message stringlengths 2 57.2k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 61 108k | cluster float64 22 22 | __index_level_0__ int64 122 217k |
|---|---|---|---|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
Holidays have finished. Thanks to the help of the hacker Leha, Noora managed to enter the university of her dreams which is located in a town Pavlopolis. It's well known that universities provide students with dormitory for the period of university studies. Consequently Noora had to leave Vičkopolis and move to Pavlopolis. Thus Leha was left completely alone in a quiet town Vičkopolis. He almost even fell into a depression from boredom!
Leha came up with a task for himself to relax a little. He chooses two integers A and B and then calculates the greatest common divisor of integers "A factorial" and "B factorial". Formally the hacker wants to find out GCD(A!, B!). It's well known that the factorial of an integer x is a product of all positive integers less than or equal to x. Thus x! = 1·2·3·...·(x - 1)·x. For example 4! = 1·2·3·4 = 24. Recall that GCD(x, y) is the largest positive integer q that divides (without a remainder) both x and y.
Leha has learned how to solve this task very effective. You are able to cope with it not worse, aren't you?
Input
The first and single line contains two integers A and B (1 ≤ A, B ≤ 109, min(A, B) ≤ 12).
Output
Print a single integer denoting the greatest common divisor of integers A! and B!.
Example
Input
4 3
Output
6
Note
Consider the sample.
4! = 1·2·3·4 = 24. 3! = 1·2·3 = 6. The greatest common divisor of integers 24 and 6 is exactly 6. | instruction | 0 | 26,152 | 22 | 52,304 |
Tags: implementation, math, number theory
Correct Solution:
```
def main():
n, m = [int(v) for v in input().split()]
d = 1
for i in range(1, min(n,m)+1):
d*=i
print(d)
if __name__ == "__main__":
main()
``` | output | 1 | 26,152 | 22 | 52,305 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Holidays have finished. Thanks to the help of the hacker Leha, Noora managed to enter the university of her dreams which is located in a town Pavlopolis. It's well known that universities provide students with dormitory for the period of university studies. Consequently Noora had to leave Vičkopolis and move to Pavlopolis. Thus Leha was left completely alone in a quiet town Vičkopolis. He almost even fell into a depression from boredom!
Leha came up with a task for himself to relax a little. He chooses two integers A and B and then calculates the greatest common divisor of integers "A factorial" and "B factorial". Formally the hacker wants to find out GCD(A!, B!). It's well known that the factorial of an integer x is a product of all positive integers less than or equal to x. Thus x! = 1·2·3·...·(x - 1)·x. For example 4! = 1·2·3·4 = 24. Recall that GCD(x, y) is the largest positive integer q that divides (without a remainder) both x and y.
Leha has learned how to solve this task very effective. You are able to cope with it not worse, aren't you?
Input
The first and single line contains two integers A and B (1 ≤ A, B ≤ 109, min(A, B) ≤ 12).
Output
Print a single integer denoting the greatest common divisor of integers A! and B!.
Example
Input
4 3
Output
6
Note
Consider the sample.
4! = 1·2·3·4 = 24. 3! = 1·2·3 = 6. The greatest common divisor of integers 24 and 6 is exactly 6. | instruction | 0 | 26,153 | 22 | 52,306 |
Tags: implementation, math, number theory
Correct Solution:
```
from math import factorial
n,m=map(int,input().split())
print(factorial(min(n,m)))
#print(' '.join([str(i) for i in a]))
``` | output | 1 | 26,153 | 22 | 52,307 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Holidays have finished. Thanks to the help of the hacker Leha, Noora managed to enter the university of her dreams which is located in a town Pavlopolis. It's well known that universities provide students with dormitory for the period of university studies. Consequently Noora had to leave Vičkopolis and move to Pavlopolis. Thus Leha was left completely alone in a quiet town Vičkopolis. He almost even fell into a depression from boredom!
Leha came up with a task for himself to relax a little. He chooses two integers A and B and then calculates the greatest common divisor of integers "A factorial" and "B factorial". Formally the hacker wants to find out GCD(A!, B!). It's well known that the factorial of an integer x is a product of all positive integers less than or equal to x. Thus x! = 1·2·3·...·(x - 1)·x. For example 4! = 1·2·3·4 = 24. Recall that GCD(x, y) is the largest positive integer q that divides (without a remainder) both x and y.
Leha has learned how to solve this task very effective. You are able to cope with it not worse, aren't you?
Input
The first and single line contains two integers A and B (1 ≤ A, B ≤ 109, min(A, B) ≤ 12).
Output
Print a single integer denoting the greatest common divisor of integers A! and B!.
Example
Input
4 3
Output
6
Note
Consider the sample.
4! = 1·2·3·4 = 24. 3! = 1·2·3 = 6. The greatest common divisor of integers 24 and 6 is exactly 6. | instruction | 0 | 26,154 | 22 | 52,308 |
Tags: implementation, math, number theory
Correct Solution:
```
a, b = input().split()
a = int(a)
b = int(b)
s = 1
if a >= b:
for i in range(1,b+1):
s = s * i
else:
for i in range(1,a+1):
s = s * i
print(s)
``` | output | 1 | 26,154 | 22 | 52,309 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Holidays have finished. Thanks to the help of the hacker Leha, Noora managed to enter the university of her dreams which is located in a town Pavlopolis. It's well known that universities provide students with dormitory for the period of university studies. Consequently Noora had to leave Vičkopolis and move to Pavlopolis. Thus Leha was left completely alone in a quiet town Vičkopolis. He almost even fell into a depression from boredom!
Leha came up with a task for himself to relax a little. He chooses two integers A and B and then calculates the greatest common divisor of integers "A factorial" and "B factorial". Formally the hacker wants to find out GCD(A!, B!). It's well known that the factorial of an integer x is a product of all positive integers less than or equal to x. Thus x! = 1·2·3·...·(x - 1)·x. For example 4! = 1·2·3·4 = 24. Recall that GCD(x, y) is the largest positive integer q that divides (without a remainder) both x and y.
Leha has learned how to solve this task very effective. You are able to cope with it not worse, aren't you?
Input
The first and single line contains two integers A and B (1 ≤ A, B ≤ 109, min(A, B) ≤ 12).
Output
Print a single integer denoting the greatest common divisor of integers A! and B!.
Example
Input
4 3
Output
6
Note
Consider the sample.
4! = 1·2·3·4 = 24. 3! = 1·2·3 = 6. The greatest common divisor of integers 24 and 6 is exactly 6. | instruction | 0 | 26,155 | 22 | 52,310 |
Tags: implementation, math, number theory
Correct Solution:
```
from math import factorial
a,b=tuple(input().split())
a=int(a)
b=int(b)
print(factorial(min(a,b)))
``` | output | 1 | 26,155 | 22 | 52,311 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Holidays have finished. Thanks to the help of the hacker Leha, Noora managed to enter the university of her dreams which is located in a town Pavlopolis. It's well known that universities provide students with dormitory for the period of university studies. Consequently Noora had to leave Vičkopolis and move to Pavlopolis. Thus Leha was left completely alone in a quiet town Vičkopolis. He almost even fell into a depression from boredom!
Leha came up with a task for himself to relax a little. He chooses two integers A and B and then calculates the greatest common divisor of integers "A factorial" and "B factorial". Formally the hacker wants to find out GCD(A!, B!). It's well known that the factorial of an integer x is a product of all positive integers less than or equal to x. Thus x! = 1·2·3·...·(x - 1)·x. For example 4! = 1·2·3·4 = 24. Recall that GCD(x, y) is the largest positive integer q that divides (without a remainder) both x and y.
Leha has learned how to solve this task very effective. You are able to cope with it not worse, aren't you?
Input
The first and single line contains two integers A and B (1 ≤ A, B ≤ 109, min(A, B) ≤ 12).
Output
Print a single integer denoting the greatest common divisor of integers A! and B!.
Example
Input
4 3
Output
6
Note
Consider the sample.
4! = 1·2·3·4 = 24. 3! = 1·2·3 = 6. The greatest common divisor of integers 24 and 6 is exactly 6.
Submitted Solution:
```
s = input()
a, b = [int(x) for x in s.split()]
num = min(a, b)
f = 1
for i in range(1, num + 1):
f *= i
print(f)
``` | instruction | 0 | 26,156 | 22 | 52,312 |
Yes | output | 1 | 26,156 | 22 | 52,313 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Holidays have finished. Thanks to the help of the hacker Leha, Noora managed to enter the university of her dreams which is located in a town Pavlopolis. It's well known that universities provide students with dormitory for the period of university studies. Consequently Noora had to leave Vičkopolis and move to Pavlopolis. Thus Leha was left completely alone in a quiet town Vičkopolis. He almost even fell into a depression from boredom!
Leha came up with a task for himself to relax a little. He chooses two integers A and B and then calculates the greatest common divisor of integers "A factorial" and "B factorial". Formally the hacker wants to find out GCD(A!, B!). It's well known that the factorial of an integer x is a product of all positive integers less than or equal to x. Thus x! = 1·2·3·...·(x - 1)·x. For example 4! = 1·2·3·4 = 24. Recall that GCD(x, y) is the largest positive integer q that divides (without a remainder) both x and y.
Leha has learned how to solve this task very effective. You are able to cope with it not worse, aren't you?
Input
The first and single line contains two integers A and B (1 ≤ A, B ≤ 109, min(A, B) ≤ 12).
Output
Print a single integer denoting the greatest common divisor of integers A! and B!.
Example
Input
4 3
Output
6
Note
Consider the sample.
4! = 1·2·3·4 = 24. 3! = 1·2·3 = 6. The greatest common divisor of integers 24 and 6 is exactly 6.
Submitted Solution:
```
a, b= input().split()
a=int(a)
b=int(b)
aa=1
bb=1
if a < b:
for i in range(1,a+1):
aa= aa*i
print(aa)
if b <= a:
for i in range(1,b+1):
bb= bb*i
print(bb)
``` | instruction | 0 | 26,157 | 22 | 52,314 |
Yes | output | 1 | 26,157 | 22 | 52,315 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Holidays have finished. Thanks to the help of the hacker Leha, Noora managed to enter the university of her dreams which is located in a town Pavlopolis. It's well known that universities provide students with dormitory for the period of university studies. Consequently Noora had to leave Vičkopolis and move to Pavlopolis. Thus Leha was left completely alone in a quiet town Vičkopolis. He almost even fell into a depression from boredom!
