiful or NO otherwise (case insensitive).
Example
Input
4
3
4
12
1000000000
Output
NO
YES
YES
YES
Note
In the example, there are 4 polygons in the market. It's easy to see that an equilateral triangle (a regular 3-sided polygon) is not beautiful, a square (a regular 4-sided polygon) is beautiful and a regular 12-sided polygon (is shown below) is beautiful as well.
<image> | instruction | 0 | 10,943 | 23 | 21,886 |
Tags: geometry, math
Correct Solution:
```
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
"""
:license: GPLv3 --- Copyright (C) 2020 Olivier Pirson
:author: Olivier Pirson --- http://www.opimedia.be/
"""
import sys
#
# Main
######
def main():
"""Run main work."""
nb = int(sys.stdin.readline())
for _ in range(nb):
n = int(sys.stdin.readline())
print('YES' if n % 4 == 0
else 'NO')
if __name__ == '__main__':
main()
``` | output | 1 | 10,943 | 23 | 21,887 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Lee is going to fashionably decorate his house for a party, using some regular convex polygons...
Lee thinks a regular n-sided (convex) polygon is beautiful if and only if he can rotate it in such a way that at least one of its edges is parallel to the OX-axis and at least one of its edges is parallel to the OY-axis at the same time.
Recall that a regular n-sided polygon is a convex polygon with n vertices such that all the edges and angles are equal.
Now he is shopping: the market has t regular polygons. For each of them print YES if it is beautiful and NO otherwise.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of polygons in the market.
Each of the next t lines contains a single integer n_i (3 ≤ n_i ≤ 10^9): it means that the i-th polygon is a regular n_i-sided polygon.
Output
For each polygon, print YES if it's beautiful or NO otherwise (case insensitive).
Example
Input
4
3
4
12
1000000000
Output
NO
YES
YES
YES
Note
In the example, there are 4 polygons in the market. It's easy to see that an equilateral triangle (a regular 3-sided polygon) is not beautiful, a square (a regular 4-sided polygon) is beautiful and a regular 12-sided polygon (is shown below) is beautiful as well.
<image> | instruction | 0 | 10,944 | 23 | 21,888 |
Tags: geometry, math
Correct Solution:
```
from collections import defaultdict as dd
import math
import sys
import string
input=sys.stdin.readline
def nn():
return int(input())
def li():
return list(input())
def mi():
return map(int, input().split())
def lm():
return list(map(int, input().split()))
q=nn()
for _ in range(q):
n = nn()
if n%4==0:
print("YES")
else:
print("NO")
``` | output | 1 | 10,944 | 23 | 21,889 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Lee is going to fashionably decorate his house for a party, using some regular convex polygons...
Lee thinks a regular n-sided (convex) polygon is beautiful if and only if he can rotate it in such a way that at least one of its edges is parallel to the OX-axis and at least one of its edges is parallel to the OY-axis at the same time.
Recall that a regular n-sided polygon is a convex polygon with n vertices such that all the edges and angles are equal.
Now he is shopping: the market has t regular polygons. For each of them print YES if it is beautiful and NO otherwise.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of polygons in the market.
Each of the next t lines contains a single integer n_i (3 ≤ n_i ≤ 10^9): it means that the i-th polygon is a regular n_i-sided polygon.
Output
For each polygon, print YES if it's beautiful or NO otherwise (case insensitive).
Example
Input
4
3
4
12
1000000000
Output
NO
YES
YES
YES
Note
In the example, there are 4 polygons in the market. It's easy to see that an equilateral triangle (a regular 3-sided polygon) is not beautiful, a square (a regular 4-sided polygon) is beautiful and a regular 12-sided polygon (is shown below) is beautiful as well.
<image> | instruction | 0 | 10,945 | 23 | 21,890 |
Tags: geometry, math
Correct Solution:
```
t = int(input())
for _ in range(t):
n = int(input())
if(not n%4):print("Yes")
else:print("No")
``` | output | 1 | 10,945 | 23 | 21,891 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Lee is going to fashionably decorate his house for a party, using some regular convex polygons...
Lee thinks a regular n-sided (convex) polygon is beautiful if and only if he can rotate it in such a way that at least one of its edges is parallel to the OX-axis and at least one of its edges is parallel to the OY-axis at the same time.
Recall that a regular n-sided polygon is a convex polygon with n vertices such that all the edges and angles are equal.
Now he is shopping: the market has t regular polygons. For each of them print YES if it is beautiful and NO otherwise.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of polygons in the market.
Each of the next t lines contains a single integer n_i (3 ≤ n_i ≤ 10^9): it means that the i-th polygon is a regular n_i-sided polygon.
Output
For each polygon, print YES if it's beautiful or NO otherwise (case insensitive).
Example
Input
4
3
4
12
1000000000
Output
NO
YES
YES
YES
Note
In the example, there are 4 polygons in the market. It's easy to see that an equilateral triangle (a regular 3-sided polygon) is not beautiful, a square (a regular 4-sided polygon) is beautiful and a regular 12-sided polygon (is shown below) is beautiful as well.
<image> | instruction | 0 | 10,946 | 23 | 21,892 |
Tags: geometry, math
Correct Solution:
```
T=int(input())
for _ in range(T):
N=int(input())
if N%4==0:
print("YES")
else:
print("NO")
``` | output | 1 | 10,946 | 23 | 21,893 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Lee is going to fashionably decorate his house for a party, using some regular convex polygons...
Lee thinks a regular n-sided (convex) polygon is beautiful if and only if he can rotate it in such a way that at least one of its edges is parallel to the OX-axis and at least one of its edges is parallel to the OY-axis at the same time.
Recall that a regular n-sided polygon is a convex polygon with n vertices such that all the edges and angles are equal.
Now he is shopping: the market has t regular polygons. For each of them print YES if it is beautiful and NO otherwise.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of polygons in the market.
Each of the next t lines contains a single integer n_i (3 ≤ n_i ≤ 10^9): it means that the i-th polygon is a regular n_i-sided polygon.
Output
For each polygon, print YES if it's beautiful or NO otherwise (case insensitive).
Example
Input
4
3
4
12
1000000000
Output
NO
YES
YES
YES
Note
In the example, there are 4 polygons in the market. It's easy to see that an equilateral triangle (a regular 3-sided polygon) is not beautiful, a square (a regular 4-sided polygon) is beautiful and a regular 12-sided polygon (is shown below) is beautiful as well.
<image> | instruction | 0 | 10,947 | 23 | 21,894 |
Tags: geometry, math
Correct Solution:
```
#!/usr/bin/env python
import os
import sys
from io import BytesIO, IOBase
import threading
from math import floor
from bisect import bisect_right
from collections import Counter
from math import gcd
mod=998244353
def main():
for _ in range(int(input())):
n=int(input())
# k=180*(n-2)//n
if n%4==0:
print('YES')
else:
print('NO')
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# endregion
if __name__ == "__main__":
main()
``` | output | 1 | 10,947 | 23 | 21,895 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Lee is going to fashionably decorate his house for a party, using some regular convex polygons...
Lee thinks a regular n-sided (convex) polygon is beautiful if and only if he can rotate it in such a way that at least one of its edges is parallel to the OX-axis and at least one of its edges is parallel to the OY-axis at the same time.
Recall that a regular n-sided polygon is a convex polygon with n vertices such that all the edges and angles are equal.
Now he is shopping: the market has t regular polygons. For each of them print YES if it is beautiful and NO otherwise.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of polygons in the market.
Each of the next t lines contains a single integer n_i (3 ≤ n_i ≤ 10^9): it means that the i-th polygon is a regular n_i-sided polygon.
Output
For each polygon, print YES if it's beautiful or NO otherwise (case insensitive).
Example
Input
4
3
4
12
1000000000
Output
NO
YES
YES
YES
Note
In the example, there are 4 polygons in the market. It's easy to see that an equilateral triangle (a regular 3-sided polygon) is not beautiful, a square (a regular 4-sided polygon) is beautiful and a regular 12-sided polygon (is shown below) is beautiful as well.
<image> | instruction | 0 | 10,948 | 23 | 21,896 |
Tags: geometry, math
Correct Solution:
```
import sys
input = sys.stdin.readline
t = int(input())
for _ in range(t):
n = int(input())
if n % 4 == 0:
print("YES")
else:
print("NO")
``` | output | 1 | 10,948 | 23 | 21,897 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Lee is going to fashionably decorate his house for a party, using some regular convex polygons...
Lee thinks a regular n-sided (convex) polygon is beautiful if and only if he can rotate it in such a way that at least one of its edges is parallel to the OX-axis and at least one of its edges is parallel to the OY-axis at the same time.
Recall that a regular n-sided polygon is a convex polygon with n vertices such that all the edges and angles are equal.
Now he is shopping: the market has t regular polygons. For each of them print YES if it is beautiful and NO otherwise.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of polygons in the market.
Each of the next t lines contains a single integer n_i (3 ≤ n_i ≤ 10^9): it means that the i-th polygon is a regular n_i-sided polygon.
Output
For each polygon, print YES if it's beautiful or NO otherwise (case insensitive).
Example
Input
4
3
4
12
1000000000
Output
NO
YES
YES
YES
Note
In the example, there are 4 polygons in the market. It's easy to see that an equilateral triangle (a regular 3-sided polygon) is not beautiful, a square (a regular 4-sided polygon) is beautiful and a regular 12-sided polygon (is shown below) is beautiful as well.
<image> | instruction | 0 | 10,949 | 23 | 21,898 |
Tags: geometry, math
Correct Solution:
```
tc = int(input())
for i in range(tc):
N = int(input())
if N%4 == 0:
print("yEs")
else:
print("nO")
if N == 69420:
print("bing bong")
``` | output | 1 | 10,949 | 23 | 21,899 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Lee is going to fashionably decorate his house for a party, using some regular convex polygons...
Lee thinks a regular n-sided (convex) polygon is beautiful if and only if he can rotate it in such a way that at least one of its edges is parallel to the OX-axis and at least one of its edges is parallel to the OY-axis at the same time.
Recall that a regular n-sided polygon is a convex polygon with n vertices such that all the edges and angles are equal.
Now he is shopping: the market has t regular polygons. For each of them print YES if it is beautiful and NO otherwise.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of polygons in the market.
Each of the next t lines contains a single integer n_i (3 ≤ n_i ≤ 10^9): it means that the i-th polygon is a regular n_i-sided polygon.
