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 |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given are positive integers A and B.
Let us choose some number of positive common divisors of A and B.
Here, any two of the chosen divisors must be coprime.
At most, how many divisors can we choose?
Definition of common divisor
An integer d is said to be a common divisor of integers x and y when d divides both x and y.
Definition of being coprime
Integers x and y are said to be coprime when x and y have no positive common divisors other than 1.
Definition of dividing
An integer x is said to divide another integer y when there exists an integer \alpha such that y = \alpha x.
Constraints
* All values in input are integers.
* 1 \leq A, B \leq 10^{12}
Input
Input is given from Standard Input in the following format:
A B
Output
Print the maximum number of divisors that can be chosen to satisfy the condition.
Examples
Input
12 18
Output
3
Input
420 660
Output
4
Input
1 2019
Output
1
Submitted Solution:
```
from fractions import gcd
if __name__ == '__main__':
n = gcd(*map(int, input().split()))
a = i = 1
while i * i < n:
i += 1
a += n % i < 1
while n % i < 1:
n /= i
print(a + (n > 1))
``` | instruction | 0 | 73,985 | 22 | 147,970 |
Yes | output | 1 | 73,985 | 22 | 147,971 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given are positive integers A and B.
Let us choose some number of positive common divisors of A and B.
Here, any two of the chosen divisors must be coprime.
At most, how many divisors can we choose?
Definition of common divisor
An integer d is said to be a common divisor of integers x and y when d divides both x and y.
Definition of being coprime
Integers x and y are said to be coprime when x and y have no positive common divisors other than 1.
Definition of dividing
An integer x is said to divide another integer y when there exists an integer \alpha such that y = \alpha x.
Constraints
* All values in input are integers.
* 1 \leq A, B \leq 10^{12}
Input
Input is given from Standard Input in the following format:
A B
Output
Print the maximum number of divisors that can be chosen to satisfy the condition.
Examples
Input
12 18
Output
3
Input
420 660
Output
4
Input
1 2019
Output
1
Submitted Solution:
```
a, b = map(int, input().split())
from fractions import gcd
n = gcd(a, b)
i = 2
primes = []
while True:
if i > 10**6 + 1:
break
if n % i == 0:
n //= i
primes.append(i)
else:
i += 1
print(len(set(primes)) + (n != 1) + 1)
``` | instruction | 0 | 73,986 | 22 | 147,972 |
Yes | output | 1 | 73,986 | 22 | 147,973 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given are positive integers A and B.
Let us choose some number of positive common divisors of A and B.
Here, any two of the chosen divisors must be coprime.
At most, how many divisors can we choose?
Definition of common divisor
An integer d is said to be a common divisor of integers x and y when d divides both x and y.
Definition of being coprime
Integers x and y are said to be coprime when x and y have no positive common divisors other than 1.
Definition of dividing
An integer x is said to divide another integer y when there exists an integer \alpha such that y = \alpha x.
Constraints
* All values in input are integers.
* 1 \leq A, B \leq 10^{12}
Input
Input is given from Standard Input in the following format:
A B
Output
Print the maximum number of divisors that can be chosen to satisfy the condition.
Examples
Input
12 18
Output
3
Input
420 660
Output
4
Input
1 2019
Output
1
Submitted Solution:
```
s=set
def f(n):
a=2;t=[]
while a*a<=n:
while n%a==0:
n//=a;t.append(a)
a+=1
if n>1:
t.append(n)
return t
a,b=map(int,input().split())
print(len(s(f(a))&s(f(b)))+1)
``` | instruction | 0 | 73,987 | 22 | 147,974 |
Yes | output | 1 | 73,987 | 22 | 147,975 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given are positive integers A and B.
Let us choose some number of positive common divisors of A and B.
Here, any two of the chosen divisors must be coprime.
At most, how many divisors can we choose?
Definition of common divisor
An integer d is said to be a common divisor of integers x and y when d divides both x and y.
Definition of being coprime
Integers x and y are said to be coprime when x and y have no positive common divisors other than 1.
Definition of dividing
An integer x is said to divide another integer y when there exists an integer \alpha such that y = \alpha x.
Constraints
* All values in input are integers.
* 1 \leq A, B \leq 10^{12}
Input
Input is given from Standard Input in the following format:
A B
Output
Print the maximum number of divisors that can be chosen to satisfy the condition.