Leha came up with a task for himself to relax a little. He chooses two integers A and B and then calculates the greatest common divisor of integers "A factorial" and "B factorial". Formally the hacker wants to find out GCD(A!, B!). It's well known that the factorial of an integer x is a product of all positive integers less than or equal to x. Thus x! = 1·2·3·...·(x - 1)·x. For example 4! = 1·2·3·4 = 24. Recall that GCD(x, y) is the largest positive integer q that divides (without a remainder) both x and y.
Leha has learned how to solve this task very effective. You are able to cope with it not worse, aren't you?
Input
The first and single line contains two integers A and B (1 ≤ A, B ≤ 109, min(A, B) ≤ 12).
Output
Print a single integer denoting the greatest common divisor of integers A! and B!.
Example
Input
4 3
Output
6
Note
Consider the sample.
4! = 1·2·3·4 = 24. 3! = 1·2·3 = 6. The greatest common divisor of integers 24 and 6 is exactly 6.
Submitted Solution:
```
import sys
""" sys.stdin = open("/Users/isym444/Desktop/PythonCP/CP1/Codewars/Practice/input.txt", "r")
sys.stdout = open("/Users/isym444/Desktop/PythonCP/CP1/Codewars/Practice/output.txt", "w")
"""
def fact(x):
"""
docstring
"""
result = 1
for i in range(1, x + 1):
result = result * i
return result
a = list(map(int, input().split()))
print(fact(min(a)))
``` | instruction | 0 | 26,158 | 22 | 52,316 |
Yes | output | 1 | 26,158 | 22 | 52,317 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Holidays have finished. Thanks to the help of the hacker Leha, Noora managed to enter the university of her dreams which is located in a town Pavlopolis. It's well known that universities provide students with dormitory for the period of university studies. Consequently Noora had to leave Vičkopolis and move to Pavlopolis. Thus Leha was left completely alone in a quiet town Vičkopolis. He almost even fell into a depression from boredom!
Leha came up with a task for himself to relax a little. He chooses two integers A and B and then calculates the greatest common divisor of integers "A factorial" and "B factorial". Formally the hacker wants to find out GCD(A!, B!). It's well known that the factorial of an integer x is a product of all positive integers less than or equal to x. Thus x! = 1·2·3·...·(x - 1)·x. For example 4! = 1·2·3·4 = 24. Recall that GCD(x, y) is the largest positive integer q that divides (without a remainder) both x and y.
Leha has learned how to solve this task very effective. You are able to cope with it not worse, aren't you?
Input
The first and single line contains two integers A and B (1 ≤ A, B ≤ 109, min(A, B) ≤ 12).
Output
Print a single integer denoting the greatest common divisor of integers A! and B!.
Example
Input
4 3
Output
6
Note
Consider the sample.
4! = 1·2·3·4 = 24. 3! = 1·2·3 = 6. The greatest common divisor of integers 24 and 6 is exactly 6.
Submitted Solution:
```
from math import factorial
a, b = [int(x) for x in input().split()]
print(factorial(min(a, b)))
``` | instruction | 0 | 26,159 | 22 | 52,318 |
Yes | output | 1 | 26,159 | 22 | 52,319 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Holidays have finished. Thanks to the help of the hacker Leha, Noora managed to enter the university of her dreams which is located in a town Pavlopolis. It's well known that universities provide students with dormitory for the period of university studies. Consequently Noora had to leave Vičkopolis and move to Pavlopolis. Thus Leha was left completely alone in a quiet town Vičkopolis. He almost even fell into a depression from boredom!
Leha came up with a task for himself to relax a little. He chooses two integers A and B and then calculates the greatest common divisor of integers "A factorial" and "B factorial". Formally the hacker wants to find out GCD(A!, B!). It's well known that the factorial of an integer x is a product of all positive integers less than or equal to x. Thus x! = 1·2·3·...·(x - 1)·x. For example 4! = 1·2·3·4 = 24. Recall that GCD(x, y) is the largest positive integer q that divides (without a remainder) both x and y.
Leha has learned how to solve this task very effective. You are able to cope with it not worse, aren't you?
Input
The first and single line contains two integers A and B (1 ≤ A, B ≤ 109, min(A, B) ≤ 12).
Output
Print a single integer denoting the greatest common divisor of integers A! and B!.
Example
Input
4 3
Output
6
Note
Consider the sample.
4! = 1·2·3·4 = 24. 3! = 1·2·3 = 6. The greatest common divisor of integers 24 and 6 is exactly 6.
Submitted Solution:
```
a,b=map(int,input().split(" "))
# print(a,b)
if a<b:
i=a-1
while i>1:
af=a*i
i=i-1
print(af)
elif a>=b:
i=b-1
while i>1:
bf=b*i
i=i-1
print(bf)
``` | instruction | 0 | 26,160 | 22 | 52,320 |
No | output | 1 | 26,160 | 22 | 52,321 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Holidays have finished. Thanks to the help of the hacker Leha, Noora managed to enter the university of her dreams which is located in a town Pavlopolis. It's well known that universities provide students with dormitory for the period of university studies. Consequently Noora had to leave Vičkopolis and move to Pavlopolis. Thus Leha was left completely alone in a quiet town Vičkopolis. He almost even fell into a depression from boredom!
Leha came up with a task for himself to relax a little. He chooses two integers A and B and then calculates the greatest common divisor of integers "A factorial" and "B factorial". Formally the hacker wants to find out GCD(A!, B!). It's well known that the factorial of an integer x is a product of all positive integers less than or equal to x. Thus x! = 1·2·3·...·(x - 1)·x. For example 4! = 1·2·3·4 = 24. Recall that GCD(x, y) is the largest positive integer q that divides (without a remainder) both x and y.
Leha has learned how to solve this task very effective. You are able to cope with it not worse, aren't you?
Input
The first and single line contains two integers A and B (1 ≤ A, B ≤ 109, min(A, B) ≤ 12).
Output
Print a single integer denoting the greatest common divisor of integers A! and B!.
Example
Input
4 3
Output
6
Note
Consider the sample.
4! = 1·2·3·4 = 24. 3! = 1·2·3 = 6. The greatest common divisor of integers 24 and 6 is exactly 6.
Submitted Solution:
```
import math
x,y=map(int,input().split())
xfac=1
yfac=1
for i in range(1,x+1):
xfac*=i
for i in range(1,x+1):
yfac*=i
gcd=math.gcd(xfac,yfac)
print(gcd)
``` | instruction | 0 | 26,161 | 22 | 52,322 |
No | output | 1 | 26,161 | 22 | 52,323 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Holidays have finished. Thanks to the help of the hacker Leha, Noora managed to enter the university of her dreams which is located in a town Pavlopolis. It's well known that universities provide students with dormitory for the period of university studies. Consequently Noora had to leave Vičkopolis and move to Pavlopolis. Thus Leha was left completely alone in a quiet town Vičkopolis. He almost even fell into a depression from boredom!
Leha came up with a task for himself to relax a little. He chooses two integers A and B and then calculates the greatest common divisor of integers "A factorial" and "B factorial". Formally the hacker wants to find out GCD(A!, B!). It's well known that the factorial of an integer x is a product of all positive integers less than or equal to x. Thus x! = 1·2·3·...·(x - 1)·x. For example 4! = 1·2·3·4 = 24. Recall that GCD(x, y) is the largest positive integer q that divides (without a remainder) both x and y.
Leha has learned how to solve this task very effective. You are able to cope with it not worse, aren't you?
Input
The first and single line contains two integers A and B (1 ≤ A, B ≤ 109, min(A, B) ≤ 12).
Output
Print a single integer denoting the greatest common divisor of integers A! and B!.
Example
Input
4 3
Output
6
Note
Consider the sample.
4! = 1·2·3·4 = 24. 3! = 1·2·3 = 6. The greatest common divisor of integers 24 and 6 is exactly 6.
Submitted Solution:
```
a = input().split()
small = (min(int(a[0]), int(a[1])))
print((small*(small-1))//2)
``` | instruction | 0 | 26,162 | 22 | 52,324 |
No | output | 1 | 26,162 | 22 | 52,325 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Holidays have finished. Thanks to the help of the hacker Leha, Noora managed to enter the university of her dreams which is located in a town Pavlopolis. It's well known that universities provide students with dormitory for the period of university studies. Consequently Noora had to leave Vičkopolis and move to Pavlopolis. Thus Leha was left completely alone in a quiet town Vičkopolis. He almost even fell into a depression from boredom!
Leha came up with a task for himself to relax a little. He chooses two integers A and B and then calculates the greatest common divisor of integers "A factorial" and "B factorial". Formally the hacker wants to find out GCD(A!, B!). It's well known that the factorial of an integer x is a product of all positive integers less than or equal to x. Thus x! = 1·2·3·...·(x - 1)·x. For example 4! = 1·2·3·4 = 24. Recall that GCD(x, y) is the largest positive integer q that divides (without a remainder) both x and y.
Leha has learned how to solve this task very effective. You are able to cope with it not worse, aren't you?
Input
The first and single line contains two integers A and B (1 ≤ A, B ≤ 109, min(A, B) ≤ 12).
Output
Print a single integer denoting the greatest common divisor of integers A! and B!.
Example
Input
4 3
Output
6
Note
Consider the sample.
4! = 1·2·3·4 = 24. 3! = 1·2·3 = 6. The greatest common divisor of integers 24 and 6 is exactly 6.
Submitted Solution:
```
def gcd(a,b):
if a==0:
return b
else:
return gcd(b%a,a)
n, m = map(int, input().split())
nn=n
mm=m
ansn = 1
x= min(nn,mm)
while(x!=1):
ansn=ansn*x
x=x-1
print(gcd(ansn,max(nn,mm)))
``` | instruction | 0 | 26,163 | 22 | 52,326 |
No | output | 1 | 26,163 | 22 | 52,327 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a consisting of n integers a_1, a_2, ..., a_n.
Your problem is to find such pair of indices i, j (1 ≤ i < j ≤ n) that lcm(a_i, a_j) is minimum possible.
lcm(x, y) is the least common multiple of x and y (minimum positive number such that both x and y are divisors of this number).
Input
The first line of the input contains one integer n (2 ≤ n ≤ 10^6) — the number of elements in a.
The second line of the input contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^7), where a_i is the i-th element of a.
Output
Print two integers i and j (1 ≤ i < j ≤ n) such that the value of lcm(a_i, a_j) is minimum among all valid pairs i, j. If there are multiple answers, you can print any.