Output
For each polygon, print YES if it's beautiful or NO otherwise (case insensitive).
Example
Input
4
3
4
12
1000000000
Output
NO
YES
YES
YES
Note
In the example, there are 4 polygons in the market. It's easy to see that an equilateral triangle (a regular 3-sided polygon) is not beautiful, a square (a regular 4-sided polygon) is beautiful and a regular 12-sided polygon (is shown below) is beautiful as well.
<image>
Submitted Solution:
```
for _ in range(int(input())):
p = int(input())
if p %4 != 0 :
print('NO')
else:
print('YES')
``` | instruction | 0 | 10,950 | 23 | 21,900 |
Yes | output | 1 | 10,950 | 23 | 21,901 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Lee is going to fashionably decorate his house for a party, using some regular convex polygons...
Lee thinks a regular n-sided (convex) polygon is beautiful if and only if he can rotate it in such a way that at least one of its edges is parallel to the OX-axis and at least one of its edges is parallel to the OY-axis at the same time.
Recall that a regular n-sided polygon is a convex polygon with n vertices such that all the edges and angles are equal.
Now he is shopping: the market has t regular polygons. For each of them print YES if it is beautiful and NO otherwise.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of polygons in the market.
Each of the next t lines contains a single integer n_i (3 ≤ n_i ≤ 10^9): it means that the i-th polygon is a regular n_i-sided polygon.
Output
For each polygon, print YES if it's beautiful or NO otherwise (case insensitive).
Example
Input
4
3
4
12
1000000000
Output
NO
YES
YES
YES
Note
In the example, there are 4 polygons in the market. It's easy to see that an equilateral triangle (a regular 3-sided polygon) is not beautiful, a square (a regular 4-sided polygon) is beautiful and a regular 12-sided polygon (is shown below) is beautiful as well.
<image>
Submitted Solution:
```
t = int(input())
# print('t', t)
for _ in range(t):
n = int(input())
# print('n', n)
if n%4==0:
print('YES')
else:
print('NO')
``` | instruction | 0 | 10,951 | 23 | 21,902 |
Yes | output | 1 | 10,951 | 23 | 21,903 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Lee is going to fashionably decorate his house for a party, using some regular convex polygons...
Lee thinks a regular n-sided (convex) polygon is beautiful if and only if he can rotate it in such a way that at least one of its edges is parallel to the OX-axis and at least one of its edges is parallel to the OY-axis at the same time.
Recall that a regular n-sided polygon is a convex polygon with n vertices such that all the edges and angles are equal.
Now he is shopping: the market has t regular polygons. For each of them print YES if it is beautiful and NO otherwise.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of polygons in the market.
Each of the next t lines contains a single integer n_i (3 ≤ n_i ≤ 10^9): it means that the i-th polygon is a regular n_i-sided polygon.
Output
For each polygon, print YES if it's beautiful or NO otherwise (case insensitive).
Example
Input
4
3
4
12
1000000000
Output
NO
YES
YES
YES
Note
In the example, there are 4 polygons in the market. It's easy to see that an equilateral triangle (a regular 3-sided polygon) is not beautiful, a square (a regular 4-sided polygon) is beautiful and a regular 12-sided polygon (is shown below) is beautiful as well.
<image>
Submitted Solution:
```
import sys, math
if __name__ == "__main__":
t = int(input())
for i in range(t):
ni = int(input())
if ni % 4 == 0:
print("YES")
else:
print("NO")
``` | instruction | 0 | 10,952 | 23 | 21,904 |
Yes | output | 1 | 10,952 | 23 | 21,905 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Lee is going to fashionably decorate his house for a party, using some regular convex polygons...
Lee thinks a regular n-sided (convex) polygon is beautiful if and only if he can rotate it in such a way that at least one of its edges is parallel to the OX-axis and at least one of its edges is parallel to the OY-axis at the same time.
Recall that a regular n-sided polygon is a convex polygon with n vertices such that all the edges and angles are equal.
Now he is shopping: the market has t regular polygons. For each of them print YES if it is beautiful and NO otherwise.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of polygons in the market.
Each of the next t lines contains a single integer n_i (3 ≤ n_i ≤ 10^9): it means that the i-th polygon is a regular n_i-sided polygon.
Output
For each polygon, print YES if it's beautiful or NO otherwise (case insensitive).
Example
Input
4
3
4
12
1000000000
Output
NO
YES
YES
YES
Note
In the example, there are 4 polygons in the market. It's easy to see that an equilateral triangle (a regular 3-sided polygon) is not beautiful, a square (a regular 4-sided polygon) is beautiful and a regular 12-sided polygon (is shown below) is beautiful as well.
<image>
Submitted Solution:
```
for j in range(int(input())):
n=int(input())
if n%4!=0:
print("NO")
else:
print("YES")
``` | instruction | 0 | 10,953 | 23 | 21,906 |
Yes | output | 1 | 10,953 | 23 | 21,907 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Lee is going to fashionably decorate his house for a party, using some regular convex polygons...
Lee thinks a regular n-sided (convex) polygon is beautiful if and only if he can rotate it in such a way that at least one of its edges is parallel to the OX-axis and at least one of its edges is parallel to the OY-axis at the same time.
Recall that a regular n-sided polygon is a convex polygon with n vertices such that all the edges and angles are equal.
Now he is shopping: the market has t regular polygons. For each of them print YES if it is beautiful and NO otherwise.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of polygons in the market.
Each of the next t lines contains a single integer n_i (3 ≤ n_i ≤ 10^9): it means that the i-th polygon is a regular n_i-sided polygon.
Output
For each polygon, print YES if it's beautiful or NO otherwise (case insensitive).
Example
Input
4
3
4
12
1000000000
Output
NO
YES
YES
YES
Note
In the example, there are 4 polygons in the market. It's easy to see that an equilateral triangle (a regular 3-sided polygon) is not beautiful, a square (a regular 4-sided polygon) is beautiful and a regular 12-sided polygon (is shown below) is beautiful as well.
<image>
Submitted Solution:
```
import math
import datetime
import collections
import statistics
import itertools
def is_prime(num):
for i in range(2, int(math.sqrt(num)) + 1):
if num % i == 0:
return False
return True
def input_list():
ll = list(map(int, input().split(" ")))
return ll
tc = int(input())
for _ in range(tc):
n = int(input())
if n % 2 == 0:
print("YES")
else:
print("NO")
``` | instruction | 0 | 10,954 | 23 | 21,908 |
No | output | 1 | 10,954 | 23 | 21,909 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Lee is going to fashionably decorate his house for a party, using some regular convex polygons...
Lee thinks a regular n-sided (convex) polygon is beautiful if and only if he can rotate it in such a way that at least one of its edges is parallel to the OX-axis and at least one of its edges is parallel to the OY-axis at the same time.
Recall that a regular n-sided polygon is a convex polygon with n vertices such that all the edges and angles are equal.
Now he is shopping: the market has t regular polygons. For each of them print YES if it is beautiful and NO otherwise.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of polygons in the market.
Each of the next t lines contains a single integer n_i (3 ≤ n_i ≤ 10^9): it means that the i-th polygon is a regular n_i-sided polygon.
Output
For each polygon, print YES if it's beautiful or NO otherwise (case insensitive).
Example
Input
4
3
4
12
1000000000
Output
NO
YES
YES
YES
Note
In the example, there are 4 polygons in the market. It's easy to see that an equilateral triangle (a regular 3-sided polygon) is not beautiful, a square (a regular 4-sided polygon) is beautiful and a regular 12-sided polygon (is shown below) is beautiful as well.
<image>
Submitted Solution:
```
t=int(input())
for i in range(t):
n=int(input())
if(n==3):
print("NO")
else:
print("YES")
``` | instruction | 0 | 10,955 | 23 | 21,910 |
No | output | 1 | 10,955 | 23 | 21,911 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Lee is going to fashionably decorate his house for a party, using some regular convex polygons...
Lee thinks a regular n-sided (convex) polygon is beautiful if and only if he can rotate it in such a way that at least one of its edges is parallel to the OX-axis and at least one of its edges is parallel to the OY-axis at the same time.
Recall that a regular n-sided polygon is a convex polygon with n vertices such that all the edges and angles are equal.
Now he is shopping: the market has t regular polygons. For each of them print YES if it is beautiful and NO otherwise.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of polygons in the market.
Each of the next t lines contains a single integer n_i (3 ≤ n_i ≤ 10^9): it means that the i-th polygon is a regular n_i-sided polygon.
Output
For each polygon, print YES if it's beautiful or NO otherwise (case insensitive).
Example
Input
4
3
4
12
1000000000
Output
NO
YES
YES
YES
Note
In the example, there are 4 polygons in the market. It's easy to see that an equilateral triangle (a regular 3-sided polygon) is not beautiful, a square (a regular 4-sided polygon) is beautiful and a regular 12-sided polygon (is shown below) is beautiful as well.
<image>
Submitted Solution:
```
def f():
n=int(input(""))
if n>3:
print("YES")
else:
print("NO")
n=int(input(""))
for i in range(n):
f()
``` | instruction | 0 | 10,956 | 23 | 21,912 |
No | output | 1 | 10,956 | 23 | 21,913 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Lee is going to fashionably decorate his house for a party, using some regular convex polygons...
Lee thinks a regular n-sided (convex) polygon is beautiful if and only if he can rotate it in such a way that at least one of its edges is parallel to the OX-axis and at least one of its edges is parallel to the OY-axis at the same time.
Recall that a regular n-sided polygon is a convex polygon with n vertices such that all the edges and angles are equal.
Now he is shopping: the market has t regular polygons. For each of them print YES if it is beautiful and NO otherwise.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of polygons in the market.
Each of the next t lines contains a single integer n_i (3 ≤ n_i ≤ 10^9): it means that the i-th polygon is a regular n_i-sided polygon.
Output
For each polygon, print YES if it's beautiful or NO otherwise (case insensitive).
Example
Input
4
3
4
12
1000000000
Output
NO
YES
YES
YES
Note
In the example, there are 4 polygons in the market. It's easy to see that an equilateral triangle (a regular 3-sided polygon) is not beautiful, a square (a regular 4-sided polygon) is beautiful and a regular 12-sided polygon (is shown below) is beautiful as well.