Examples
Input
12 18
Output
3
Input
420 660
Output
4
Input
1 2019
Output
1
Submitted Solution:
```
import numpy as np
def factorization(n):
arr = []
temp = n
for i in range(2, int(-(-n**0.5//1))+1):
if temp%i==0:
cnt=0
while temp%i==0:
cnt+=1
temp //= i
arr.append([i, cnt])
if temp!=1:
arr.append([temp, 1])
if arr==[]:
arr.append([n, 1])
return arr
a, b = map(int, input().split())
x = np.array(factorization(a))
y = np.array(factorization(b))
num = set(x[:, 0]) & set(y[:, 0])
print(len(num) + 1)
``` | instruction | 0 | 73,988 | 22 | 147,976 |
No | output | 1 | 73,988 | 22 | 147,977 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given are positive integers A and B.
Let us choose some number of positive common divisors of A and B.
Here, any two of the chosen divisors must be coprime.
At most, how many divisors can we choose?
Definition of common divisor
An integer d is said to be a common divisor of integers x and y when d divides both x and y.
Definition of being coprime
Integers x and y are said to be coprime when x and y have no positive common divisors other than 1.
Definition of dividing
An integer x is said to divide another integer y when there exists an integer \alpha such that y = \alpha x.
Constraints
* All values in input are integers.
* 1 \leq A, B \leq 10^{12}
Input
Input is given from Standard Input in the following format:
A B
Output
Print the maximum number of divisors that can be chosen to satisfy the condition.
Examples
Input
12 18
Output
3
Input
420 660
Output
4
Input
1 2019
Output
1
Submitted Solution:
```
a, b = map(int, input().split())
ans = 1
n = 2
while n <= int(max(a, b)**0.5)+1:
if a%n == 0 and b%n == 0:
while a%n == 0:
a = a//n
while b%n == 0:
b = b//n
ans += 1
n += 1
print(ans)
``` | instruction | 0 | 73,989 | 22 | 147,978 |
No | output | 1 | 73,989 | 22 | 147,979 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given are positive integers A and B.
Let us choose some number of positive common divisors of A and B.
Here, any two of the chosen divisors must be coprime.
At most, how many divisors can we choose?
Definition of common divisor
An integer d is said to be a common divisor of integers x and y when d divides both x and y.
Definition of being coprime
Integers x and y are said to be coprime when x and y have no positive common divisors other than 1.
Definition of dividing
An integer x is said to divide another integer y when there exists an integer \alpha such that y = \alpha x.
Constraints
* All values in input are integers.
* 1 \leq A, B \leq 10^{12}
Input
Input is given from Standard Input in the following format:
A B
Output
Print the maximum number of divisors that can be chosen to satisfy the condition.
Examples
Input
12 18
Output
3
Input
420 660
Output
4
Input
1 2019
Output
1
Submitted Solution:
```
import math
def get_prime(n):
prime = [2]
limit = int(n**0.5)
data = [i + 1 for i in range(2, n, 2)]
while True:
p = data[0]
if limit <= p:
return prime + data
prime.append(p)
data = [e for e in data if e % p != 0]
def main():
A,B = map(int,input().split())
m = min(A,B)
prime_count = 0
# lim = m / 2
lim = math.sqrt(m)
if (m > 7) :
prime_list = get_prime(int(lim))
else :
prime_list = [2,3,5,7]
# print(prime_list)
# for i in range(1,m+1) :
for i in prime_list :
# print(i)
# if (m % i == 0) :
if (A % i == 0 and B % i == 0) :
if (i in prime_list) :
# print(i)
prime_count += 1
# 1を含むので+1しておく
# a = (prime_count+1) * (prime_count)
# print(int(a/2))
# 問題読み違えてた！！！！！！
print(prime_count+1)
# メモ
# 案1
# まずは公約数を列挙する
# 素数かどうかを調べる
# 素数のうち、組み合わせを数える
# 間に合わない気がする
# 案2
# A,Bのちいさいほうの半分までの素数を列挙
# 小さい方の公約数を求める
# nC2を出す
main()
``` | instruction | 0 | 73,990 | 22 | 147,980 |
No | output | 1 | 73,990 | 22 | 147,981 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given are positive integers A and B.
Let us choose some number of positive common divisors of A and B.
Here, any two of the chosen divisors must be coprime.