Examples
Input
5
2 4 8 3 6
Output
1 2
Input
5
5 2 11 3 7
Output
2 4
Input
6
2 5 10 1 10 2
Output
1 4 | instruction | 0 | 26,614 | 22 | 53,228 |
Tags: brute force, greedy, math, number theory
Correct Solution:
```
# by the authority of GOD author: manhar singh sachdev #
import os,sys
from io import BytesIO, IOBase
from array import array
def main():
INF = 10**18
n = int(input())
a = list(map(int,input().split()))
lim = max(a)+1
counter = array('i',[0]*lim)
for i in a:
counter[i] += 1
mini = INF
for i in range(1,lim):
m,m1 = INF,INF
for j in range(i,lim,i):
if counter[j] >= 2:
m,m1 = min(m,j),j
break
if counter[j] == 1:
m,m1 = j,m
if m1 != INF:
break
z = m*m1//i
if z < mini:
mini,n1,n2 = z,m,m1
ind1,ind2 = -1,-1
for i in range(n):
if ind1 == -1 and n1 == a[i]:
ind1 = i+1
elif ind2 == -1 and n2 == a[i]:
ind2 = i+1
if ind1 > ind2:
ind1,ind2 = ind2,ind1
print(ind1,ind2)
# Fast IO Region
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")
if __name__ == "__main__":
main()
``` | output | 1 | 26,614 | 22 | 53,229 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a consisting of n integers a_1, a_2, ..., a_n.
Your problem is to find such pair of indices i, j (1 ≤ i < j ≤ n) that lcm(a_i, a_j) is minimum possible.
lcm(x, y) is the least common multiple of x and y (minimum positive number such that both x and y are divisors of this number).
Input
The first line of the input contains one integer n (2 ≤ n ≤ 10^6) — the number of elements in a.
The second line of the input contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^7), where a_i is the i-th element of a.
Output
Print two integers i and j (1 ≤ i < j ≤ n) such that the value of lcm(a_i, a_j) is minimum among all valid pairs i, j. If there are multiple answers, you can print any.
Examples
Input
5
2 4 8 3 6
Output
1 2
Input
5
5 2 11 3 7
Output
2 4
Input
6
2 5 10 1 10 2
Output
1 4 | instruction | 0 | 26,615 | 22 | 53,230 |
Tags: brute force, greedy, math, number theory
Correct Solution:
```
import math
from collections import defaultdict
import sys
input = sys.stdin.readline
def main():
n = int(input())
a = list(map(int, input().split()))
MAX = 10**7 + 1
res = MAX * MAX
#MAX_P = int(math.sqrt(MAX))
MAX_P = 3163
primes = []
p = 2
sieve = [True] * (MAX_P+1)
while p < MAX_P:
if sieve[p]:
primes.append(p)
k = 2
while k * p < MAX_P:
sieve[k * p] = False
k += 1
p += 1
np = len(primes)
cand1 = {}
cand2 = {}
ind1 = {}
ind2 = {}
res = MAX * MAX
for index in range(n):
val = a[index]
if val >= res:
continue
divisors = [1]
p = 0
while val > 0 and p < np:
while val % primes[p] == 0:
divisors += [d * primes[p] for d in divisors]
val //= primes[p]
p += 1
if val > 1:
divisors += [d * val for d in divisors]
for d in set(divisors):
if d not in cand1:
cand1[d] = a[index]
ind1[d] = index
else:
if d not in cand2:
if a[index] < cand1[d]:
cand2[d] = cand1[d]
ind2[d] = ind1[d]
cand1[d] = a[index]
ind1[d] = index
else:
cand2[d] = a[index]
ind2[d] = index
else:
if a[index] < cand1[d]:
cand2[d] = cand1[d]
ind2[d] = ind1[d]
cand1[d] = a[index]
ind1[d] = index
elif a[index] < cand2[d]:
cand2[d] = a[index]
ind2[d] = index
else:
continue
if res > cand1[d] // d * cand2[d]:
x, y = ind1[d], ind2[d]
res = cand1[d] // d * cand2[d]
print(min(x+1, y+1), max(x+1, y+1))
if __name__ == '__main__':
main()
``` | output | 1 | 26,615 | 22 | 53,231 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a consisting of n integers a_1, a_2, ..., a_n.
Your problem is to find such pair of indices i, j (1 ≤ i < j ≤ n) that lcm(a_i, a_j) is minimum possible.
lcm(x, y) is the least common multiple of x and y (minimum positive number such that both x and y are divisors of this number).
Input
The first line of the input contains one integer n (2 ≤ n ≤ 10^6) — the number of elements in a.
The second line of the input contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^7), where a_i is the i-th element of a.
Output
Print two integers i and j (1 ≤ i < j ≤ n) such that the value of lcm(a_i, a_j) is minimum among all valid pairs i, j. If there are multiple answers, you can print any.
Examples
Input
5
2 4 8 3 6
Output
1 2
Input
5
5 2 11 3 7
Output
2 4
Input
6
2 5 10 1 10 2
Output
1 4
Submitted Solution:
```
from collections import defaultdict
def gcd(a, b):
return b if a % b == 0 else gcd(b, a % b)
def lcm(a, b):
return a * b // gcd(a, b)
input()
nums = map(int, input().split())
cnt = defaultdict(lambda : [])
for i, v in enumerate(nums):
cnt[v] += [i]
maxv = max(cnt.keys())
done = False
ans_lcm = 1e20
ans = None
for i in range(1, maxv + 1):
prev = 0
prev_i = 0
for j in range(i, maxv + 1, i):
if j >= ans_lcm:
break
if j in cnt:
if len(cnt[j]) >= 2:
if j < ans_lcm:
ans_lcm = j
ans = cnt[j][:2]
break
elif prev:
candidate = lcm(j, prev)
if candidate < ans_lcm:
ans_lcm = candidate
ans = [prev_i, cnt[j][0]]
break
else:
prev = j
prev_i = cnt[j][0]
print(ans[0] + 1, ans[1] + 1)
``` | instruction | 0 | 26,617 | 22 | 53,234 |
No | output | 1 | 26,617 | 22 | 53,235 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a consisting of n integers a_1, a_2, ..., a_n.
Your problem is to find such pair of indices i, j (1 ≤ i < j ≤ n) that lcm(a_i, a_j) is minimum possible.
lcm(x, y) is the least common multiple of x and y (minimum positive number such that both x and y are divisors of this number).
Input
The first line of the input contains one integer n (2 ≤ n ≤ 10^6) — the number of elements in a.
The second line of the input contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^7), where a_i is the i-th element of a.
Output
Print two integers i and j (1 ≤ i < j ≤ n) such that the value of lcm(a_i, a_j) is minimum among all valid pairs i, j. If there are multiple answers, you can print any.
Examples
Input
5
2 4 8 3 6
Output
1 2
Input
5
5 2 11 3 7
Output
2 4
Input
6
2 5 10 1 10 2
Output
1 4
Submitted Solution:
```
def hcf(a,b):
if a==0:
return b
else:
return hcf(b%a,a)
def lcm(a,b):
return a*b//(hcf(a,b))
def get(list):
l=[0,0]
z=True
mi=lcm(list[0],list[1])
k=[]
for i in range(len(list)):
k.append(list[i])
list.sort()
i=1
while i<len(list) and z:
if list[i]>=mi:
z=False
else:
j=0
while j<i and z:
if lcm(min(list[i],list[j]),max(list[i],list[j]))<mi:
mi=lcm(min(list[i],list[j]),max(list[i],list[j]))
l[0],l[1]=list[j],list[i]
else:
pass
j+=1
i+=1
a,b=True,True
for i in range(len(k)):
if k[i]==l[0] and a:
l[0]=i+1
a=False
elif k[i]==l[1] and b:
l[1]=i+1
b=False
l.sort()
return l
n=int(input())
l=list(map(int,input().strip().split()))
print(" ".join(str(x) for x in get(l)))
``` | instruction | 0 | 26,618 | 22 | 53,236 |
No | output | 1 | 26,618 | 22 | 53,237 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a consisting of n integers a_1, a_2, ..., a_n.
Your problem is to find such pair of indices i, j (1 ≤ i < j ≤ n) that lcm(a_i, a_j) is minimum possible.
lcm(x, y) is the least common multiple of x and y (minimum positive number such that both x and y are divisors of this number).
Input
The first line of the input contains one integer n (2 ≤ n ≤ 10^6) — the number of elements in a.
The second line of the input contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^7), where a_i is the i-th element of a.
Output
Print two integers i and j (1 ≤ i < j ≤ n) such that the value of lcm(a_i, a_j) is minimum among all valid pairs i, j. If there are multiple answers, you can print any.
Examples
Input
5
2 4 8 3 6
Output
1 2
Input
5
5 2 11 3 7
Output
2 4
Input
6
2 5 10 1 10 2
Output
1 4
Submitted Solution:
```
# from bisect import bisect_left as bl
N = int(input())
M = 10**7+5
A = [int(a) for a in input().split()]
X = [0 for _ in range(M)]
for i in range(N):
X[A[i]] += 1
mi = 10**100
a, b = -1, -1
for i in range(1, M):
if X[i] >= 2:
if i < mi:
mi = i
a, b = i, i
else:
fi = 0
se = 0
for j in range(i, M, i):
if X[j]:
if fi == 0:
fi = j
else:
se = j
if fi*se//i < mi:
mi = fi*se//i
a, b = fi, se
break
i1, i2 = 0, 0
if a == b:
for i in range(N):
if A[i] == a:
if i1 == 0:
i1 = i + 1
else:
i2 = i + 1
print(i1, i2)
else:
for i in range(N):
if A[i] == a:
i1 = i + 1
if A[i] == b:
i2 = i + 1
print(min(i1, i2), max(i1, i2))
``` | instruction | 0 | 26,619 | 22 | 53,238 |
No | output | 1 | 26,619 | 22 | 53,239 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a consisting of n integers a_1, a_2, ..., a_n.
Your problem is to find such pair of indices i, j (1 ≤ i < j ≤ n) that lcm(a_i, a_j) is minimum possible.
lcm(x, y) is the least common multiple of x and y (minimum positive number such that both x and y are divisors of this number).
Input
The first line of the input contains one integer n (2 ≤ n ≤ 10^6) — the number of elements in a.
The second line of the input contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^7), where a_i is the i-th element of a.