<image>
Submitted Solution:
```
import io
import os
from collections import Counter, defaultdict, deque
def solve(N, ):
if N % 2 == 0:
return "YES"
return "NO"
if __name__ == "__main__":
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
T = int(input())
for t in range(T):
N, = [int(x) for x in input().split()]
ans = solve(N, )
print(ans)
``` | instruction | 0 | 10,957 | 23 | 21,914 |
No | output | 1 | 10,957 | 23 | 21,915 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Imagine you have an infinite 2D plane with Cartesian coordinate system. Some of the integral points are blocked, and others are not. Two integral points A and B on the plane are 4-connected if and only if:
* the Euclidean distance between A and B is one unit and neither A nor B is blocked;
* or there is some integral point C, such that A is 4-connected with C, and C is 4-connected with B.
Let's assume that the plane doesn't contain blocked points. Consider all the integral points of the plane whose Euclidean distance from the origin is no more than n, we'll name these points special. Chubby Yang wants to get the following property: no special point is 4-connected to some non-special point. To get the property she can pick some integral points of the plane and make them blocked. What is the minimum number of points she needs to pick?
Input
The first line contains an integer n (0 ≤ n ≤ 4·107).
Output
Print a single integer — the minimum number of points that should be blocked.
Examples
Input
1
Output
4
Input
2
Output
8
Input
3
Output
16 | instruction | 0 | 11,108 | 23 | 22,216 |
Tags: math
Correct Solution:
```
import math
n = int(input())
if n == 0:
print(1)
else:
print(4 * int(n * math.sqrt(2)))
``` | output | 1 | 11,108 | 23 | 22,217 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Imagine you have an infinite 2D plane with Cartesian coordinate system. Some of the integral points are blocked, and others are not. Two integral points A and B on the plane are 4-connected if and only if:
* the Euclidean distance between A and B is one unit and neither A nor B is blocked;
* or there is some integral point C, such that A is 4-connected with C, and C is 4-connected with B.
Let's assume that the plane doesn't contain blocked points. Consider all the integral points of the plane whose Euclidean distance from the origin is no more than n, we'll name these points special. Chubby Yang wants to get the following property: no special point is 4-connected to some non-special point. To get the property she can pick some integral points of the plane and make them blocked. What is the minimum number of points she needs to pick?
Input
The first line contains an integer n (0 ≤ n ≤ 4·107).
Output
Print a single integer — the minimum number of points that should be blocked.
Examples
Input
1
Output
4
Input
2
Output
8
Input
3
Output
16 | instruction | 0 | 11,109 | 23 | 22,218 |
Tags: math
Correct Solution:
```
from math import sqrt
n = int(input())
if n == 0:
print(1)
else:
print(4 * int(sqrt(2) * n))
``` | output | 1 | 11,109 | 23 | 22,219 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Imagine you have an infinite 2D plane with Cartesian coordinate system. Some of the integral points are blocked, and others are not. Two integral points A and B on the plane are 4-connected if and only if:
* the Euclidean distance between A and B is one unit and neither A nor B is blocked;
* or there is some integral point C, such that A is 4-connected with C, and C is 4-connected with B.
Let's assume that the plane doesn't contain blocked points. Consider all the integral points of the plane whose Euclidean distance from the origin is no more than n, we'll name these points special. Chubby Yang wants to get the following property: no special point is 4-connected to some non-special point. To get the property she can pick some integral points of the plane and make them blocked. What is the minimum number of points she needs to pick?
Input
The first line contains an integer n (0 ≤ n ≤ 4·107).
Output
Print a single integer — the minimum number of points that should be blocked.
Examples
Input
1
Output
4
Input
2
Output
8
Input
3
Output
16 | instruction | 0 | 11,110 | 23 | 22,220 |
Tags: math
Correct Solution:
```
print(max(1, 4 * int(int(input()) * 2 ** 0.5)))
``` | output | 1 | 11,110 | 23 | 22,221 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Imagine you have an infinite 2D plane with Cartesian coordinate system. Some of the integral points are blocked, and others are not. Two integral points A and B on the plane are 4-connected if and only if:
* the Euclidean distance between A and B is one unit and neither A nor B is blocked;
* or there is some integral point C, such that A is 4-connected with C, and C is 4-connected with B.
Let's assume that the plane doesn't contain blocked points. Consider all the integral points of the plane whose Euclidean distance from the origin is no more than n, we'll name these points special. Chubby Yang wants to get the following property: no special point is 4-connected to some non-special point. To get the property she can pick some integral points of the plane and make them blocked. What is the minimum number of points she needs to pick?
Input
The first line contains an integer n (0 ≤ n ≤ 4·107).
Output
Print a single integer — the minimum number of points that should be blocked.
Examples
Input
1
Output
4
Input
2
Output
8
Input
3
Output
16 | instruction | 0 | 11,111 | 23 | 22,222 |
Tags: math
Correct Solution:
```
guFat = int(input(""))
if guFat == 0:
print(1)
else:
print(4*int(guFat*2**(1/2)))
``` | output | 1 | 11,111 | 23 | 22,223 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Imagine you have an infinite 2D plane with Cartesian coordinate system. Some of the integral points are blocked, and others are not. Two integral points A and B on the plane are 4-connected if and only if:
* the Euclidean distance between A and B is one unit and neither A nor B is blocked;
* or there is some integral point C, such that A is 4-connected with C, and C is 4-connected with B.
Let's assume that the plane doesn't contain blocked points. Consider all the integral points of the plane whose Euclidean distance from the origin is no more than n, we'll name these points special. Chubby Yang wants to get the following property: no special point is 4-connected to some non-special point. To get the property she can pick some integral points of the plane and make them blocked. What is the minimum number of points she needs to pick?
Input
The first line contains an integer n (0 ≤ n ≤ 4·107).
Output
Print a single integer — the minimum number of points that should be blocked.
Examples
Input
1
Output
4
Input
2
Output
8
Input
3
Output
16 | instruction | 0 | 11,112 | 23 | 22,224 |
Tags: math
Correct Solution:
```
from math import sqrt
n = int(input())
if n == 0:
print(1)
else:
print(4 * int(n * sqrt(2)))
``` | output | 1 | 11,112 | 23 | 22,225 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Imagine you have an infinite 2D plane with Cartesian coordinate system. Some of the integral points are blocked, and others are not. Two integral points A and B on the plane are 4-connected if and only if:
* the Euclidean distance between A and B is one unit and neither A nor B is blocked;
* or there is some integral point C, such that A is 4-connected with C, and C is 4-connected with B.
Let's assume that the plane doesn't contain blocked points. Consider all the integral points of the plane whose Euclidean distance from the origin is no more than n, we'll name these points special. Chubby Yang wants to get the following property: no special point is 4-connected to some non-special point. To get the property she can pick some integral points of the plane and make them blocked. What is the minimum number of points she needs to pick?
Input
The first line contains an integer n (0 ≤ n ≤ 4·107).
Output
Print a single integer — the minimum number of points that should be blocked.
Examples
Input
1
Output
4
Input
2
Output
8
Input
3
Output
16 | instruction | 0 | 11,113 | 23 | 22,226 |
Tags: math
Correct Solution:
```
n = int(input())
x = 0
y = n
count = 0
while y>x:
check = (n+x+1)*(n-x-1)
if y*y <= check:
x+=1
elif (y-1)*(y-1) <= check:
x+=1
y-=1
else:
y-=1
count+=1
# print(x,y)
if n == 0:
print(1)
else:
if x == y:
print(count*8)
else:
print(count*8-4)
``` | output | 1 | 11,113 | 23 | 22,227 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Imagine you have an infinite 2D plane with Cartesian coordinate system. Some of the integral points are blocked, and others are not. Two integral points A and B on the plane are 4-connected if and only if:
* the Euclidean distance between A and B is one unit and neither A nor B is blocked;
* or there is some integral point C, such that A is 4-connected with C, and C is 4-connected with B.
Let's assume that the plane doesn't contain blocked points. Consider all the integral points of the plane whose Euclidean distance from the origin is no more than n, we'll name these points special. Chubby Yang wants to get the following property: no special point is 4-connected to some non-special point. To get the property she can pick some integral points of the plane and make them blocked. What is the minimum number of points she needs to pick?
Input
The first line contains an integer n (0 ≤ n ≤ 4·107).
Output
Print a single integer — the minimum number of points that should be blocked.
Examples
Input
1
Output
4
Input
2
Output
8
Input
3
Output
16 | instruction | 0 | 11,114 | 23 | 22,228 |
Tags: math
Correct Solution:
```
import math
n=int(input())
if(n==0):print(1)
elif(n==1):print(4)
elif(n==2):print(8)
elif(n==3):print(16)
else:
k=4+8*(int(math.sqrt(n*n/2)))
p=int(math.sqrt(n*n/2))
if(p*p+(p+1)*(p+1)>n*n):k-=4
print(k)
``` | output | 1 | 11,114 | 23 | 22,229 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Imagine you have an infinite 2D plane with Cartesian coordinate system. Some of the integral points are blocked, and others are not. Two integral points A and B on the plane are 4-connected if and only if:
* the Euclidean distance between A and B is one unit and neither A nor B is blocked;
* or there is some integral point C, such that A is 4-connected with C, and C is 4-connected with B.
Let's assume that the plane doesn't contain blocked points. Consider all the integral points of the plane whose Euclidean distance from the origin is no more than n, we'll name these points special. Chubby Yang wants to get the following property: no special point is 4-connected to some non-special point. To get the property she can pick some integral points of the plane and make them blocked. What is the minimum number of points she needs to pick?
Input
The first line contains an integer n (0 ≤ n ≤ 4·107).
Output
Print a single integer — the minimum number of points that should be blocked.
Examples
Input
1
Output
4
Input
2
Output
8
Input
3
Output
16 | instruction | 0 | 11,115 | 23 | 22,230 |
Tags: math
Correct Solution:
```
from math import sqrt, floor
def calc(n):
if n == 0:
return 1
# y = n
# x = 1
# c = 0
# while x - y < 0:
# if x ** 2 + y ** 2 <= n ** 2:
# c += 1
# x += 1
# continue
# if x ** 2 + y ** 2 > n ** 2:
# y -= 1
x = floor(sqrt(n**2/2))
y = floor(sqrt(n**2-x**2))
#print(x, y)
if x == y:
return x*8
else:
return x * 8 +4
n = int(input())
print(calc(n))
``` | output | 1 | 11,115 | 23 | 22,231 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Imagine you have an infinite 2D plane with Cartesian coordinate system. Some of the integral points are blocked, and others are not. Two integral points A and B on the plane are 4-connected if and only if:
* the Euclidean distance between A and B is one unit and neither A nor B is blocked;
* or there is some integral point C, such that A is 4-connected with C, and C is 4-connected with B.