At most, how many divisors can we choose?
Definition of common divisor
An integer d is said to be a common divisor of integers x and y when d divides both x and y.
Definition of being coprime
Integers x and y are said to be coprime when x and y have no positive common divisors other than 1.
Definition of dividing
An integer x is said to divide another integer y when there exists an integer \alpha such that y = \alpha x.
Constraints
* All values in input are integers.
* 1 \leq A, B \leq 10^{12}
Input
Input is given from Standard Input in the following format:
A B
Output
Print the maximum number of divisors that can be chosen to satisfy the condition.
Examples
Input
12 18
Output
3
Input
420 660
Output
4
Input
1 2019
Output
1
Submitted Solution:
```
import math
d = math.gcd(*map(int, input().split()))
r = 1
i = 2
while i * i <= d:
if d % i == 0:
d //= i
r += 1
while d % i == 0:
d //= i
i += 1
print(r + (d > 1))
``` | instruction | 0 | 73,991 | 22 | 147,982 |
No | output | 1 | 73,991 | 22 | 147,983 |
Provide a correct Python 3 solution for this coding contest problem.
Given n integers a1, a2,…, an and n integers p1, p2,…, pn, integer m. The operation of selecting the kth integer ak with a probability of pk [%] is performed for each k (1 ≤ k ≤ n), and 0 or more and n or less integers are selected. Find the expected number of integers between 1 and m that are divisible by at least one of the selected integers.
Constraints
* 1 ≤ n ≤ 20
* 1 ≤ m ≤ 1018
* 1 ≤ ak ≤ 1018 (1 ≤ k ≤ n)
* 1 ≤ pk ≤ 99 (1 ≤ k ≤ n)
Input
The input is given in the following format.
> n m
> a1 a2… an
> p1 p2… pn
>
Output
Print the solution to the problem on one line. The output must not have an absolute or relative error of more than 0.0000001 (= 10-7).
Examples
Input
2 15
3 5
50 50
Output
3.75
Input
4 100
2 3 5 7
80 60 40 20
Output
57.352
Input
4 210
2 3 5 7
80 60 40 20
Output
119.9136 | instruction | 0 | 74,131 | 22 | 148,262 |
"Correct Solution:
```
def GCD(x, y):
if y == 0:
return x
return GCD(y, x % y)
def CountBitFlag(T):
T = (T & 0x55555555) + ((T >> 1) & 0x55555555)
T = (T & 0x33333333) + ((T >> 2) & 0x33333333)
T = (T & 0x0F0F0F0F) + ((T >> 4) & 0x0F0F0F0F)
T = (T & 0x00FF00FF) + ((T >> 8) & 0x00FF00FF)
T = (T & 0x0000FFFF) + ((T >> 16) & 0x0000FFFF)
return T
N, M = map(int, input().split())
A = tuple(map(int, input().split()))
P = tuple(map(int, input().split()))
Prob = [1.0] * (1 << N)
LCM = [1] * (1 << N)
ans = 0
for bit in range(1, 1 << N):
top = (bit & -bit).bit_length() - 1
l = bit & (bit - 1)
Prob[bit] = Prob[l] * P[top] / 100
LCM[bit] = LCM[l] // GCD(LCM[l], A[top]) * A[top]
if CountBitFlag(bit) % 2 == 1:
ans += M // LCM[bit] * Prob[bit]
else:
ans -= M // LCM[bit] * Prob[bit]
print("{:.8f}".format(ans))
``` | output | 1 | 74,131 | 22 | 148,263 |
Provide a correct Python 3 solution for this coding contest problem.
Given n integers a1, a2,…, an and n integers p1, p2,…, pn, integer m. The operation of selecting the kth integer ak with a probability of pk [%] is performed for each k (1 ≤ k ≤ n), and 0 or more and n or less integers are selected. Find the expected number of integers between 1 and m that are divisible by at least one of the selected integers.
Constraints
* 1 ≤ n ≤ 20
* 1 ≤ m ≤ 1018
* 1 ≤ ak ≤ 1018 (1 ≤ k ≤ n)
* 1 ≤ pk ≤ 99 (1 ≤ k ≤ n)
Input
The input is given in the following format.
> n m
> a1 a2… an
> p1 p2… pn
>
Output
Print the solution to the problem on one line. The output must not have an absolute or relative error of more than 0.0000001 (= 10-7).