Output
Print two integers i and j (1 ≤ i < j ≤ n) such that the value of lcm(a_i, a_j) is minimum among all valid pairs i, j. If there are multiple answers, you can print any.
Examples
Input
5
2 4 8 3 6
Output
1 2
Input
5
5 2 11 3 7
Output
2 4
Input
6
2 5 10 1 10 2
Output
1 4
Submitted Solution:
```
def lcm(a, b):
return a * b / gcd(a, b)
def gcd(a,b):
while b > 0: a, b = b, a % b
return a
import io, os
# input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
# input = io.StringIO(os.read(0, os.fstat(0).st_size).decode()).readline
ii = lambda:int(input())
kk=lambda:map(int,input().split())
# k2=lambda:map(lambda x:int(x)-1, input().split())
ll=lambda:list(kk())
n,ls=ii(),ll()
l2 = sorted(ls)
minLCM = lcm(l2[0], l2[1])
minv = (l2[0], l2[1])
testers = [l2[0]]
for i in range(1, n):
if l2[i] > minLCM: break
for t in testers:
lc = lcm(t, l2[i])
if lc < minLCM:
minLCM, minv = lc, (t, l2[i])
if lc == l2[i]:
break
else:
testers.append(t)
p1 = p2 = -1
for i in range(n):
if p1 == -1 and ls[i] == minv[0]: p1 = i+1
elif p2 == -1 and ls[i] == minv[1]: p2 = i+1
if p1 != -1 and p2 != -1: break
print(max(p1, p2), min(p1, p2))
``` | instruction | 0 | 26,620 | 22 | 53,240 |
No | output | 1 | 26,620 | 22 | 53,241 |
Provide tags and a correct Python 3 solution for this coding contest problem.
It is the hard version of the problem. The only difference is that in this version 3 ≤ k ≤ n.
You are given a positive integer n. Find k positive integers a_1, a_2, …, a_k, such that:
* a_1 + a_2 + … + a_k = n
* LCM(a_1, a_2, …, a_k) ≤ n/2
Here LCM is the [least common multiple](https://en.wikipedia.org/wiki/Least_common_multiple) of numbers a_1, a_2, …, a_k.
We can show that for given constraints the answer always exists.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The only line of each test case contains two integers n, k (3 ≤ n ≤ 10^9, 3 ≤ k ≤ n).
It is guaranteed that the sum of k over all test cases does not exceed 10^5.
Output
For each test case print k positive integers a_1, a_2, …, a_k, for which all conditions are satisfied.
Example
Input
2
6 4
9 5
Output
1 2 2 1
1 3 3 1 1 | instruction | 0 | 26,758 | 22 | 53,516 |
Tags: constructive algorithms, math
Correct Solution:
```
for _ in range(int(input())):
n,k=map(int,input().split())
b=[1]*(k-3)
n-=(k-3)
if n%3==0:
b.append(n//3)
b.append(n//3)
b.append(n//3)
else:
if n%2!=0:
b.append(1)
b.append((n-1)//2)
b.append((n-1)//2)
else:
a=n//2
if a%2==0:
a-=1
while(max(a,(n-2*a))%min(a,(n-2*a))!=0):
a-=1
else:
a-=1
b.append(a)
b.append(a)
b.append(n-2*a)
print(*b)
``` | output | 1 | 26,758 | 22 | 53,517 |
Provide tags and a correct Python 3 solution for this coding contest problem.
It is the hard version of the problem. The only difference is that in this version 3 ≤ k ≤ n.
You are given a positive integer n. Find k positive integers a_1, a_2, …, a_k, such that:
* a_1 + a_2 + … + a_k = n
* LCM(a_1, a_2, …, a_k) ≤ n/2
Here LCM is the [least common multiple](https://en.wikipedia.org/wiki/Least_common_multiple) of numbers a_1, a_2, …, a_k.
We can show that for given constraints the answer always exists.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The only line of each test case contains two integers n, k (3 ≤ n ≤ 10^9, 3 ≤ k ≤ n).
It is guaranteed that the sum of k over all test cases does not exceed 10^5.
Output
For each test case print k positive integers a_1, a_2, …, a_k, for which all conditions are satisfied.
Example
Input
2
6 4
9 5
Output
1 2 2 1
1 3 3 1 1 | instruction | 0 | 26,759 | 22 | 53,518 |
Tags: constructive algorithms, math
Correct Solution:
```
for nt in range(int(input())):
n, k = map(int,input().split())
ans = [1]*(k-3)
n -= (k-3)
if n%2:
ans.extend([1, n//2, n//2])
else:
if n//2%2:
ans.extend([2, n//2-1, n//2-1])
else:
ans.extend([n//2, n//4, n//4])
print (*ans)
``` | output | 1 | 26,759 | 22 | 53,519 |
Provide tags and a correct Python 3 solution for this coding contest problem.
It is the hard version of the problem. The only difference is that in this version 3 ≤ k ≤ n.
You are given a positive integer n. Find k positive integers a_1, a_2, …, a_k, such that:
* a_1 + a_2 + … + a_k = n
* LCM(a_1, a_2, …, a_k) ≤ n/2
Here LCM is the [least common multiple](https://en.wikipedia.org/wiki/Least_common_multiple) of numbers a_1, a_2, …, a_k.
We can show that for given constraints the answer always exists.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The only line of each test case contains two integers n, k (3 ≤ n ≤ 10^9, 3 ≤ k ≤ n).
It is guaranteed that the sum of k over all test cases does not exceed 10^5.
Output
For each test case print k positive integers a_1, a_2, …, a_k, for which all conditions are satisfied.
Example
Input
2
6 4
9 5
Output
1 2 2 1
1 3 3 1 1 | instruction | 0 | 26,760 | 22 | 53,520 |
Tags: constructive algorithms, math
Correct Solution:
```
import sys
from sys import stdout
import bisect as bi
import math
from collections import defaultdict as dd
from types import GeneratorType
##import queue
##from heapq import heapify, heappush, heappop
##import itertools
##import io
##import os
##import operator
import random
##sys.setrecursionlimit(10**7)
input=sys.stdin.readline
##input = io.BytesIsoO(os.read(0, os.fstat(0).st_size)).readline
##fo=open("output2.txt","w")
##fi=open("input2.txt","w")
mo= 10**9+7
MOD=998244353
def cin():return map(int,sin().split())
def ain():return list(map(int,sin().split()))
def sin():return input().strip()
def inin():return int(input())
##----------------------------------------------------------------------------------------------------------------------
##for _ in range(inin()):
## n=inin()
## d=dd(int)
## l=ain()
## for i in l:
## d[i]+=1
## ke=list(d.keys())
## ke.sort()
## ans=ke[::]
## for i in ke:
## if(d[i]>1):
## ans+=[i]*(d[i]-1)
##
## print(*ans)
##for _ in range(inin()):
## n,m=cin()
## l=ain()
## d=dd(int)
## for i in l:
## d[i%m]+=1
## ans=0
## if(d[0]):ans+=1
## for i in range(1,m//2+1):
## if(d[i]>0):
## if(m-i!=i):
## if(d[m-i]):
## ans+=1
## mi=min(d[i],d[m-i])
## if(mi==d[i]):
## ans+=max(d[m-i]-mi-1,0)
## else:
## ans+=max(d[i]-mi-1,0)
## else:
## ans+=d[i]
## elif(m-i==i):
## ans+=1
## elif(d[m-i]):
## ans+=d[m-i]
## print(ans)
##for _ in range(inin()):
## n,k=cin()
## if(n%2):
## print(n//2,n//2,1)
## else:
## if(n%4==0):
## print(n//2,n//4,n//4)
## else:
## print(n//2-1,n//2-1,2)
def solve(n,k):
if(n%2):
return [n//2,n//2,1]
else:
if(n%4==0):
return [n//2,n//4,n//4]
else:
return [n//2-1,n//2-1,2]
for _ in range(inin()):
n,k=cin()
ans=[]
newn,newk=n-(k-3),3
print(*(solve(newn,newk)+[1]*(k-3)))
##-----------------------------------------------------------------------------------------------------------------------#
def msb(n):n|=n>>1;n|=n>>2;n|=n>>4;n|=n>>8;n|=n>>16;n|=n>>32;n|=n>>64;return n-(n>>1)
##2 ki power
def pref(a,n,f):
pre=[0]*n
if(f==0): ##from beginning
pre[0]=a[0]
for i in range(1,n):
pre[i]=a[i]+pre[i-1]
else: ##from end
pre[-1]=a[-1]
for i in range(n-2,-1,-1):
pre[i]=pre[i+1]+a[i]
return pre
# in given set of integers
def kadane(A):
maxSoFar = maxEndingHere = start = end = beg = 0
for i in range(len(A)):
maxEndingHere = maxEndingHere + A[i]
if maxEndingHere < 0:
maxEndingHere = 0;beg = i + 1
if maxSoFar < maxEndingHere:
maxSoFar = maxEndingHere
start = beg
end = i
return (maxSoFar,start,end) #max subarray sum and its range
def modFact(n, p):
if(n<0):return 0
if n >= p: return 0
result = 1
for i in range(1, n + 1):result = (result * i) % p
return result
def ncr(n, r, p):
if(n<r or n<0): return 0
num = den = 1
for i in range(r):
num = (num * (n - i)) % p ;den = (den * (i + 1)) % p
return (num * pow(den,p - 2, p)) % p
##https://github.com/cheran-senthil/PyRival/blob/master/pyrival/misc/bootstrap.py
##write @bootstrap before recursive func
def bootstrap(f, stack=[]):
def wrappedfunc(*args, **kwargs):
if stack:
return f(*args, **kwargs)
else:
to = f(*args, **kwargs)
while True:
if type(to) is GeneratorType:
stack.append(to)
to = next(to)
else:
stack.pop()
if not stack:
break
to = stack[-1].send(to)
return to
return wrappedfunc
``` | output | 1 | 26,760 | 22 | 53,521 |
Provide tags and a correct Python 3 solution for this coding contest problem.
It is the hard version of the problem. The only difference is that in this version 3 ≤ k ≤ n.
You are given a positive integer n. Find k positive integers a_1, a_2, …, a_k, such that:
* a_1 + a_2 + … + a_k = n
* LCM(a_1, a_2, …, a_k) ≤ n/2
Here LCM is the [least common multiple](https://en.wikipedia.org/wiki/Least_common_multiple) of numbers a_1, a_2, …, a_k.