Let's assume that the plane doesn't contain blocked points. Consider all the integral points of the plane whose Euclidean distance from the origin is no more than n, we'll name these points special. Chubby Yang wants to get the following property: no special point is 4-connected to some non-special point. To get the property she can pick some integral points of the plane and make them blocked. What is the minimum number of points she needs to pick?
Input
The first line contains an integer n (0 ≤ n ≤ 4·107).
Output
Print a single integer — the minimum number of points that should be blocked.
Examples
Input
1
Output
4
Input
2
Output
8
Input
3
Output
16
Submitted Solution:
```
print(max(1, 4 * int(int(input()) * 2 ** 0.5)))
# Made By Mostafa_Khaled
``` | instruction | 0 | 11,116 | 23 | 22,232 |
Yes | output | 1 | 11,116 | 23 | 22,233 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Imagine you have an infinite 2D plane with Cartesian coordinate system. Some of the integral points are blocked, and others are not. Two integral points A and B on the plane are 4-connected if and only if:
* the Euclidean distance between A and B is one unit and neither A nor B is blocked;
* or there is some integral point C, such that A is 4-connected with C, and C is 4-connected with B.
Let's assume that the plane doesn't contain blocked points. Consider all the integral points of the plane whose Euclidean distance from the origin is no more than n, we'll name these points special. Chubby Yang wants to get the following property: no special point is 4-connected to some non-special point. To get the property she can pick some integral points of the plane and make them blocked. What is the minimum number of points she needs to pick?
Input
The first line contains an integer n (0 ≤ n ≤ 4·107).
Output
Print a single integer — the minimum number of points that should be blocked.
Examples
Input
1
Output
4
Input
2
Output
8
Input
3
Output
16
Submitted Solution:
```
import math
n = int(input())
if n == 0:
print('1')
else:
print(4*int(math.sqrt(2.0)*n))
``` | instruction | 0 | 11,117 | 23 | 22,234 |
Yes | output | 1 | 11,117 | 23 | 22,235 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Imagine you have an infinite 2D plane with Cartesian coordinate system. Some of the integral points are blocked, and others are not. Two integral points A and B on the plane are 4-connected if and only if:
* the Euclidean distance between A and B is one unit and neither A nor B is blocked;
* or there is some integral point C, such that A is 4-connected with C, and C is 4-connected with B.
Let's assume that the plane doesn't contain blocked points. Consider all the integral points of the plane whose Euclidean distance from the origin is no more than n, we'll name these points special. Chubby Yang wants to get the following property: no special point is 4-connected to some non-special point. To get the property she can pick some integral points of the plane and make them blocked. What is the minimum number of points she needs to pick?
Input
The first line contains an integer n (0 ≤ n ≤ 4·107).
Output
Print a single integer — the minimum number of points that should be blocked.
Examples
Input
1
Output
4
Input
2
Output
8
Input
3
Output
16
Submitted Solution:
```
import math
a = int(input())
n = 0
if a >0:
b = int(math.sqrt((a**2)/2))
if b**2 + (b+1)**2 <= a**2:
n = b * 8 + 4
else:
n = b * 8
print(n)
else:
print('1')
``` | instruction | 0 | 11,118 | 23 | 22,236 |
Yes | output | 1 | 11,118 | 23 | 22,237 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Imagine you have an infinite 2D plane with Cartesian coordinate system. Some of the integral points are blocked, and others are not. Two integral points A and B on the plane are 4-connected if and only if:
* the Euclidean distance between A and B is one unit and neither A nor B is blocked;
* or there is some integral point C, such that A is 4-connected with C, and C is 4-connected with B.
Let's assume that the plane doesn't contain blocked points. Consider all the integral points of the plane whose Euclidean distance from the origin is no more than n, we'll name these points special. Chubby Yang wants to get the following property: no special point is 4-connected to some non-special point. To get the property she can pick some integral points of the plane and make them blocked. What is the minimum number of points she needs to pick?
Input
The first line contains an integer n (0 ≤ n ≤ 4·107).
Output
Print a single integer — the minimum number of points that should be blocked.
Examples
Input
1
Output
4
Input
2
Output
8
Input
3
Output
16
Submitted Solution:
```
# Made By Mostafa_Khaled
bot = True
print(max(1, 4 * int(int(input()) * 2 ** 0.5)))
# Made By Mostafa_Khaled
``` | instruction | 0 | 11,119 | 23 | 22,238 |
Yes | output | 1 | 11,119 | 23 | 22,239 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Imagine you have an infinite 2D plane with Cartesian coordinate system. Some of the integral points are blocked, and others are not. Two integral points A and B on the plane are 4-connected if and only if:
* the Euclidean distance between A and B is one unit and neither A nor B is blocked;
* or there is some integral point C, such that A is 4-connected with C, and C is 4-connected with B.
Let's assume that the plane doesn't contain blocked points. Consider all the integral points of the plane whose Euclidean distance from the origin is no more than n, we'll name these points special. Chubby Yang wants to get the following property: no special point is 4-connected to some non-special point. To get the property she can pick some integral points of the plane and make them blocked. What is the minimum number of points she needs to pick?
Input
The first line contains an integer n (0 ≤ n ≤ 4·107).
Output
Print a single integer — the minimum number of points that should be blocked.
Examples
Input
1
Output
4
Input
2
Output
8
Input
3
Output
16
Submitted Solution:
```
N = int(input())
print(pow(2,N+1))
``` | instruction | 0 | 11,120 | 23 | 22,240 |
No | output | 1 | 11,120 | 23 | 22,241 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Imagine you have an infinite 2D plane with Cartesian coordinate system. Some of the integral points are blocked, and others are not. Two integral points A and B on the plane are 4-connected if and only if:
* the Euclidean distance between A and B is one unit and neither A nor B is blocked;
* or there is some integral point C, such that A is 4-connected with C, and C is 4-connected with B.
Let's assume that the plane doesn't contain blocked points. Consider all the integral points of the plane whose Euclidean distance from the origin is no more than n, we'll name these points special. Chubby Yang wants to get the following property: no special point is 4-connected to some non-special point. To get the property she can pick some integral points of the plane and make them blocked. What is the minimum number of points she needs to pick?
Input
The first line contains an integer n (0 ≤ n ≤ 4·107).
Output
Print a single integer — the minimum number of points that should be blocked.
Examples
Input
1
Output
4
Input
2
Output
8
Input
3
Output
16
Submitted Solution:
```
import math
import fileinput
for line in fileinput.input():
n = int(line)
def countPoints(n):
if n==0:
return 1
elif n==1:
return 4
else:
X = math.floor(math.sqrt(2)/2*n)
if (X**2 + (X+1)**2 == n**2):
return 8*X +4
else:
return 8*X
print(countPoints(n))
``` | instruction | 0 | 11,121 | 23 | 22,242 |
No | output | 1 | 11,121 | 23 | 22,243 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Imagine you have an infinite 2D plane with Cartesian coordinate system. Some of the integral points are blocked, and others are not. Two integral points A and B on the plane are 4-connected if and only if:
* the Euclidean distance between A and B is one unit and neither A nor B is blocked;
* or there is some integral point C, such that A is 4-connected with C, and C is 4-connected with B.
Let's assume that the plane doesn't contain blocked points. Consider all the integral points of the plane whose Euclidean distance from the origin is no more than n, we'll name these points special. Chubby Yang wants to get the following property: no special point is 4-connected to some non-special point. To get the property she can pick some integral points of the plane and make them blocked. What is the minimum number of points she needs to pick?
Input
The first line contains an integer n (0 ≤ n ≤ 4·107).
Output
Print a single integer — the minimum number of points that should be blocked.
Examples
Input
1
Output
4
Input
2
Output
8
Input
3
Output
16
Submitted Solution:
```
print(pow(2,int(input())+1))
``` | instruction | 0 | 11,122 | 23 | 22,244 |
No | output | 1 | 11,122 | 23 | 22,245 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Imagine you have an infinite 2D plane with Cartesian coordinate system. Some of the integral points are blocked, and others are not. Two integral points A and B on the plane are 4-connected if and only if:
* the Euclidean distance between A and B is one unit and neither A nor B is blocked;
* or there is some integral point C, such that A is 4-connected with C, and C is 4-connected with B.
Let's assume that the plane doesn't contain blocked points. Consider all the integral points of the plane whose Euclidean distance from the origin is no more than n, we'll name these points special. Chubby Yang wants to get the following property: no special point is 4-connected to some non-special point. To get the property she can pick some integral points of the plane and make them blocked. What is the minimum number of points she needs to pick?
Input
The first line contains an integer n (0 ≤ n ≤ 4·107).
Output
Print a single integer — the minimum number of points that should be blocked.
Examples
Input
1
Output
4
Input
2
Output
8
Input
3
Output
16
Submitted Solution:
```
x=input("")
x=int(x)
if x==4:
print(20)
else:
x+=1
print(2**x)
``` | instruction | 0 | 11,123 | 23 | 22,246 |
No | output | 1 | 11,123 | 23 | 22,247 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In Medieval times existed the tradition of burning witches at steaks together with their pets, black cats. By the end of the 15-th century the population of black cats ceased to exist. The difficulty of the situation led to creating the EIC - the Emergency Inquisitory Commission.
The resolution #666 says that a white cat is considered black when and only when the perimeter of its black spots exceeds the acceptable norm. But what does the acceptable norm equal to? Every inquisitor will choose it himself depending on the situation. And your task is to find the perimeter of black spots on the cat's fur.
The very same resolution says that the cat's fur is a white square with the length of 105. During the measurement of spots it is customary to put the lower left corner of the fur into the origin of axes (0;0) and the upper right one — to the point with coordinates (105;105). The cats' spots are nondegenerate triangles. The spots can intersect and overlap with each other, but it is guaranteed that each pair of the triangular spots' sides have no more than one common point.
We'll regard the perimeter in this problem as the total length of the boarders where a cat's fur changes color.
Input
The first input line contains a single integer n (0 ≤ n ≤ 100). It is the number of spots on the cat's fur. The i-th of the last n lines contains 6 integers: x1i, y1i, x2i, y2i, x3i, y3i. They are the coordinates of the i-th triangular spot (0 < xji, yji < 105).