Examples
Input
2 15
3 5
50 50
Output
3.75
Input
4 100
2 3 5 7
80 60 40 20
Output
57.352
Input
4 210
2 3 5 7
80 60 40 20
Output
119.9136 | instruction | 0 | 74,132 | 22 | 148,264 |
"Correct Solution:
```
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time,copy,functools
sys.setrecursionlimit(10**7)
inf = 10**20
eps = 1.0 / 10**10
mod = 10**9+7
dd = [(-1,0),(0,1),(1,0),(0,-1)]
ddn = [(-1,0),(-1,1),(0,1),(1,1),(1,0),(1,-1),(0,-1),(-1,-1)]
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LI_(): return [int(x)-1 for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def S(): return input()
def pf(s): return print(s, flush=True)
def main():
n,m = LI()
a = LI()
p = [c / 100.0 for c in LI()]
d = collections.defaultdict(lambda: 1.0)
e = collections.defaultdict(lambda: 1)
r = 0
for k in range(1,1<<n):
t = [i for i in range(n) if k & (1 << i)]
l = k & (k-1)
d[k] = d[l] * p[t[0]]
e[k] = e[l] * a[t[0]] // fractions.gcd(a[t[0]], e[l])
o = m // e[k]
if len(t) % 2 == 1:
r += o * d[k]
else:
r -= o * d[k]
return '{:0.8f}'.format(r)
print(main())
``` | output | 1 | 74,132 | 22 | 148,265 |
Provide a correct Python 3 solution for this coding contest problem.
Given n integers a1, a2,…, an and n integers p1, p2,…, pn, integer m. The operation of selecting the kth integer ak with a probability of pk [%] is performed for each k (1 ≤ k ≤ n), and 0 or more and n or less integers are selected. Find the expected number of integers between 1 and m that are divisible by at least one of the selected integers.
Constraints
* 1 ≤ n ≤ 20
* 1 ≤ m ≤ 1018
* 1 ≤ ak ≤ 1018 (1 ≤ k ≤ n)
* 1 ≤ pk ≤ 99 (1 ≤ k ≤ n)
Input
The input is given in the following format.
> n m
> a1 a2… an
> p1 p2… pn
>
Output
Print the solution to the problem on one line. The output must not have an absolute or relative error of more than 0.0000001 (= 10-7).
Examples
Input
2 15
3 5
50 50
Output
3.75
Input
4 100
2 3 5 7
80 60 40 20
Output
57.352
Input
4 210
2 3 5 7
80 60 40 20
Output
119.9136 | instruction | 0 | 74,133 | 22 | 148,266 |
"Correct Solution:
```
import sys
readline = sys.stdin.readline
write = sys.stdout.write
def solve():
N, M = map(int, readline().split())
*A, = map(int, readline().split())
*P, = map(int, readline().split())
def lcm(m, n):
m0 = m; n0 = n
while n:
m, n = n, m % n
return m0//m*n0
N2 = 1 << N
Q = [1]*N2
L = [1]*N2
C = [0]*N2
for i in range(N):
L[1 << i] = A[i]
Q[1 << i] = P[i] / 100
ans = 0
for state in range(1, N2):
b = state & -state
L[state] = lcm(L[state ^ b], L[b])
C[state] = C[state ^ b] + 1
Q[state] = Q[state ^ b] * Q[b]
if C[state] % 2:
ans += Q[state] * (M // L[state])
else:
ans -= Q[state] * (M // L[state])
write("%.16f\n" % ans)
solve()
``` | output | 1 | 74,133 | 22 | 148,267 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given n integers a1, a2,…, an and n integers p1, p2,…, pn, integer m. The operation of selecting the kth integer ak with a probability of pk [%] is performed for each k (1 ≤ k ≤ n), and 0 or more and n or less integers are selected. Find the expected number of integers between 1 and m that are divisible by at least one of the selected integers.
Constraints
* 1 ≤ n ≤ 20
* 1 ≤ m ≤ 1018
* 1 ≤ ak ≤ 1018 (1 ≤ k ≤ n)
* 1 ≤ pk ≤ 99 (1 ≤ k ≤ n)
Input
The input is given in the following format.
> n m
> a1 a2… an
> p1 p2… pn
>
Output
Print the solution to the problem on one line. The output must not have an absolute or relative error of more than 0.0000001 (= 10-7).