We can show that for given constraints the answer always exists.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The only line of each test case contains two integers n, k (3 ≤ n ≤ 10^9, 3 ≤ k ≤ n).
It is guaranteed that the sum of k over all test cases does not exceed 10^5.
Output
For each test case print k positive integers a_1, a_2, …, a_k, for which all conditions are satisfied.
Example
Input
2
6 4
9 5
Output
1 2 2 1
1 3 3 1 1 | instruction | 0 | 26,761 | 22 | 53,522 |
Tags: constructive algorithms, math
Correct Solution:
```
for _ in range(int(input())):
n, m = map(int, input().split())
n -= (m - 3)
if n % 2 == 1:
print(n // 2, n // 2, 1, end=' ')
elif n % 4 == 2:
print(n // 2 - 1, n // 2 - 1, 2, end=' ')
else:
print(n // 2, n // 4, n // 4, end=' ')
for i in range(m - 3):
print(1, end=' ')
print()
``` | output | 1 | 26,761 | 22 | 53,523 |
Provide tags and a correct Python 3 solution for this coding contest problem.
It is the hard version of the problem. The only difference is that in this version 3 ≤ k ≤ n.
You are given a positive integer n. Find k positive integers a_1, a_2, …, a_k, such that:
* a_1 + a_2 + … + a_k = n
* LCM(a_1, a_2, …, a_k) ≤ n/2
Here LCM is the [least common multiple](https://en.wikipedia.org/wiki/Least_common_multiple) of numbers a_1, a_2, …, a_k.
We can show that for given constraints the answer always exists.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The only line of each test case contains two integers n, k (3 ≤ n ≤ 10^9, 3 ≤ k ≤ n).
It is guaranteed that the sum of k over all test cases does not exceed 10^5.
Output
For each test case print k positive integers a_1, a_2, …, a_k, for which all conditions are satisfied.
Example
Input
2
6 4
9 5
Output
1 2 2 1
1 3 3 1 1 | instruction | 0 | 26,762 | 22 | 53,524 |
Tags: constructive algorithms, math
Correct Solution:
```
import sys
input = sys.stdin.readline
for _ in range(int(input())):
n, k = map(int, input().split())
n -= (k - 3)
if n % 2:
print(n // 2, n // 2, 1, *[1 for i in range(k - 3)])
else:
if n % 4:
print(n // 2 - 1, n // 2 - 1, 2, *[1 for i in range(k - 3)])
else:
print(n // 2, n // 4, n // 4, *[1 for i in range(k - 3)])
``` | output | 1 | 26,762 | 22 | 53,525 |
Provide tags and a correct Python 3 solution for this coding contest problem.
It is the hard version of the problem. The only difference is that in this version 3 ≤ k ≤ n.
You are given a positive integer n. Find k positive integers a_1, a_2, …, a_k, such that:
* a_1 + a_2 + … + a_k = n
* LCM(a_1, a_2, …, a_k) ≤ n/2
Here LCM is the [least common multiple](https://en.wikipedia.org/wiki/Least_common_multiple) of numbers a_1, a_2, …, a_k.
We can show that for given constraints the answer always exists.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The only line of each test case contains two integers n, k (3 ≤ n ≤ 10^9, 3 ≤ k ≤ n).
It is guaranteed that the sum of k over all test cases does not exceed 10^5.
Output
For each test case print k positive integers a_1, a_2, …, a_k, for which all conditions are satisfied.
Example
Input
2
6 4
9 5
Output
1 2 2 1
1 3 3 1 1 | instruction | 0 | 26,763 | 22 | 53,526 |
Tags: constructive algorithms, math
Correct Solution:
```
from sys import stdin
# input=stdin.buffer.readline
input=lambda : stdin.readline().strip()
lin=lambda :list(map(int,input().split()))
iin=lambda :int(input())
main=lambda :map(int,input().split())
from math import ceil,sqrt,factorial,log
from collections import deque
from bisect import bisect_left
mod=998244353
mod=1000000007
def gcd(a,b):
a,b=max(a,b),min(a,b)
while a%b!=0:
a,b=b,a%b
return b
def moduloinverse(a):
return(pow(a,mod-2,mod))
def solve(we):
n,k=main()
d=[]
for i in range(k-3):
d.append(1)
n=n-sum(d)
z=[]
if n==3:
z=[1,1,1]
elif n==4:
z=[1,1,2]
else:
if n%2!=0:
z=[n//2,n//2,1]
else:
t=n//2
if t%2==0:
if n%3==0:
z=[n//3,n//3,n//3]
elif n%4==0:
z=[n//2,t//2,t//2]
else:
z=[t-1,t-1,2]
print(*d,*z)
qwe=iin()
for _ in range(qwe):
solve(_+1)
``` | output | 1 | 26,763 | 22 | 53,527 |
Provide tags and a correct Python 3 solution for this coding contest problem.
It is the hard version of the problem. The only difference is that in this version 3 ≤ k ≤ n.
You are given a positive integer n. Find k positive integers a_1, a_2, …, a_k, such that:
* a_1 + a_2 + … + a_k = n
* LCM(a_1, a_2, …, a_k) ≤ n/2
Here LCM is the [least common multiple](https://en.wikipedia.org/wiki/Least_common_multiple) of numbers a_1, a_2, …, a_k.
We can show that for given constraints the answer always exists.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The only line of each test case contains two integers n, k (3 ≤ n ≤ 10^9, 3 ≤ k ≤ n).
It is guaranteed that the sum of k over all test cases does not exceed 10^5.
Output
For each test case print k positive integers a_1, a_2, …, a_k, for which all conditions are satisfied.
Example
Input
2
6 4
9 5
Output
1 2 2 1
1 3 3 1 1 | instruction | 0 | 26,764 | 22 | 53,528 |
Tags: constructive algorithms, math
Correct Solution:
```
import sys
from functools import lru_cache, cmp_to_key
from heapq import merge, heapify, heappop, heappush
from math import *
from collections import defaultdict as dd, deque, Counter as C
from itertools import combinations as comb, permutations as perm
from bisect import bisect_left as bl, bisect_right as br, bisect, insort
from time import perf_counter
from fractions import Fraction
import copy
from copy import deepcopy
import time
starttime = time.time()
mod = int(pow(10, 9) + 7)
mod2 = 998244353
def data(): return sys.stdin.readline().strip()
def out(*var, end="\n"): sys.stdout.write(' '.join(map(str, var))+end)
def L(): return list(sp())
def sl(): return list(ssp())
def sp(): return map(int, data().split())
def ssp(): return map(str, data().split())
def l1d(n, val=0): return [val for i in range(n)]
def l2d(n, m, val=0): return [l1d(n, val) for j in range(m)]
try:
# sys.setrecursionlimit(int(pow(10,6)))
sys.stdin = open("input.txt", "r")
# sys.stdout = open("../output.txt", "w")
except:
pass
def pmat(A):
for ele in A:
print(*ele,end="\n")
for _ in range(L()[0]):
n,k=L()
A=[1]*(k-3)
n-=(k-3)
if n%4==0:
A+=[n//2,n//4,n//4]
elif n%2==0:
A+=[2,(n-2)//2,(n-2)//2]
else:
A+=[1,(n-1)//2,(n-1)//2]
print(*A)
endtime = time.time()
# print(f"Runtime of the program is {endtime - starttime}")
``` | output | 1 | 26,764 | 22 | 53,529 |
Provide tags and a correct Python 3 solution for this coding contest problem.
It is the hard version of the problem. The only difference is that in this version 3 ≤ k ≤ n.
You are given a positive integer n. Find k positive integers a_1, a_2, …, a_k, such that:
* a_1 + a_2 + … + a_k = n
* LCM(a_1, a_2, …, a_k) ≤ n/2
Here LCM is the [least common multiple](https://en.wikipedia.org/wiki/Least_common_multiple) of numbers a_1, a_2, …, a_k.
We can show that for given constraints the answer always exists.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The only line of each test case contains two integers n, k (3 ≤ n ≤ 10^9, 3 ≤ k ≤ n).
It is guaranteed that the sum of k over all test cases does not exceed 10^5.
Output
For each test case print k positive integers a_1, a_2, …, a_k, for which all conditions are satisfied.
Example
Input
2
6 4
9 5
Output
1 2 2 1
1 3 3 1 1 | instruction | 0 | 26,765 | 22 | 53,530 |
Tags: constructive algorithms, math
Correct Solution:
```
from collections import Counter
t = int(input())
for _ in range(t):
n,k = map(int, input().split())
for i in range(k-3):
print(1,end=" ")
n = n- (k-3)
k =3
if(n%2 == 0):
q = n//2
if(q %2 != 0):
print(2,q-1,q-1)
else:
print(q//2,q//2,q)
else:
q = n//2
w = n- (q*2)
print(w,q,q)
``` | output | 1 | 26,765 | 22 | 53,531 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It is the hard version of the problem. The only difference is that in this version 3 ≤ k ≤ n.
You are given a positive integer n. Find k positive integers a_1, a_2, …, a_k, such that:
* a_1 + a_2 + … + a_k = n
* LCM(a_1, a_2, …, a_k) ≤ n/2
Here LCM is the [least common multiple](https://en.wikipedia.org/wiki/Least_common_multiple) of numbers a_1, a_2, …, a_k.
We can show that for given constraints the answer always exists.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The only line of each test case contains two integers n, k (3 ≤ n ≤ 10^9, 3 ≤ k ≤ n).
It is guaranteed that the sum of k over all test cases does not exceed 10^5.
Output
For each test case print k positive integers a_1, a_2, …, a_k, for which all conditions are satisfied.