Output
Print a single number, the answer to the problem, perimeter of the union of triangles. Your answer should differ from the correct one in no more than 10 - 6.
Examples
Input
1
1 1 2 1 1 2
Output
3.4142135624
Input
3
3 3 10 3 3 10
1 1 9 4 5 6
2 2 11 7 6 11
Output
37.7044021497
Submitted Solution:
```
#not completed yet
a=int(input())
if a==1:
print("3.4142135624")
else:
print("37.7044021497")
``` | instruction | 0 | 11,222 | 23 | 22,444 |
No | output | 1 | 11,222 | 23 | 22,445 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Young Teodor enjoys drawing. His favourite hobby is drawing segments with integer borders inside his huge [1;m] segment. One day Teodor noticed that picture he just drawn has one interesting feature: there doesn't exist an integer point, that belongs each of segments in the picture. Having discovered this fact, Teodor decided to share it with Sasha.
Sasha knows that Teodor likes to show off so he never trusts him. Teodor wants to prove that he can be trusted sometimes, so he decided to convince Sasha that there is no such integer point in his picture, which belongs to each segment. However Teodor is lazy person and neither wills to tell Sasha all coordinates of segments' ends nor wills to tell him their amount, so he suggested Sasha to ask him series of questions 'Given the integer point xi, how many segments in Fedya's picture contain that point?', promising to tell correct answers for this questions.
Both boys are very busy studying and don't have much time, so they ask you to find out how many questions can Sasha ask Teodor, that having only answers on his questions, Sasha can't be sure that Teodor isn't lying to him. Note that Sasha doesn't know amount of segments in Teodor's picture. Sure, Sasha is smart person and never asks about same point twice.
Input
First line of input contains two integer numbers: n and m (1 ≤ n, m ≤ 100 000) — amount of segments of Teodor's picture and maximal coordinate of point that Sasha can ask about.
ith of next n lines contains two integer numbers li and ri (1 ≤ li ≤ ri ≤ m) — left and right ends of ith segment in the picture. Note that that left and right ends of segment can be the same point.
It is guaranteed that there is no integer point, that belongs to all segments.
Output
Single line of output should contain one integer number k – size of largest set (xi, cnt(xi)) where all xi are different, 1 ≤ xi ≤ m, and cnt(xi) is amount of segments, containing point with coordinate xi, such that one can't be sure that there doesn't exist point, belonging to all of segments in initial picture, if he knows only this set(and doesn't know n).
Examples
Input
2 4
1 2
3 4
Output
4
Input
4 6
1 3
2 3
4 6
5 6
Output
5
Note
First example shows situation where Sasha can never be sure that Teodor isn't lying to him, because even if one knows cnt(xi) for each point in segment [1;4], he can't distinguish this case from situation Teodor has drawn whole [1;4] segment.
In second example Sasha can ask about 5 points e.g. 1, 2, 3, 5, 6, still not being sure if Teodor haven't lied to him. But once he knows information about all points in [1;6] segment, Sasha can be sure that Teodor haven't lied to him. | instruction | 0 | 11,365 | 23 | 22,730 |
Tags: data structures, dp
Correct Solution:
```
# ---------------------------iye ha aam zindegi---------------------------------------------
import math
import random
import heapq, bisect
import sys
from collections import deque, defaultdict
from fractions import Fraction
import sys
import threading
from collections import defaultdict
#threading.stack_size(10**8)
mod = 10 ** 9 + 7
mod1 = 998244353
# ------------------------------warmup----------------------------
import os
import sys
from io import BytesIO, IOBase
#sys.setrecursionlimit(300000)
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# -------------------game starts now----------------------------------------------------import math
class TreeNode:
def __init__(self, k, v):
self.key = k
self.value = v
self.left = None
self.right = None
self.parent = None
self.height = 1
self.num_left = 1
self.num_total = 1
class AvlTree:
def __init__(self):
self._tree = None
def add(self, k, v):
if not self._tree:
self._tree = TreeNode(k, v)
return
node = self._add(k, v)
if node:
self._rebalance(node)
def _add(self, k, v):
node = self._tree
while node:
if k < node.key:
if node.left:
node = node.left
else:
node.left = TreeNode(k, v)
node.left.parent = node
return node.left
elif node.key < k:
if node.right:
node = node.right
else:
node.right = TreeNode(k, v)
node.right.parent = node
return node.right
else:
node.value = v
return
@staticmethod
def get_height(x):
return x.height if x else 0
@staticmethod
def get_num_total(x):
return x.num_total if x else 0
def _rebalance(self, node):
n = node
while n:
lh = self.get_height(n.left)
rh = self.get_height(n.right)
n.height = max(lh, rh) + 1
balance_factor = lh - rh
n.num_total = 1 + self.get_num_total(n.left) + self.get_num_total(n.right)
n.num_left = 1 + self.get_num_total(n.left)
if balance_factor > 1:
if self.get_height(n.left.left) < self.get_height(n.left.right):
self._rotate_left(n.left)
self._rotate_right(n)
elif balance_factor < -1:
if self.get_height(n.right.right) < self.get_height(n.right.left):
self._rotate_right(n.right)
self._rotate_left(n)
else:
n = n.parent
def _remove_one(self, node):
"""
Side effect!!! Changes node. Node should have exactly one child
"""
replacement = node.left or node.right
if node.parent:
if AvlTree._is_left(node):
node.parent.left = replacement
else:
node.parent.right = replacement
replacement.parent = node.parent
node.parent = None
else:
self._tree = replacement
replacement.parent = None
node.left = None
node.right = None
node.parent = None
self._rebalance(replacement)
def _remove_leaf(self, node):
if node.parent:
if AvlTree._is_left(node):
node.parent.left = None
else:
node.parent.right = None
self._rebalance(node.parent)
else:
self._tree = None
node.parent = None
node.left = None
node.right = None
def remove(self, k):
node = self._get_node(k)
if not node:
return
if AvlTree._is_leaf(node):
self._remove_leaf(node)
return
if node.left and node.right:
nxt = AvlTree._get_next(node)
node.key = nxt.key
node.value = nxt.value
if self._is_leaf(nxt):
self._remove_leaf(nxt)
else:
self._remove_one(nxt)
self._rebalance(node)
else:
self._remove_one(node)
def get(self, k):
node = self._get_node(k)
return node.value if node else -1
def _get_node(self, k):
if not self._tree:
return None
node = self._tree
while node:
if k < node.key:
node = node.left
elif node.key < k:
node = node.right
else:
return node
return None
def get_at(self, pos):
x = pos + 1
node = self._tree
while node:
if x < node.num_left:
node = node.left
elif node.num_left < x:
x -= node.num_left
node = node.right
else:
return (node.key, node.value)
raise IndexError("Out of ranges")
@staticmethod
def _is_left(node):
return node.parent.left and node.parent.left == node
@staticmethod
def _is_leaf(node):
return node.left is None and node.right is None
def _rotate_right(self, node):
if not node.parent:
self._tree = node.left
node.left.parent = None
elif AvlTree._is_left(node):
node.parent.left = node.left
node.left.parent = node.parent
else:
node.parent.right = node.left
node.left.parent = node.parent
bk = node.left.right
node.left.right = node
node.parent = node.left
node.left = bk
if bk:
bk.parent = node
node.height = max(self.get_height(node.left), self.get_height(node.right)) + 1
node.num_total = 1 + self.get_num_total(node.left) + self.get_num_total(node.right)
node.num_left = 1 + self.get_num_total(node.left)
def _rotate_left(self, node):
if not node.parent:
self._tree = node.right
node.right.parent = None
elif AvlTree._is_left(node):
node.parent.left = node.right
node.right.parent = node.parent
else:
node.parent.right = node.right
node.right.parent = node.parent
bk = node.right.left
node.right.left = node
node.parent = node.right
node.right = bk
if bk:
bk.parent = node
node.height = max(self.get_height(node.left), self.get_height(node.right)) + 1
node.num_total = 1 + self.get_num_total(node.left) + self.get_num_total(node.right)
node.num_left = 1 + self.get_num_total(node.left)
@staticmethod
def _get_next(node):
if not node.right:
return node.parent
n = node.right
while n.left:
n = n.left
return n
# -----------------------------------------------binary seacrh tree---------------------------------------
class SegmentTree1:
def __init__(self, data, default=2**51, func=lambda a, b: a & b):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
# -------------------game starts now----------------------------------------------------import math
class SegmentTree:
def __init__(self, data, default=0, func=lambda a, b: max(a , b)):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
# -------------------------------iye ha chutiya zindegi-------------------------------------
class Factorial:
def __init__(self, MOD):
self.MOD = MOD
self.factorials = [1, 1]
self.invModulos = [0, 1]
self.invFactorial_ = [1, 1]
def calc(self, n):
if n <= -1:
print("Invalid argument to calculate n!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.factorials):
return self.factorials[n]
nextArr = [0] * (n + 1 - len(self.factorials))
initialI = len(self.factorials)
prev = self.factorials[-1]
m = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = prev * i % m
self.factorials += nextArr
return self.factorials[n]
def inv(self, n):
if n <= -1:
print("Invalid argument to calculate n^(-1)")
print("n must be non-negative value. But the argument was " + str(n))
exit()
p = self.MOD
pi = n % p
if pi < len(self.invModulos):
return self.invModulos[pi]
nextArr = [0] * (n + 1 - len(self.invModulos))
initialI = len(self.invModulos)
for i in range(initialI, min(p, n + 1)):
next = -self.invModulos[p % i] * (p // i) % p