Examples
Input
2 15
3 5
50 50
Output
3.75
Input
4 100
2 3 5 7
80 60 40 20
Output
57.352
Input
4 210
2 3 5 7
80 60 40 20
Output
119.9136
Submitted Solution:
```
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time,copy,functools
sys.setrecursionlimit(10**7)
inf = 10**20
eps = 1.0 / 10**10
mod = 10**9+7
dd = [(-1,0),(0,1),(1,0),(0,-1)]
ddn = [(-1,0),(-1,1),(0,1),(1,1),(1,0),(1,-1),(0,-1),(-1,-1)]
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LI_(): return [int(x)-1 for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def S(): return input()
def pf(s): return print(s, flush=True)
def main():
n,m = LI()
a = LI()
p = LI()
fm = {}
def f(k):
if k in fm:
return fm[k]
g = 1
for i in range(n):
if k & (1 << i):
g *= a[i] // fractions.gcd(g, a[i])
fm[k] = m // g
return fm[k]
r = 0
for k in range(1,1<<n):
t = []
b = 1.0
for i in range(n):
if k & (1 << i):
t.append(1 << i)
b *= p[i]
else:
b *= 100 - p[i]
b /= 100
c = 0
tl = len(t)
for l in range(1,1<<tl):
o = 0
lc = 0
for j in range(tl):
if l & (1 << j):
o |= t[j]
lc += 1
if lc % 2 == 1:
c += f(o)
else:
c -= f(o)
r += b * c
# print(fm)
return '{:0.8f}'.format(r)
print(main())
``` | instruction | 0 | 74,134 | 22 | 148,268 |
No | output | 1 | 74,134 | 22 | 148,269 |
Provide a correct Python 3 solution for this coding contest problem.
Ikta loves fast programs. Recently, I'm trying to speed up the division program. However, it doesn't get much faster, so I thought it would be better to make it faster only for "common sense and typical" inputs. The problem Ikta is trying to solve is as follows.
For a given non-negative integer n, divide p (n) − 1-digit positive integer 11 ... 1 by p (n) in decimal notation. However, p (n) represents the smallest prime number larger than 22 {... 2} (n 2). Let p (0) = 2.
Your job is to complete the program faster than Ikta.
Input
The input is given in the following format.
n
Given the non-negative integer n of the input in question.
Constraints
Each variable being input satisfies the following constraints.
* 0 ≤ n <1000
Output
Print the solution to the problem on one line.
Examples
Input
0
Output
1
Input
1
Output
2
Input
2
Output
1 | instruction | 0 | 74,135 | 22 | 148,270 |
"Correct Solution:
```
# AOJ 2610: Fast Division
# Python3 2018.6.30 bal4u
ans = [1, 2, 1, 0]
n = int(input())
if n > 2: n = 3
print(ans[n])
``` | output | 1 | 74,135 | 22 | 148,271 |
Provide a correct Python 3 solution for this coding contest problem.
Ikta loves fast programs. Recently, I'm trying to speed up the division program. However, it doesn't get much faster, so I thought it would be better to make it faster only for "common sense and typical" inputs. The problem Ikta is trying to solve is as follows.
For a given non-negative integer n, divide p (n) − 1-digit positive integer 11 ... 1 by p (n) in decimal notation. However, p (n) represents the smallest prime number larger than 22 {... 2} (n 2). Let p (0) = 2.
Your job is to complete the program faster than Ikta.
Input
The input is given in the following format.
n
Given the non-negative integer n of the input in question.
Constraints
Each variable being input satisfies the following constraints.
* 0 ≤ n <1000
Output
Print the solution to the problem on one line.
Examples
Input
0
Output
1
Input
1
Output
2
Input
2
Output
1 | instruction | 0 | 74,136 | 22 | 148,272 |
"Correct Solution:
```
n=int(input())
if n>3:n=3
print([1,2,1,0][n])
``` | output | 1 | 74,136 | 22 | 148,273 |
Provide a correct Python 3 solution for this coding contest problem.
Ikta loves fast programs. Recently, I'm trying to speed up the division program. However, it doesn't get much faster, so I thought it would be better to make it faster only for "common sense and typical" inputs. The problem Ikta is trying to solve is as follows.