Example
Input
2
6 4
9 5
Output
1 2 2 1
1 3 3 1 1
Submitted Solution:
```
"""
#If FastIO not needed, use this and don't forget to strip
#import sys, math
#input = sys.stdin.readline
"""
import os, sys, heapq as h, time
from io import BytesIO, IOBase
from types import GeneratorType
from bisect import bisect_left, bisect_right
from collections import defaultdict as dd, deque as dq, Counter as dc
import math, string
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
import os
self.os = os
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 = self.os.read(self._fd, max(self.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 = self.os.read(self._fd, max(self.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:
self.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")
#start_time = time.time()
def getInt(): return int(input())
def getStrs(): return input().split()
def getInts(): return list(map(int,input().split()))
def getStr(): return input()
def listStr(): return list(input())
def getMat(n): return [getInts() for _ in range(n)]
def isInt(s): return '0' <= s[0] <= '9'
MOD = 10**9 + 7
"""
What if N = 4
4
1 1 1 1
5
1 1 1 2
6
1 1 2 2
7
1 2 2 2
8
2 2 2 2
9
4 2 2 1
10
4 2 2 2
11
4 4 2 1
12
4 4 2 2
13
6 3 3 1
14
6 3 3 2
15
6 6 2 1
16
6 6 3 1
17
8 4 4 1
18
8 4 4 2
19
8 8 2 1
20
8 8 2 2
21
10 5 5 1
22
10 5 5 2
23
10 10 2 1
24
10 10 2 2
11, 4
2.75 <= big <= 5.5
So we have 3,4,5 to work with
The biggest number must be >= N/K and <= N/2
The smallest number must be <= N/K
2 1 1 1 1
2 2 1 1 1
2 2 2 1 1
2 2 2 2 1
2 2 2 2 2
4 2 2 2 1
4 2 2 2 2
4 4 2 2 1
4 4 2 2 2
4 4 4 2 1
Biggest number should be math.floor((N-1)/K)*2
We're going to sum up the numbers as follows
X, X, .., X, X/2, ..., X/2, 1, ..., 1
If it's even, we can make it out of 2s and some other number
If it's odd, we can make it out of a 1, 2s and some other number
1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 2
1 1 1 1 1 1 2 2
1 1 1 1 1 2 2 2
...
2 2 2 2 2 2 2 2
1 2 2 2 2 2 2 4
33 7
(33-1)/7 = 4
Highest power of 2 I can use
100001
00241
111111
63
1
111110
As an extension of the three problem, it's
K-3 ones, then solve the original
"""
def solve():
N, K = getInts()
ans = [1]*(K-3)
N -= (K-3)
if N % 2 == 1:
ans2 = [1,N//2,N//2]
elif N % 4 == 0:
ans2 = [N//4,N//4,N//2]
else:
ans2 = [(N-2)//2,(N-2)//2,2]
print(*(ans+ans2))
return
for _ in range(getInt()):
#print(solve())
solve()
#print(time.time()-start_time)
``` | instruction | 0 | 26,769 | 22 | 53,538 |
Yes | output | 1 | 26,769 | 22 | 53,539 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It is the hard version of the problem. The only difference is that in this version 3 ≤ k ≤ n.
You are given a positive integer n. Find k positive integers a_1, a_2, …, a_k, such that:
* a_1 + a_2 + … + a_k = n
* LCM(a_1, a_2, …, a_k) ≤ n/2
Here LCM is the [least common multiple](https://en.wikipedia.org/wiki/Least_common_multiple) of numbers a_1, a_2, …, a_k.
We can show that for given constraints the answer always exists.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The only line of each test case contains two integers n, k (3 ≤ n ≤ 10^9, 3 ≤ k ≤ n).
It is guaranteed that the sum of k over all test cases does not exceed 10^5.
Output
For each test case print k positive integers a_1, a_2, …, a_k, for which all conditions are satisfied.
Example
Input
2
6 4
9 5
Output
1 2 2 1
1 3 3 1 1
Submitted Solution:
```
import sys
import math
from itertools import permutations
from bisect import bisect_left
import heapq
from collections import deque
def II():
return int(sys.stdin.readline())
def LI():
return list(map(int, sys.stdin.readline().split()))
def MI():
return map(int, sys.stdin.readline().split())
def SI():
return sys.stdin.readline().strip()
def FACT(n, mod):
s = 1
facts = [1]
for i in range(1,n+1):
s*=i
s%=mod
facts.append(s)
return facts[n]
def C(n, k, mod):
return (FACT(n,mod) * pow((FACT(k,mod)*FACT(n-k,mod))%mod,mod-2, mod))%mod
def lcm(a,b):
return abs(a*b) // math.gcd(a, b)
for _ in range(II()):
n,k = MI()
ans = []
if n%2 == k%2:
ans = [1 for i in range(k-2)]
ans+=[(n-k+2)//2]*2
elif (n-k-1)%2 == 0:
ans = [2]
for i in range(k-3):
ans.append(1)
ans+=[(n-sum(ans))//2]*2
else:
ans = [2, 2]
for i in range(k-4):
ans.append(1)
ans+=[(n-sum(ans))//2]*2
print(*ans)
``` | instruction | 0 | 26,770 | 22 | 53,540 |
No | output | 1 | 26,770 | 22 | 53,541 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a of n integers.
You want to make all elements of a equal to zero by doing the following operation exactly three times:
* Select a segment, for each number in this segment we can add a multiple of len to it, where len is the length of this segment (added integers can be different).
It can be proven that it is always possible to make all elements of a equal to zero.
Input
The first line contains one integer n (1 ≤ n ≤ 100 000): the number of elements of the array.
The second line contains n elements of an array a separated by spaces: a_1, a_2, ..., a_n (-10^9 ≤ a_i ≤ 10^9).
Output
The output should contain six lines representing three operations.
For each operation, print two lines:
* The first line contains two integers l, r (1 ≤ l ≤ r ≤ n): the bounds of the selected segment.
* The second line contains r-l+1 integers b_l, b_{l+1}, ..., b_r (-10^{18} ≤ b_i ≤ 10^{18}): the numbers to add to a_l, a_{l+1}, …, a_r, respectively; b_i should be divisible by r - l + 1.
Example
Input
4
1 3 2 4
Output
1 1
-1
3 4
4 2
2 4
-3 -6 -6 | instruction | 0 | 27,608 | 22 | 55,216 |
Tags: constructive algorithms, greedy, number theory
Correct Solution:
```
import sys
#comment these out later
#sys.stdin = open("in.in", "r")
#sys.stdout = open("out.out", "w")
import operator as op
from functools import reduce
def gcd(x, y):
"""greatest common divisor of x and y"""
while y:
x, y = y, x % y
return x
def chinese_remainder(a, p):
"""returns x s.t. x = a[i] (mod p[i]) where p[i] is prime for all i"""
prod = reduce(op.mul, p, 1)
x = [prod // pi for pi in p]
return sum(a[i] * pow(x[i], p[i] - 2, p[i]) * x[i] for i in range(len(a))) % prod
def extended_gcd(a, b):
"""returns gcd(a, b), s, r s.t. a * s + b * r == gcd(a, b)"""
s, old_s = 0, 1
r, old_r = b, a
while r:
q = old_r // r
old_r, r = r, old_r - q * r
old_s, s = s, old_s - q * s
return old_r, old_s, (old_r - old_s * a) // b if b else 0
def composite_crt(b, m):
"""returns x s.t. x = b[i] (mod m[i]) for all i"""
x, m_prod = 0, 1
for bi, mi in zip(b, m):
g, s, _ = extended_gcd(m_prod, mi)
if ((bi - x) % mi) % g:
return None
x += m_prod * (s * ((bi - x) % mi) // g)
m_prod = (m_prod * mi) // gcd(m_prod, mi)
return x % m_prod
inp = [int(x) for x in sys.stdin.read().split()]; ii = 0
n = inp[ii]; ii += 1
p = n-1
ar = inp[ii:ii+n]
if n == 1:
x = ar[0]
print(1, 1)
print(0)
print(1, 1)
print(0)
print(1, 1)
print(-x)
exit()
#first operation
toprint1 = ["0"]
toprint2 = []
for x in ar[1:]:
target = composite_crt([x, 0], [n, n-1])
toadd = target - x
toprint1.append(str(toadd))
toprint2.append(str(-target))
print(1, n)
print(" ".join(toprint1))
#second operation
print(1, 1)
print(-ar[0])
#third operation
print(2, n)
print(" ".join(toprint2))
``` | output | 1 | 27,608 | 22 | 55,217 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Unfortunately, Vasya can only sum pairs of integers (a, b), such that for any decimal place at least one number has digit 0 in this place. For example, Vasya can sum numbers 505 and 50, but he cannot sum 1 and 4.
Vasya has a set of k distinct non-negative integers d1, d2, ..., dk.
Vasya wants to choose some integers from this set so that he could sum any two chosen numbers. What maximal number of integers can he choose in the required manner?
Input
The first input line contains integer k (1 ≤ k ≤ 100) — the number of integers.
The second line contains k distinct space-separated integers d1, d2, ..., dk (0 ≤ di ≤ 100).
Output
In the first line print a single integer n the maximum number of the chosen integers. In the second line print n distinct non-negative integers — the required integers.
If there are multiple solutions, print any of them. You can print the numbers in any order.
Examples
Input
4
100 10 1 0
Output
4
0 1 10 100
Input
3
2 70 3
Output
2
2 70
Submitted Solution:
```
import sys
import math
from sys import stdin, stdout
# Most Frequently Used Number Theory Concepts
# SOE(helper function of get gcd)
def sieve(N):
primeNumbers = [True]*(N+1)
primeNumbers[0] = False
primeNumbers[1] = False
i = 2
while i*i <= N:
j = i
if primeNumbers[j]:
while j*i <= N:
primeNumbers[j*i] = False
j += 1
i += 1
return primeNumbers
# get prime number form 1 to N
def getPrime(N):
primes = sieve(N)
result = []
for i in range(len(primes)):
if primes[i]:
result.append(i)
return result
# factors of n
def factor(N):
factors = []
i = 1
while i*i <= N:
if N % i == 0:
factors.append(i)
if i != N//i:
factors.append(N//i)
i += 1
return sorted(factors)
# reduce a string in number under range(1000000007)
def reduceB(b) :
# Initialize result
mod = 0
# Calculating mod of b with a
# to make b like 0 <= b < a
for i in range(0, len(b)) :
mod = (mod * 10 + ord(b[i])) % 1000000007
# print(b[i], ord(b[i]), mod)
return mod # return modulo
# gcd of two numbers
def gcd(a, b):
if a < b:
return gcd(b, a)
if b == 0:
return a
return gcd(b, a % b)
# TAKE INPUT
def get_ints_in_variables():
return map(int, sys.stdin.readline().strip().split())
def get_int(): return int(input())
def get_ints_in_list(): return list(
map(int, sys.stdin.readline().strip().split()))
def get_list_of_list(n): return [list(
map(int, sys.stdin.readline().strip().split())) for _ in range(n)]
def get_string(): return sys.stdin.readline().strip()
def containsZero(a):
if a == 0:
return True
while a != 0:
diff = a % 10
if diff == 0:
return True
a = int(a/10)
return False
def main():
# Write Your Code Here
k = int(input())
d = get_ints_in_list()
res = []
for i in range(0, 5):
res.append(-1)
for i in range(0, k):
# if value is 0 always include
if d[i] == 0:
res[0] = d[i]
continue
# if value is 100 always include
if d[i] == 100:
res[1] = d[i]
continue
# if value is divisible by 10 include
if d[i] % 10 == 0:
res[2] = d[i]
continue
# if value is less than 10 include
if d[i] < 10:
res[3] = d[i]
continue
# if value is greater than 10 include
res[4] = d[i]
# print(res)
ans = []
if res[0] != -1:
ans.append(res[0])
if res[1] != -1:
ans.append(res[1])
if res[2] != -1:
ans.append(res[2])
if res[3] != -1:
ans.append(res[3])
if res[2] == -1 and res[3] == -1 and res[4] != -1:
ans.append(res[4])
print(len(ans))
for i in range(0, len(ans)):
print(ans[i], end=" ")
print()
# for printing format
# print("Case #{}: {}".format(t+1, ans))
# calling main Function
if __name__ == "__main__":
main()
``` | instruction | 0 | 27,772 | 22 | 55,544 |
Yes | output | 1 | 27,772 | 22 | 55,545 |
Provide a correct Python 3 solution for this coding contest problem.