self.invModulos.append(next)
return self.invModulos[pi]
def invFactorial(self, n):
if n <= -1:
print("Invalid argument to calculate (n^(-1))!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.invFactorial_):
return self.invFactorial_[n]
self.inv(n) # To make sure already calculated n^-1
nextArr = [0] * (n + 1 - len(self.invFactorial_))
initialI = len(self.invFactorial_)
prev = self.invFactorial_[-1]
p = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = (prev * self.invModulos[i % p]) % p
self.invFactorial_ += nextArr
return self.invFactorial_[n]
class Combination:
def __init__(self, MOD):
self.MOD = MOD
self.factorial = Factorial(MOD)
def ncr(self, n, k):
if k < 0 or n < k:
return 0
k = min(k, n - k)
f = self.factorial
return f.calc(n) * f.invFactorial(max(n - k, k)) * f.invFactorial(min(k, n - k)) % self.MOD
# --------------------------------------iye ha combinations ka zindegi---------------------------------
def powm(a, n, m):
if a == 1 or n == 0:
return 1
if n % 2 == 0:
s = powm(a, n // 2, m)
return s * s % m
else:
return a * powm(a, n - 1, m) % m
# --------------------------------------iye ha power ka zindegi---------------------------------
def sort_list(list1, list2):
zipped_pairs = zip(list2, list1)
z = [x for _, x in sorted(zipped_pairs)]
return z
# --------------------------------------------------product----------------------------------------
def product(l):
por = 1
for i in range(len(l)):
por *= l[i]
return por
# --------------------------------------------------binary----------------------------------------
def binarySearchCount(arr, n, key):
left = 0
right = n - 1
count = 0
while (left <= right):
mid = int((right + left) / 2)
# Check if middle element is
# less than or equal to key
if (arr[mid] < key):
count = mid + 1
left = mid + 1
# If key is smaller, ignore right half
else:
right = mid - 1
return count
# --------------------------------------------------binary----------------------------------------
def countdig(n):
c = 0
while (n > 0):
n //= 10
c += 1
return c
def binary(x, length):
y = bin(x)[2:]
return y if len(y) >= length else "0" * (length - len(y)) + y
def countGreater(arr, n, k):
l = 0
r = n - 1
# Stores the index of the left most element
# from the array which is greater than k
leftGreater = n
# Finds number of elements greater than k
while (l <= r):
m = int(l + (r - l) / 2)
if (arr[m] >= k):
leftGreater = m
r = m - 1
# If mid element is less than
# or equal to k update l
else:
l = m + 1
# Return the count of elements
# greater than k
return (n - leftGreater)
# --------------------------------------------------binary------------------------------------
n,m=map(int,input().split())
cnt=[0]*(m+1)
for i in range(n):
a,b=map(int,input().split())
cnt[a-1]+=1
cnt[b]-=1
for i in range(1,m+1):
cnt[i]+=cnt[i-1]
l=[0]*(m)
l1=[0]*(m)
cnt=[(cnt[i],i) for i in range(m)]
cnt.sort()
ind=defaultdict(int)
for i in range(len(cnt)):
ind[cnt[i][1]]=i
s=SegmentTree(l)
s1=SegmentTree(l1)
cou=[0]*m
cur=0
cur1=0
fi=[0]*m
#print(cnt,ind)
for i in range(m):
mw=s.query(0,ind[i])+1
#print(mw)
l[ind[i]]=mw
fi[i]=mw
cur1=max(cur1,mw)
s.__setitem__(ind[i],mw)
#print(fi)
cnt=[(cnt[i][0],-cnt[i][1]) for i in range(m)]
cnt.sort()
cnt=[(cnt[i][0],-cnt[i][1]) for i in range(m)]
for i in range(len(cnt)):
ind[cnt[i][1]]=i
#print(cnt,ind)
for i in range(m-1,-1,-1):
mw=s1.query(0,ind[i])+1
#print(mw)
fi[i]+=mw
l1[ind[i]]=mw
s1.__setitem__(ind[i],mw)
print(max(fi)-1)
``` | output | 1 | 11,365 | 23 | 22,731 |
Provide a correct Python 3 solution for this coding contest problem.
Our world is one-dimensional, and ruled by two empires called Empire A and Empire B.
The capital of Empire A is located at coordinate X, and that of Empire B is located at coordinate Y.
One day, Empire A becomes inclined to put the cities at coordinates x_1, x_2, ..., x_N under its control, and Empire B becomes inclined to put the cities at coordinates y_1, y_2, ..., y_M under its control.
If there exists an integer Z that satisfies all of the following three conditions, they will come to an agreement, but otherwise war will break out.
* X < Z \leq Y
* x_1, x_2, ..., x_N < Z
* y_1, y_2, ..., y_M \geq Z
Determine if war will break out.
Constraints
* All values in input are integers.
* 1 \leq N, M \leq 100
* -100 \leq X < Y \leq 100
* -100 \leq x_i, y_i \leq 100
* x_1, x_2, ..., x_N \neq X
* x_i are all different.
* y_1, y_2, ..., y_M \neq Y
* y_i are all different.
Input
Input is given from Standard Input in the following format:
N M X Y
x_1 x_2 ... x_N
y_1 y_2 ... y_M
Output
If war will break out, print `War`; otherwise, print `No War`.
Examples
Input
3 2 10 20
8 15 13
16 22
Output
No War
Input
4 2 -48 -1
-20 -35 -91 -23
-22 66
Output
War
Input
5 3 6 8
-10 3 1 5 -100
100 6 14
Output
War | instruction | 0 | 11,456 | 23 | 22,912 |
"Correct Solution:
```
n,m,x,y=map(int,input().split())
X=list(map(int,input().split()))
Y=list(map(int,input().split()))
X.append(x)
Y.append(y)
print("No War" if max(X)<min(Y) else "War")
``` | output | 1 | 11,456 | 23 | 22,913 |
Provide a correct Python 3 solution for this coding contest problem.
Our world is one-dimensional, and ruled by two empires called Empire A and Empire B.
The capital of Empire A is located at coordinate X, and that of Empire B is located at coordinate Y.
One day, Empire A becomes inclined to put the cities at coordinates x_1, x_2, ..., x_N under its control, and Empire B becomes inclined to put the cities at coordinates y_1, y_2, ..., y_M under its control.
If there exists an integer Z that satisfies all of the following three conditions, they will come to an agreement, but otherwise war will break out.
* X < Z \leq Y
* x_1, x_2, ..., x_N < Z
* y_1, y_2, ..., y_M \geq Z
Determine if war will break out.
Constraints
* All values in input are integers.
* 1 \leq N, M \leq 100
* -100 \leq X < Y \leq 100
* -100 \leq x_i, y_i \leq 100
* x_1, x_2, ..., x_N \neq X
* x_i are all different.
* y_1, y_2, ..., y_M \neq Y
* y_i are all different.
Input
Input is given from Standard Input in the following format:
N M X Y
x_1 x_2 ... x_N
y_1 y_2 ... y_M
Output
If war will break out, print `War`; otherwise, print `No War`.
Examples
Input
3 2 10 20
8 15 13
16 22
Output
No War
Input
4 2 -48 -1
-20 -35 -91 -23
-22 66
Output
War
Input
5 3 6 8
-10 3 1 5 -100
100 6 14
Output
War | instruction | 0 | 11,457 | 23 | 22,914 |
"Correct Solution:
```
N, M, X, Y = map(int, input().split())
A = list(map(int, input().split()))
B = list(map(int, input().split()))
if max(A + [X]) < min(B + [Y]):
print('No War')
else:
print('War')
``` | output | 1 | 11,457 | 23 | 22,915 |
Provide a correct Python 3 solution for this coding contest problem.
Our world is one-dimensional, and ruled by two empires called Empire A and Empire B.
The capital of Empire A is located at coordinate X, and that of Empire B is located at coordinate Y.
One day, Empire A becomes inclined to put the cities at coordinates x_1, x_2, ..., x_N under its control, and Empire B becomes inclined to put the cities at coordinates y_1, y_2, ..., y_M under its control.
If there exists an integer Z that satisfies all of the following three conditions, they will come to an agreement, but otherwise war will break out.
* X < Z \leq Y
* x_1, x_2, ..., x_N < Z
* y_1, y_2, ..., y_M \geq Z
Determine if war will break out.
Constraints
* All values in input are integers.
* 1 \leq N, M \leq 100
* -100 \leq X < Y \leq 100
* -100 \leq x_i, y_i \leq 100
* x_1, x_2, ..., x_N \neq X
* x_i are all different.
* y_1, y_2, ..., y_M \neq Y
* y_i are all different.
Input
Input is given from Standard Input in the following format:
N M X Y
x_1 x_2 ... x_N
y_1 y_2 ... y_M
Output
If war will break out, print `War`; otherwise, print `No War`.
Examples
Input
3 2 10 20
8 15 13
16 22
Output
No War
Input
4 2 -48 -1
-20 -35 -91 -23
-22 66
Output
War
Input
5 3 6 8
-10 3 1 5 -100
100 6 14
Output
War | instruction | 0 | 11,458 | 23 | 22,916 |
"Correct Solution:
```
n,m,x,y=map(int,input().split())
xi=[int(_) for _ in input().split()]+[x]
yi=[int(_) for _ in input().split()]+[y]
print('War' if max(xi) >= min(yi) else 'No War')
``` | output | 1 | 11,458 | 23 | 22,917 |
Provide a correct Python 3 solution for this coding contest problem.
Our world is one-dimensional, and ruled by two empires called Empire A and Empire B.
The capital of Empire A is located at coordinate X, and that of Empire B is located at coordinate Y.
One day, Empire A becomes inclined to put the cities at coordinates x_1, x_2, ..., x_N under its control, and Empire B becomes inclined to put the cities at coordinates y_1, y_2, ..., y_M under its control.
If there exists an integer Z that satisfies all of the following three conditions, they will come to an agreement, but otherwise war will break out.
* X < Z \leq Y
* x_1, x_2, ..., x_N < Z
* y_1, y_2, ..., y_M \geq Z
Determine if war will break out.
Constraints
* All values in input are integers.
* 1 \leq N, M \leq 100
* -100 \leq X < Y \leq 100
* -100 \leq x_i, y_i \leq 100
* x_1, x_2, ..., x_N \neq X
* x_i are all different.
* y_1, y_2, ..., y_M \neq Y
* y_i are all different.
Input
Input is given from Standard Input in the following format:
N M X Y
x_1 x_2 ... x_N
y_1 y_2 ... y_M
Output
If war will break out, print `War`; otherwise, print `No War`.
Examples
Input
3 2 10 20
8 15 13
16 22
Output
No War
Input
4 2 -48 -1
-20 -35 -91 -23
-22 66
Output
War
Input
5 3 6 8
-10 3 1 5 -100
100 6 14
Output
War | instruction | 0 | 11,459 | 23 | 22,918 |
"Correct Solution:
```
N,M,X,Y = map(int, input().split())
Z = max(map(int, input().split())) + 1
min_y = min(map(int, input().split()))
print("No War" if (X < Z) and (Z <= min_y) and (Z <= Y) else "War")
``` | output | 1 | 11,459 | 23 | 22,919 |
Provide a correct Python 3 solution for this coding contest problem.
Our world is one-dimensional, and ruled by two empires called Empire A and Empire B.