For a given non-negative integer n, divide p (n) − 1-digit positive integer 11 ... 1 by p (n) in decimal notation. However, p (n) represents the smallest prime number larger than 22 {... 2} (n 2). Let p (0) = 2.
Your job is to complete the program faster than Ikta.
Input
The input is given in the following format.
n
Given the non-negative integer n of the input in question.
Constraints
Each variable being input satisfies the following constraints.
* 0 ≤ n <1000
Output
Print the solution to the problem on one line.
Examples
Input
0
Output
1
Input
1
Output
2
Input
2
Output
1 | instruction | 0 | 74,137 | 22 | 148,274 |
"Correct Solution:
```
#!usr/bin/env python3
from collections import defaultdict
from collections import deque
from heapq import heappush, heappop
import sys
import math
import bisect
import random
def LI(): return list(map(int, sys.stdin.readline().split()))
def I(): return int(sys.stdin.readline())
def LS():return list(map(list, sys.stdin.readline().split()))
def S(): return list(sys.stdin.readline())[:-1]
def IR(n):
l = [None for i in range(n)]
for i in range(n):l[i] = I()
return l
def LIR(n):
l = [None for i in range(n)]
for i in range(n):l[i] = LI()
return l
def SR(n):
l = [None for i in range(n)]
for i in range(n):l[i] = S()
return l
def LSR(n):
l = [None for i in range(n)]
for i in range(n):l[i] = LS()
return l
sys.setrecursionlimit(1000000)
mod = 1000000007
#A
def A():
while 1:
n = I()
if n == 0:
quit()
a = LI()
m = sum(a)/n
c = 0
for i in range(n):
if a[i] <= m:
c += 1
print(c)
return
#B
def B():
while 1:
n = I()
if n == 0:
quit()
t = [0 for i in range(24*3600)]
for i in range(n):
x,y = input().split()
x = list(map(int, x.split(":")))
y = list(map(int, y.split(":")))
l = x[0]*3600+x[1]*60+x[2]
r = y[0]*3600+y[1]*60+y[2]
t[l] += 1
t[r] -= 1
ans = 0
for i in range(24*3600-1):
t[i+1] += t[i]
ans = max(ans, t[i])
print(ans)
return
#C
def C():
n = I()
if n == 0:
print(1)
elif n == 1:
print(2)
elif n == 2:
print(1)
else:
print(0)
return
#D
def D():
return
#E
def E():
return
#F
def F():
return
#G
def G():
return
#H
def H():
return
#I
def I_():
return
#J
def J():
return
#Solve
if __name__ == "__main__":
C()
``` | output | 1 | 74,137 | 22 | 148,275 |
Provide a correct Python 3 solution for this coding contest problem.
Ikta loves fast programs. Recently, I'm trying to speed up the division program. However, it doesn't get much faster, so I thought it would be better to make it faster only for "common sense and typical" inputs. The problem Ikta is trying to solve is as follows.
For a given non-negative integer n, divide p (n) − 1-digit positive integer 11 ... 1 by p (n) in decimal notation. However, p (n) represents the smallest prime number larger than 22 {... 2} (n 2). Let p (0) = 2.
Your job is to complete the program faster than Ikta.
Input
The input is given in the following format.
n
Given the non-negative integer n of the input in question.
Constraints
Each variable being input satisfies the following constraints.
* 0 ≤ n <1000
Output
Print the solution to the problem on one line.
Examples
Input
0
Output
1
Input
1
Output
2
Input
2
Output
1 | instruction | 0 | 74,138 | 22 | 148,276 |
"Correct Solution:
```
n=int(input())
if n==0 or n==2:print(1)
else: print(2 if n==1 else 0)
``` | output | 1 | 74,138 | 22 | 148,277 |
Provide a correct Python 3 solution for this coding contest problem.
Ikta loves fast programs. Recently, I'm trying to speed up the division program. However, it doesn't get much faster, so I thought it would be better to make it faster only for "common sense and typical" inputs. The problem Ikta is trying to solve is as follows.
For a given non-negative integer n, divide p (n) − 1-digit positive integer 11 ... 1 by p (n) in decimal notation. However, p (n) represents the smallest prime number larger than 22 {... 2} (n 2). Let p (0) = 2.
Your job is to complete the program faster than Ikta.
Input
The input is given in the following format.
n
Given the non-negative integer n of the input in question.
Constraints
Each variable being input satisfies the following constraints.