How Many Divisors?
Write a program which reads three integers a, b and c, and prints the number of divisors of c between a and b.
Constraints
* 1 ≤ a, b, c ≤ 10000
* a ≤ b
Input
Three integers a, b and c are given in a line separated by a single space.
Output
Print the number of divisors in a line.
Example
Input
5 14 80
Output
3 | instruction | 0 | 28,236 | 22 | 56,472 |
"Correct Solution:
```
a,b,c=map(int,input().split())
yaku=0
for i in range(a,b+1):
if(c%i==0):
yaku+=1
print(yaku)
``` | output | 1 | 28,236 | 22 | 56,473 |
Provide a correct Python 3 solution for this coding contest problem.
How Many Divisors?
Write a program which reads three integers a, b and c, and prints the number of divisors of c between a and b.
Constraints
* 1 ≤ a, b, c ≤ 10000
* a ≤ b
Input
Three integers a, b and c are given in a line separated by a single space.
Output
Print the number of divisors in a line.
Example
Input
5 14 80
Output
3 | instruction | 0 | 28,237 | 22 | 56,474 |
"Correct Solution:
```
a, b, num = map(int, input().split())
print(len([i for i in range(a, b+1) if num%i == 0]))
``` | output | 1 | 28,237 | 22 | 56,475 |
Provide a correct Python 3 solution for this coding contest problem.
How Many Divisors?
Write a program which reads three integers a, b and c, and prints the number of divisors of c between a and b.
Constraints
* 1 ≤ a, b, c ≤ 10000
* a ≤ b
Input
Three integers a, b and c are given in a line separated by a single space.
Output
Print the number of divisors in a line.
Example
Input
5 14 80
Output
3 | instruction | 0 | 28,238 | 22 | 56,476 |
"Correct Solution:
```
a,b,c=map(int,input().split())
print(len([x for x in range(a,b+1)if c%x==0]))
``` | output | 1 | 28,238 | 22 | 56,477 |
Provide a correct Python 3 solution for this coding contest problem.
How Many Divisors?
Write a program which reads three integers a, b and c, and prints the number of divisors of c between a and b.
Constraints
* 1 ≤ a, b, c ≤ 10000
* a ≤ b
Input
Three integers a, b and c are given in a line separated by a single space.
Output
Print the number of divisors in a line.
Example
Input
5 14 80
Output
3 | instruction | 0 | 28,239 | 22 | 56,478 |
"Correct Solution:
```
a,b,c=map(int,input().split())
n=0
for i in range(a,b+1):
if (c%i)==0:
n+=1
print(n)
``` | output | 1 | 28,239 | 22 | 56,479 |
Provide a correct Python 3 solution for this coding contest problem.
How Many Divisors?
Write a program which reads three integers a, b and c, and prints the number of divisors of c between a and b.
Constraints
* 1 ≤ a, b, c ≤ 10000
* a ≤ b
Input
Three integers a, b and c are given in a line separated by a single space.
Output
Print the number of divisors in a line.
Example
Input
5 14 80
Output
3 | instruction | 0 | 28,240 | 22 | 56,480 |
"Correct Solution:
```
cnt=0
a,b,c=map(int,input().split())
for i in range(a,b+1):
if (c%i==0):
cnt+=1
print(cnt)
``` | output | 1 | 28,240 | 22 | 56,481 |
Provide a correct Python 3 solution for this coding contest problem.
How Many Divisors?
Write a program which reads three integers a, b and c, and prints the number of divisors of c between a and b.
Constraints
* 1 ≤ a, b, c ≤ 10000
* a ≤ b
Input
Three integers a, b and c are given in a line separated by a single space.
Output
Print the number of divisors in a line.
Example
Input
5 14 80
Output
3 | instruction | 0 | 28,241 | 22 | 56,482 |
"Correct Solution:
```
a, b, c = [int(i) for i in input().split()]
print(len([i for i in range(a, b+1) if c % i == 0]))
``` | output | 1 | 28,241 | 22 | 56,483 |
Provide a correct Python 3 solution for this coding contest problem.
How Many Divisors?
Write a program which reads three integers a, b and c, and prints the number of divisors of c between a and b.
Constraints
* 1 ≤ a, b, c ≤ 10000
* a ≤ b
Input
Three integers a, b and c are given in a line separated by a single space.
Output
Print the number of divisors in a line.
Example
Input
5 14 80
Output
3 | instruction | 0 | 28,242 | 22 | 56,484 |
"Correct Solution:
```
a, b, c = map(int, input().split())
ans = 0
for i in range(a, b+1):
ans += c // i * i == c
print(ans)
``` | output | 1 | 28,242 | 22 | 56,485 |
Provide a correct Python 3 solution for this coding contest problem.
How Many Divisors?
Write a program which reads three integers a, b and c, and prints the number of divisors of c between a and b.
Constraints
* 1 ≤ a, b, c ≤ 10000
* a ≤ b
Input
Three integers a, b and c are given in a line separated by a single space.
Output
Print the number of divisors in a line.
Example
Input
5 14 80
Output
3 | instruction | 0 | 28,243 | 22 | 56,486 |
"Correct Solution:
```
a,b,c=map(int, input().split())
print(sum(c%i==0 for i in range(a,b+1)))
``` | output | 1 | 28,243 | 22 | 56,487 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
How Many Divisors?
Write a program which reads three integers a, b and c, and prints the number of divisors of c between a and b.
Constraints
* 1 ≤ a, b, c ≤ 10000
* a ≤ b
Input
Three integers a, b and c are given in a line separated by a single space.
Output
Print the number of divisors in a line.
Example
Input
5 14 80
Output
3
Submitted Solution:
```
a,b,c=map(int,input().split(' '))
d = 0
while a<=b:
if c%a==0:
d +=1
a+=1
else:
a+=1
print(d)
``` | instruction | 0 | 28,244 | 22 | 56,488 |
Yes | output | 1 | 28,244 | 22 | 56,489 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
How Many Divisors?
Write a program which reads three integers a, b and c, and prints the number of divisors of c between a and b.
Constraints
* 1 ≤ a, b, c ≤ 10000
* a ≤ b
Input
Three integers a, b and c are given in a line separated by a single space.
Output
Print the number of divisors in a line.
Example
Input
5 14 80
Output
3
Submitted Solution:
```
A, B, C = [int(_) for _ in input().split()]
print(sum(C % i == 0 for i in range(A, B + 1)))
``` | instruction | 0 | 28,245 | 22 | 56,490 |
Yes | output | 1 | 28,245 | 22 | 56,491 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
How Many Divisors?
Write a program which reads three integers a, b and c, and prints the number of divisors of c between a and b.
Constraints
* 1 ≤ a, b, c ≤ 10000
* a ≤ b
Input
Three integers a, b and c are given in a line separated by a single space.
Output
Print the number of divisors in a line.
Example
Input
5 14 80
Output
3
Submitted Solution:
```
x=0
a,b,c = map(int, input().split())
for i in range(a, b+1):
if c%i==0:
x=x+1
print(x)
``` | instruction | 0 | 28,246 | 22 | 56,492 |
Yes | output | 1 | 28,246 | 22 | 56,493 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
How Many Divisors?
Write a program which reads three integers a, b and c, and prints the number of divisors of c between a and b.
Constraints
* 1 ≤ a, b, c ≤ 10000
* a ≤ b
Input
Three integers a, b and c are given in a line separated by a single space.
Output
Print the number of divisors in a line.
Example
Input
5 14 80
Output
3
Submitted Solution:
```
a,b,c=input().split()
i=int(a)
k=0
while i<=int(b) :
if int(c)%i==0 : k=k+1
i=i+1
print(k)
``` | instruction | 0 | 28,247 | 22 | 56,494 |
Yes | output | 1 | 28,247 | 22 | 56,495 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
How Many Divisors?
Write a program which reads three integers a, b and c, and prints the number of divisors of c between a and b.
Constraints
* 1 ≤ a, b, c ≤ 10000
* a ≤ b
Input
Three integers a, b and c are given in a line separated by a single space.
Output
Print the number of divisors in a line.
Example
Input
5 14 80
Output
3
Submitted Solution:
```
a,b,c=map(int, input().split())
k=0
for j in range(a,b):
if c%j==0:
k+=1
print(j)
``` | instruction | 0 | 28,248 | 22 | 56,496 |
No | output | 1 | 28,248 | 22 | 56,497 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
How Many Divisors?
Write a program which reads three integers a, b and c, and prints the number of divisors of c between a and b.
Constraints
* 1 ≤ a, b, c ≤ 10000
* a ≤ b
Input
Three integers a, b and c are given in a line separated by a single space.
Output
Print the number of divisors in a line.
Example
Input
5 14 80
Output
3
Submitted Solution:
```
(a, b, c) = [int(i) for i in input().split()]
count = 0
for i in range(a, b):
if (c % i) == 0:
count = count + 1
print(count)
~
``` | instruction | 0 | 28,249 | 22 | 56,498 |
No | output | 1 | 28,249 | 22 | 56,499 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
How Many Divisors?
Write a program which reads three integers a, b and c, and prints the number of divisors of c between a and b.