The capital of Empire A is located at coordinate X, and that of Empire B is located at coordinate Y.
One day, Empire A becomes inclined to put the cities at coordinates x_1, x_2, ..., x_N under its control, and Empire B becomes inclined to put the cities at coordinates y_1, y_2, ..., y_M under its control.
If there exists an integer Z that satisfies all of the following three conditions, they will come to an agreement, but otherwise war will break out.
* X < Z \leq Y
* x_1, x_2, ..., x_N < Z
* y_1, y_2, ..., y_M \geq Z
Determine if war will break out.
Constraints
* All values in input are integers.
* 1 \leq N, M \leq 100
* -100 \leq X < Y \leq 100
* -100 \leq x_i, y_i \leq 100
* x_1, x_2, ..., x_N \neq X
* x_i are all different.
* y_1, y_2, ..., y_M \neq Y
* y_i are all different.
Input
Input is given from Standard Input in the following format:
N M X Y
x_1 x_2 ... x_N
y_1 y_2 ... y_M
Output
If war will break out, print `War`; otherwise, print `No War`.
Examples
Input
3 2 10 20
8 15 13
16 22
Output
No War
Input
4 2 -48 -1
-20 -35 -91 -23
-22 66
Output
War
Input
5 3 6 8
-10 3 1 5 -100
100 6 14
Output
War | instruction | 0 | 11,460 | 23 | 22,920 |
"Correct Solution:
```
N,M,X,Y=map(int,input().split())
x=map(int,input().split())
y=map(int,input().split())
z1=max(x)
z2=min(y)
print('No War' if min(z2,Y)>max(z1,X) else 'War')
``` | output | 1 | 11,460 | 23 | 22,921 |
Provide a correct Python 3 solution for this coding contest problem.
Our world is one-dimensional, and ruled by two empires called Empire A and Empire B.
The capital of Empire A is located at coordinate X, and that of Empire B is located at coordinate Y.
One day, Empire A becomes inclined to put the cities at coordinates x_1, x_2, ..., x_N under its control, and Empire B becomes inclined to put the cities at coordinates y_1, y_2, ..., y_M under its control.
If there exists an integer Z that satisfies all of the following three conditions, they will come to an agreement, but otherwise war will break out.
* X < Z \leq Y
* x_1, x_2, ..., x_N < Z
* y_1, y_2, ..., y_M \geq Z
Determine if war will break out.
Constraints
* All values in input are integers.
* 1 \leq N, M \leq 100
* -100 \leq X < Y \leq 100
* -100 \leq x_i, y_i \leq 100
* x_1, x_2, ..., x_N \neq X
* x_i are all different.
* y_1, y_2, ..., y_M \neq Y
* y_i are all different.
Input
Input is given from Standard Input in the following format:
N M X Y
x_1 x_2 ... x_N
y_1 y_2 ... y_M
Output
If war will break out, print `War`; otherwise, print `No War`.
Examples
Input
3 2 10 20
8 15 13
16 22
Output
No War
Input
4 2 -48 -1
-20 -35 -91 -23
-22 66
Output
War
Input
5 3 6 8
-10 3 1 5 -100
100 6 14
Output
War | instruction | 0 | 11,461 | 23 | 22,922 |
"Correct Solution:
```
n,m,x,y = map(int,input().split())
print("No War" if max(max(list(map(int,input().split()))),x)<min(min(list(map(int,input().split()))),y)
else "War")
``` | output | 1 | 11,461 | 23 | 22,923 |
Provide a correct Python 3 solution for this coding contest problem.
Our world is one-dimensional, and ruled by two empires called Empire A and Empire B.
The capital of Empire A is located at coordinate X, and that of Empire B is located at coordinate Y.
One day, Empire A becomes inclined to put the cities at coordinates x_1, x_2, ..., x_N under its control, and Empire B becomes inclined to put the cities at coordinates y_1, y_2, ..., y_M under its control.
If there exists an integer Z that satisfies all of the following three conditions, they will come to an agreement, but otherwise war will break out.
* X < Z \leq Y
* x_1, x_2, ..., x_N < Z
* y_1, y_2, ..., y_M \geq Z
Determine if war will break out.
Constraints
* All values in input are integers.
* 1 \leq N, M \leq 100
* -100 \leq X < Y \leq 100
* -100 \leq x_i, y_i \leq 100
* x_1, x_2, ..., x_N \neq X
* x_i are all different.
* y_1, y_2, ..., y_M \neq Y
* y_i are all different.
Input
Input is given from Standard Input in the following format:
N M X Y
x_1 x_2 ... x_N
y_1 y_2 ... y_M
Output
If war will break out, print `War`; otherwise, print `No War`.
Examples
Input
3 2 10 20
8 15 13
16 22
Output
No War
Input
4 2 -48 -1
-20 -35 -91 -23
-22 66
Output
War
Input
5 3 6 8
-10 3 1 5 -100
100 6 14
Output
War | instruction | 0 | 11,462 | 23 | 22,924 |
"Correct Solution:
```
n,m,x,y=map(int,input().split())
a=max(list(map(int,input().split())))
b=min(list(map(int,input().split())))
print("No War" if a<b and x<b and a<y else "War")
``` | output | 1 | 11,462 | 23 | 22,925 |
Provide a correct Python 3 solution for this coding contest problem.
Our world is one-dimensional, and ruled by two empires called Empire A and Empire B.
The capital of Empire A is located at coordinate X, and that of Empire B is located at coordinate Y.
One day, Empire A becomes inclined to put the cities at coordinates x_1, x_2, ..., x_N under its control, and Empire B becomes inclined to put the cities at coordinates y_1, y_2, ..., y_M under its control.
If there exists an integer Z that satisfies all of the following three conditions, they will come to an agreement, but otherwise war will break out.
* X < Z \leq Y
* x_1, x_2, ..., x_N < Z
* y_1, y_2, ..., y_M \geq Z
Determine if war will break out.
Constraints
* All values in input are integers.
* 1 \leq N, M \leq 100
* -100 \leq X < Y \leq 100
* -100 \leq x_i, y_i \leq 100
* x_1, x_2, ..., x_N \neq X
* x_i are all different.
* y_1, y_2, ..., y_M \neq Y
* y_i are all different.
Input
Input is given from Standard Input in the following format:
N M X Y
x_1 x_2 ... x_N
y_1 y_2 ... y_M
Output
If war will break out, print `War`; otherwise, print `No War`.
Examples
Input
3 2 10 20
8 15 13
16 22
Output
No War
Input
4 2 -48 -1
-20 -35 -91 -23
-22 66
Output
War
Input
5 3 6 8
-10 3 1 5 -100
100 6 14
Output
War | instruction | 0 | 11,463 | 23 | 22,926 |
"Correct Solution:
```
a,b,x,y=map(int,input().split())
print(['War','No War'][sorted([x]+[int(i) for i in input().split()])[-1]<sorted([y]+[int(i) for i in input().split()])[0]])
``` | output | 1 | 11,463 | 23 | 22,927 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Our world is one-dimensional, and ruled by two empires called Empire A and Empire B.
The capital of Empire A is located at coordinate X, and that of Empire B is located at coordinate Y.
One day, Empire A becomes inclined to put the cities at coordinates x_1, x_2, ..., x_N under its control, and Empire B becomes inclined to put the cities at coordinates y_1, y_2, ..., y_M under its control.
If there exists an integer Z that satisfies all of the following three conditions, they will come to an agreement, but otherwise war will break out.
* X < Z \leq Y
* x_1, x_2, ..., x_N < Z
* y_1, y_2, ..., y_M \geq Z
Determine if war will break out.
Constraints
* All values in input are integers.
* 1 \leq N, M \leq 100
* -100 \leq X < Y \leq 100
* -100 \leq x_i, y_i \leq 100
* x_1, x_2, ..., x_N \neq X
* x_i are all different.
* y_1, y_2, ..., y_M \neq Y
* y_i are all different.
Input
Input is given from Standard Input in the following format:
N M X Y
x_1 x_2 ... x_N
y_1 y_2 ... y_M
Output
If war will break out, print `War`; otherwise, print `No War`.
Examples
Input
3 2 10 20
8 15 13
16 22
Output
No War
Input
4 2 -48 -1
-20 -35 -91 -23
-22 66
Output
War
Input
5 3 6 8
-10 3 1 5 -100
100 6 14
Output
War
Submitted Solution:
```
I=lambda:map(int,input().split())
N,M,X,Y=I()
print("No War"if max(list(I())+[X])<min(list(I())+[Y]) else"War")
``` | instruction | 0 | 11,464 | 23 | 22,928 |
Yes | output | 1 | 11,464 | 23 | 22,929 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Our world is one-dimensional, and ruled by two empires called Empire A and Empire B.
The capital of Empire A is located at coordinate X, and that of Empire B is located at coordinate Y.
One day, Empire A becomes inclined to put the cities at coordinates x_1, x_2, ..., x_N under its control, and Empire B becomes inclined to put the cities at coordinates y_1, y_2, ..., y_M under its control.
If there exists an integer Z that satisfies all of the following three conditions, they will come to an agreement, but otherwise war will break out.
* X < Z \leq Y
* x_1, x_2, ..., x_N < Z
* y_1, y_2, ..., y_M \geq Z
Determine if war will break out.
Constraints
* All values in input are integers.
* 1 \leq N, M \leq 100
* -100 \leq X < Y \leq 100
* -100 \leq x_i, y_i \leq 100
* x_1, x_2, ..., x_N \neq X
* x_i are all different.
* y_1, y_2, ..., y_M \neq Y
* y_i are all different.
Input
Input is given from Standard Input in the following format:
N M X Y
x_1 x_2 ... x_N
y_1 y_2 ... y_M
Output
If war will break out, print `War`; otherwise, print `No War`.
Examples
Input
3 2 10 20
8 15 13
16 22
Output
No War
Input
4 2 -48 -1
-20 -35 -91 -23
-22 66
Output
War
Input
5 3 6 8
-10 3 1 5 -100
100 6 14
Output
War
Submitted Solution:
```
N,M,X,Y=map(int,input().split())
x=list(map(int,input().split()))
y=list(map(int,input().split()))
print('No War' if max(max(x),X)<min(min(y),Y) else 'War')
``` | instruction | 0 | 11,465 | 23 | 22,930 |
Yes | output | 1 | 11,465 | 23 | 22,931 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Our world is one-dimensional, and ruled by two empires called Empire A and Empire B.