Constraints
* 1 ≤ a, b, c ≤ 10000
* a ≤ b
Input
Three integers a, b and c are given in a line separated by a single space.
Output
Print the number of divisors in a line.
Example
Input
5 14 80
Output
3
Submitted Solution:
```
i=0
a,b,c = map(int,input().split())
for x in range(a+1,b):
if c % x == 0: i+=1
print(str(i))
``` | instruction | 0 | 28,250 | 22 | 56,500 |
No | output | 1 | 28,250 | 22 | 56,501 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
How Many Divisors?
Write a program which reads three integers a, b and c, and prints the number of divisors of c between a and b.
Constraints
* 1 ≤ a, b, c ≤ 10000
* a ≤ b
Input
Three integers a, b and c are given in a line separated by a single space.
Output
Print the number of divisors in a line.
Example
Input
5 14 80
Output
3
Submitted Solution:
```
i=0
a=list(map(int,input().split()))
if(a[0]==a[1]):
if(a[2]%a[0]==0):
i+=1
for x in range(a[0],a[1]):
if(a[2]%x==0):
i+=1
print(i)
``` | instruction | 0 | 28,251 | 22 | 56,502 |
No | output | 1 | 28,251 | 22 | 56,503 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have n × n square grid and an integer k. Put an integer in each cell while satisfying the conditions below.
* All numbers in the grid should be between 1 and k inclusive.
* Minimum number of the i-th row is 1 (1 ≤ i ≤ n).
* Minimum number of the j-th column is 1 (1 ≤ j ≤ n).
Find the number of ways to put integers in the grid. Since the answer can be very large, find the answer modulo (10^{9} + 7).
<image> These are the examples of valid and invalid grid when n=k=2.
Input
The only line contains two integers n and k (1 ≤ n ≤ 250, 1 ≤ k ≤ 10^{9}).
Output
Print the answer modulo (10^{9} + 7).
Examples
Input
2 2
Output
7
Input
123 456789
Output
689974806
Note
In the first example, following 7 cases are possible.
<image>
In the second example, make sure you print the answer modulo (10^{9} + 7). | instruction | 0 | 28,356 | 22 | 56,712 |
Tags: combinatorics, dp, math
Correct Solution:
```
import sys
input = sys.stdin.readline
n,k=map(int,input().split())
mod=10**9+7
FACT=[1]
for i in range(1,250+1):
FACT.append(FACT[-1]*i%mod)
FACT_INV=[pow(FACT[-1],mod-2,mod)]
for i in range(250,0,-1):
FACT_INV.append(FACT_INV[-1]*i%mod)
FACT_INV.reverse()
def Combi(a,b):
if 0<=b<=a:
return FACT[a]*FACT_INV[b]%mod*FACT_INV[a-b]%mod
else:
return 0
COMBI=[[0]*251 for i in range(251)]
for a in range(251):
for b in range(a+1):
COMBI[a][b]=Combi(a,b)
POW_K=[1]
POW_K1=[1]
for i in range(n+1):
POW_K.append(POW_K[-1]*k%mod)
POW_K1.append(POW_K1[-1]*(k-1)%mod)
DP=[0]*(n+1)
DP[0]=1
PLUS=[[0]*(n+1) for i in range(n+1)]
for i in range(n+1):
for j in range(n+1):
if i==j:
PLUS[i][j]=(POW_K[i]*COMBI[n-i][j-i]*POW_K1[n-j]-POW_K1[n])%mod
else:
PLUS[i][j]=POW_K[i]*COMBI[n-i][j-i]*POW_K1[n-j]%mod
for yoko in range(n):
NDP=[0]*(n+1)
for i in range(n+1):
if DP[i]==0:
continue
for j in range(max(1,i),n+1):
NDP[j]=(NDP[j]+PLUS[i][j]*DP[i])%mod
DP=NDP
#print(DP)
print(DP[-1])
``` | output | 1 | 28,356 | 22 | 56,713 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have n × n square grid and an integer k. Put an integer in each cell while satisfying the conditions below.
* All numbers in the grid should be between 1 and k inclusive.
* Minimum number of the i-th row is 1 (1 ≤ i ≤ n).
* Minimum number of the j-th column is 1 (1 ≤ j ≤ n).
Find the number of ways to put integers in the grid. Since the answer can be very large, find the answer modulo (10^{9} + 7).
<image> These are the examples of valid and invalid grid when n=k=2.
Input
The only line contains two integers n and k (1 ≤ n ≤ 250, 1 ≤ k ≤ 10^{9}).
Output
Print the answer modulo (10^{9} + 7).
Examples
Input
2 2
Output
7
Input
123 456789
Output
689974806
Note
In the first example, following 7 cases are possible.
<image>
In the second example, make sure you print the answer modulo (10^{9} + 7). | instruction | 0 | 28,357 | 22 | 56,714 |
Tags: combinatorics, dp, math
Correct Solution:
```
n, k = map(int, input().split())
MOD = 10**9 + 7
Comb = [None] * (n + 1)
dp = [[None] * (n + 1) for i in range(n+1)]
for i in range(1, n + 1):
Comb[i] = [1] + [(Comb[i-1][j-1] + Comb[i-1][j]) % MOD for j in range(1, i)] + [1]
def powgen(base):
cur = 1
while True:
yield cur
cur = cur * base % MOD
gen, gen_1 = powgen(k), powgen(k - 1)
kpower = [next(gen) for i in range(n + 1)]
k1power = [next(gen_1) for i in range(n + 1)]
dp[1][0] = (kpower[n] - k1power[n] + MOD) % MOD
for i in range(1, n+1): dp[1][i] = kpower[n-i]
for r in range(2, n + 1): # row remaining
for c in range(n+1): # c means col incompleted
dp[r][c] = (dp[r-1][c] * k1power[c] * (kpower[n-c]-k1power[n-c]) + \
kpower[n-c] * sum([dp[r-1][c-i] * Comb[c][i] * k1power[c-i] for i in range(1, c+1)])) % MOD
# input 250 1000000000
print( dp[n][n])
``` | output | 1 | 28,357 | 22 | 56,715 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have n × n square grid and an integer k. Put an integer in each cell while satisfying the conditions below.
* All numbers in the grid should be between 1 and k inclusive.
* Minimum number of the i-th row is 1 (1 ≤ i ≤ n).
* Minimum number of the j-th column is 1 (1 ≤ j ≤ n).
Find the number of ways to put integers in the grid. Since the answer can be very large, find the answer modulo (10^{9} + 7).
<image> These are the examples of valid and invalid grid when n=k=2.
Input
The only line contains two integers n and k (1 ≤ n ≤ 250, 1 ≤ k ≤ 10^{9}).
Output
Print the answer modulo (10^{9} + 7).
Examples
Input
2 2
Output
7
Input
123 456789
Output
689974806
Note
In the first example, following 7 cases are possible.
<image>
In the second example, make sure you print the answer modulo (10^{9} + 7). | instruction | 0 | 28,358 | 22 | 56,716 |
Tags: combinatorics, dp, math
Correct Solution:
```
def main():
# Basic construction
MOD = 10**9 + 7 # MOD
n, k = [int(c) for c in input().split()]
f = [[None for j in range(n+1)] for i in range(n+1)]
if k==1: return 1 # Edge case
# nCr
nCr = [[0 for j in range(i+1)] for i in range(n+1)]
nCr[0][0] = 1
for i in range(1, n+1):
nCr[i] = [1] + [(nCr[i-1][j] + nCr[i-1][j-1]) % MOD for j in range(1, i)] + [1]
# k^i, (k-1)^i
def powgen(base):
this = 1
while True:
yield this
this = this*base%MOD
kpowgen, k1powgen = powgen(k), powgen(k-1)
kpower = [next(kpowgen) for i in range(n+1)]
k1power = [next(k1powgen) for i in range(n+1)]
# Edge DP
for r in range(1, n+1): f[r][0] = pow(kpower[n] - k1power[n], r, MOD)
for c in range(1, n+1): f[1][c] = kpower[n-c]
# DP
for r in range(2, n+1):
f[r] = [f[r][0]]
f[r] += [((kpower[n-c] - k1power[n-c]) * k1power[c] * f[r-1][c] +
kpower[n-c] * sum(k1power[c-c0] * nCr[c][c0] * f[r-1][c-c0] for c0 in range(1, c+1))) % MOD
for c in range(1, n+1)]
return f[n][n]
print(main())
``` | output | 1 | 28,358 | 22 | 56,717 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have n × n square grid and an integer k. Put an integer in each cell while satisfying the conditions below.
* All numbers in the grid should be between 1 and k inclusive.
* Minimum number of the i-th row is 1 (1 ≤ i ≤ n).
* Minimum number of the j-th column is 1 (1 ≤ j ≤ n).
Find the number of ways to put integers in the grid. Since the answer can be very large, find the answer modulo (10^{9} + 7).
<image> These are the examples of valid and invalid grid when n=k=2.
Input
The only line contains two integers n and k (1 ≤ n ≤ 250, 1 ≤ k ≤ 10^{9}).
Output
Print the answer modulo (10^{9} + 7).
Examples
Input
2 2
Output
7
Input
123 456789
Output
689974806
Note
In the first example, following 7 cases are possible.
<image>
In the second example, make sure you print the answer modulo (10^{9} + 7). | instruction | 0 | 28,360 | 22 | 56,720 |
Tags: combinatorics, dp, math
Correct Solution:
```
n, k = [int(item) for item in input().split()]
MOD = 10**9 + 7
MAX_N = 10**4
fac = [1] + [0] * MAX_N
fac_inv = [1] + [0] * MAX_N
for i in range(1, n+1):
fac[i] = fac[i-1] * (i) % MOD
fac_inv[i] = fac_inv[i-1] * pow(i, MOD-2, MOD) % MOD
def mod_nCr(n, r):
if n == 0 and r == 0:
return 1
if n < r or n < 0:
return 0
tmp = fac_inv[n-r] * fac_inv[r] % MOD
return tmp * fac[n] % MOD
ans = 0
for i in range(n+1):
base = pow(k, n - i, MOD) * pow(k - 1, i, MOD) - pow(k - 1, n, MOD) + MOD
base % MOD
val = pow(-1, i) * mod_nCr(n, i) * pow(base, n, MOD)
ans += val
ans %= MOD
print(ans)
``` | output | 1 | 28,360 | 22 | 56,721 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.