The capital of Empire A is located at coordinate X, and that of Empire B is located at coordinate Y.
One day, Empire A becomes inclined to put the cities at coordinates x_1, x_2, ..., x_N under its control, and Empire B becomes inclined to put the cities at coordinates y_1, y_2, ..., y_M under its control.
If there exists an integer Z that satisfies all of the following three conditions, they will come to an agreement, but otherwise war will break out.
* X < Z \leq Y
* x_1, x_2, ..., x_N < Z
* y_1, y_2, ..., y_M \geq Z
Determine if war will break out.
Constraints
* All values in input are integers.
* 1 \leq N, M \leq 100
* -100 \leq X < Y \leq 100
* -100 \leq x_i, y_i \leq 100
* x_1, x_2, ..., x_N \neq X
* x_i are all different.
* y_1, y_2, ..., y_M \neq Y
* y_i are all different.
Input
Input is given from Standard Input in the following format:
N M X Y
x_1 x_2 ... x_N
y_1 y_2 ... y_M
Output
If war will break out, print `War`; otherwise, print `No War`.
Examples
Input
3 2 10 20
8 15 13
16 22
Output
No War
Input
4 2 -48 -1
-20 -35 -91 -23
-22 66
Output
War
Input
5 3 6 8
-10 3 1 5 -100
100 6 14
Output
War
Submitted Solution:
```
x,y=input().split()[2:];print('No '*(max(map(int,input().split()+[x]))<min(map(int,input().split()+[y])))+'War')
``` | instruction | 0 | 11,466 | 23 | 22,932 |
Yes | output | 1 | 11,466 | 23 | 22,933 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Our world is one-dimensional, and ruled by two empires called Empire A and Empire B.
The capital of Empire A is located at coordinate X, and that of Empire B is located at coordinate Y.
One day, Empire A becomes inclined to put the cities at coordinates x_1, x_2, ..., x_N under its control, and Empire B becomes inclined to put the cities at coordinates y_1, y_2, ..., y_M under its control.
If there exists an integer Z that satisfies all of the following three conditions, they will come to an agreement, but otherwise war will break out.
* X < Z \leq Y
* x_1, x_2, ..., x_N < Z
* y_1, y_2, ..., y_M \geq Z
Determine if war will break out.
Constraints
* All values in input are integers.
* 1 \leq N, M \leq 100
* -100 \leq X < Y \leq 100
* -100 \leq x_i, y_i \leq 100
* x_1, x_2, ..., x_N \neq X
* x_i are all different.
* y_1, y_2, ..., y_M \neq Y
* y_i are all different.
Input
Input is given from Standard Input in the following format:
N M X Y
x_1 x_2 ... x_N
y_1 y_2 ... y_M
Output
If war will break out, print `War`; otherwise, print `No War`.
Examples
Input
3 2 10 20
8 15 13
16 22
Output
No War
Input
4 2 -48 -1
-20 -35 -91 -23
-22 66
Output
War
Input
5 3 6 8
-10 3 1 5 -100
100 6 14
Output
War
Submitted Solution:
```
N, M, X, Y = map(int, input().split())
A = [X] + [int(x) for x in input().split()]
B = [Y] + [int(x) for x in input().split()]
print('No War' if max(A) < min(B) else 'War')
``` | instruction | 0 | 11,467 | 23 | 22,934 |
Yes | output | 1 | 11,467 | 23 | 22,935 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Our world is one-dimensional, and ruled by two empires called Empire A and Empire B.
The capital of Empire A is located at coordinate X, and that of Empire B is located at coordinate Y.
One day, Empire A becomes inclined to put the cities at coordinates x_1, x_2, ..., x_N under its control, and Empire B becomes inclined to put the cities at coordinates y_1, y_2, ..., y_M under its control.
If there exists an integer Z that satisfies all of the following three conditions, they will come to an agreement, but otherwise war will break out.
* X < Z \leq Y
* x_1, x_2, ..., x_N < Z
* y_1, y_2, ..., y_M \geq Z
Determine if war will break out.
Constraints
* All values in input are integers.
* 1 \leq N, M \leq 100
* -100 \leq X < Y \leq 100
* -100 \leq x_i, y_i \leq 100
* x_1, x_2, ..., x_N \neq X
* x_i are all different.
* y_1, y_2, ..., y_M \neq Y
* y_i are all different.
Input
Input is given from Standard Input in the following format:
N M X Y
x_1 x_2 ... x_N
y_1 y_2 ... y_M
Output
If war will break out, print `War`; otherwise, print `No War`.
Examples
Input
3 2 10 20
8 15 13
16 22
Output
No War
Input
4 2 -48 -1
-20 -35 -91 -23
-22 66
Output
War
Input
5 3 6 8
-10 3 1 5 -100
100 6 14
Output
War
Submitted Solution:
```
N,M,X,Y = map(int, input().split())
x = list(map(int, input().split()))
y = list(map(int, input().split()))
if max(x)>min(y):
print("No War")
else:
print("War")
``` | instruction | 0 | 11,468 | 23 | 22,936 |
No | output | 1 | 11,468 | 23 | 22,937 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Our world is one-dimensional, and ruled by two empires called Empire A and Empire B.
The capital of Empire A is located at coordinate X, and that of Empire B is located at coordinate Y.
One day, Empire A becomes inclined to put the cities at coordinates x_1, x_2, ..., x_N under its control, and Empire B becomes inclined to put the cities at coordinates y_1, y_2, ..., y_M under its control.
If there exists an integer Z that satisfies all of the following three conditions, they will come to an agreement, but otherwise war will break out.
* X < Z \leq Y
* x_1, x_2, ..., x_N < Z
* y_1, y_2, ..., y_M \geq Z
Determine if war will break out.
Constraints
* All values in input are integers.
* 1 \leq N, M \leq 100
* -100 \leq X < Y \leq 100
* -100 \leq x_i, y_i \leq 100
* x_1, x_2, ..., x_N \neq X
* x_i are all different.
* y_1, y_2, ..., y_M \neq Y
* y_i are all different.
Input
Input is given from Standard Input in the following format:
N M X Y
x_1 x_2 ... x_N
y_1 y_2 ... y_M
Output
If war will break out, print `War`; otherwise, print `No War`.
Examples
Input
3 2 10 20
8 15 13
16 22
Output
No War
Input
4 2 -48 -1
-20 -35 -91 -23
-22 66
Output
War
Input
5 3 6 8
-10 3 1 5 -100
100 6 14
Output
War
Submitted Solution:
```
N,M,X,Y = map(int, input().split())
x = list(map(int, input().split()))
y = list(map(int, input().split()))
if X<Y or max(x)>min(y):
print("No War")
else:
print("War")
``` | instruction | 0 | 11,469 | 23 | 22,938 |
No | output | 1 | 11,469 | 23 | 22,939 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Our world is one-dimensional, and ruled by two empires called Empire A and Empire B.
The capital of Empire A is located at coordinate X, and that of Empire B is located at coordinate Y.
One day, Empire A becomes inclined to put the cities at coordinates x_1, x_2, ..., x_N under its control, and Empire B becomes inclined to put the cities at coordinates y_1, y_2, ..., y_M under its control.
If there exists an integer Z that satisfies all of the following three conditions, they will come to an agreement, but otherwise war will break out.
* X < Z \leq Y
* x_1, x_2, ..., x_N < Z
* y_1, y_2, ..., y_M \geq Z
Determine if war will break out.
Constraints
* All values in input are integers.
* 1 \leq N, M \leq 100
* -100 \leq X < Y \leq 100
* -100 \leq x_i, y_i \leq 100
* x_1, x_2, ..., x_N \neq X
* x_i are all different.
* y_1, y_2, ..., y_M \neq Y
* y_i are all different.
Input
Input is given from Standard Input in the following format:
N M X Y
x_1 x_2 ... x_N
y_1 y_2 ... y_M
Output
If war will break out, print `War`; otherwise, print `No War`.
Examples
Input
3 2 10 20
8 15 13
16 22
Output
No War
Input
4 2 -48 -1
-20 -35 -91 -23
-22 66
Output
War
Input
5 3 6 8
-10 3 1 5 -100
100 6 14
Output
War
Submitted Solution:
```
_,b,c,d,*e=map(int,open(0).read().split())
print((len(set(range(max(e[:b]),min(e[b:])+1))&set(range(c,d+1)))>0)*"No "+"War")
``` | instruction | 0 | 11,470 | 23 | 22,940 |
No | output | 1 | 11,470 | 23 | 22,941 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Our world is one-dimensional, and ruled by two empires called Empire A and Empire B.
The capital of Empire A is located at coordinate X, and that of Empire B is located at coordinate Y.
One day, Empire A becomes inclined to put the cities at coordinates x_1, x_2, ..., x_N under its control, and Empire B becomes inclined to put the cities at coordinates y_1, y_2, ..., y_M under its control.
If there exists an integer Z that satisfies all of the following three conditions, they will come to an agreement, but otherwise war will break out.
* X < Z \leq Y
* x_1, x_2, ..., x_N < Z
* y_1, y_2, ..., y_M \geq Z
Determine if war will break out.
Constraints
* All values in input are integers.
* 1 \leq N, M \leq 100
* -100 \leq X < Y \leq 100
* -100 \leq x_i, y_i \leq 100
* x_1, x_2, ..., x_N \neq X
* x_i are all different.
* y_1, y_2, ..., y_M \neq Y
* y_i are all different.
Input
Input is given from Standard Input in the following format:
N M X Y
x_1 x_2 ... x_N
y_1 y_2 ... y_M
Output
If war will break out, print `War`; otherwise, print `No War`.
Examples
Input
3 2 10 20
8 15 13
16 22
Output
No War
Input
4 2 -48 -1
-20 -35 -91 -23
-22 66
Output
War
Input
5 3 6 8
-10 3 1 5 -100
100 6 14
Output
War
Submitted Solution:
```
(n, m, x, y) = list(map(int, input().split()))
xs = list(map(int, input().split()))
ys = list(map(int, input().split()))
xs.sort()
ys.sort()
#print(n, m, x, y)
#print(xs)
#print(ys)
z = ys[0]
if(z < xs[-1]):
print("War")
elif(x < z and z <= y):
print("No War")
else:
print("War")
``` | instruction | 0 | 11,471 | 23 | 22,942 |
No | output | 1 | 11,471 | 23 | 22,943 |
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