* 0 ≤ n <1000
Output
Print the solution to the problem on one line.
Examples
Input
0
Output
1
Input
1
Output
2
Input
2
Output
1 | instruction | 0 | 74,139 | 22 | 148,278 |
"Correct Solution:
```
n=int(input())
print(1if n==0else 2//n)
``` | output | 1 | 74,139 | 22 | 148,279 |
Provide a correct Python 3 solution for this coding contest problem.
Ikta loves fast programs. Recently, I'm trying to speed up the division program. However, it doesn't get much faster, so I thought it would be better to make it faster only for "common sense and typical" inputs. The problem Ikta is trying to solve is as follows.
For a given non-negative integer n, divide p (n) − 1-digit positive integer 11 ... 1 by p (n) in decimal notation. However, p (n) represents the smallest prime number larger than 22 {... 2} (n 2). Let p (0) = 2.
Your job is to complete the program faster than Ikta.
Input
The input is given in the following format.
n
Given the non-negative integer n of the input in question.
Constraints
Each variable being input satisfies the following constraints.
* 0 ≤ n <1000
Output
Print the solution to the problem on one line.
Examples
Input
0
Output
1
Input
1
Output
2
Input
2
Output
1 | instruction | 0 | 74,140 | 22 | 148,280 |
"Correct Solution:
```
n = int(input())
if n == 0 or n == 2:
print(1)
elif n == 1:
print(2)
else:
print(0)
``` | output | 1 | 74,140 | 22 | 148,281 |
Provide a correct Python 3 solution for this coding contest problem.
Ikta loves fast programs. Recently, I'm trying to speed up the division program. However, it doesn't get much faster, so I thought it would be better to make it faster only for "common sense and typical" inputs. The problem Ikta is trying to solve is as follows.
For a given non-negative integer n, divide p (n) − 1-digit positive integer 11 ... 1 by p (n) in decimal notation. However, p (n) represents the smallest prime number larger than 22 {... 2} (n 2). Let p (0) = 2.
Your job is to complete the program faster than Ikta.
Input
The input is given in the following format.
n
Given the non-negative integer n of the input in question.
Constraints
Each variable being input satisfies the following constraints.
* 0 ≤ n <1000
Output
Print the solution to the problem on one line.
Examples
Input
0
Output
1
Input
1
Output
2
Input
2
Output
1 | instruction | 0 | 74,141 | 22 | 148,282 |
"Correct Solution:
```
n = int(input())
if n == 0:
print(1)
elif n == 1:
print(2)
elif n == 2:
print(1)
else:
print(0)
``` | output | 1 | 74,141 | 22 | 148,283 |
Provide a correct Python 3 solution for this coding contest problem.
Ikta loves fast programs. Recently, I'm trying to speed up the division program. However, it doesn't get much faster, so I thought it would be better to make it faster only for "common sense and typical" inputs. The problem Ikta is trying to solve is as follows.
For a given non-negative integer n, divide p (n) − 1-digit positive integer 11 ... 1 by p (n) in decimal notation. However, p (n) represents the smallest prime number larger than 22 {... 2} (n 2). Let p (0) = 2.
Your job is to complete the program faster than Ikta.
Input
The input is given in the following format.
n
Given the non-negative integer n of the input in question.
Constraints
Each variable being input satisfies the following constraints.
* 0 ≤ n <1000
Output
Print the solution to the problem on one line.
Examples
Input
0
Output
1
Input
1
Output
2
Input
2
Output
1 | instruction | 0 | 74,142 | 22 | 148,284 |
"Correct Solution:
```
# p(n) = 2^^n より大きい最小の素数 (ただし^^はクヌースの二重矢印記号) と定義したとき、
# x = (10^(p(n)-1)-1)/9 (mod p(n)) を求めよ
# 「フェルマーの小定理」
# pが素数ならば、a^p == a (mod p(n))
# とくに、aとpが互いに素 (gcd(a,p)=1) のとき
# a^p * a^-1 == a * a^-1 (mod p(n))
# a^(p-1) == 1 (mod p)
#
# 9とp(n)が互いに素ならば
# x * 9^-1 = 10^(p(n)-1)-1 (mod p(n))
# 10とp(n)が互いに素ならば
# 10^(p(n)-1) = 1 (mod p(n))
# ゆえに
# x * 9^-1 = 0 (mod p(n))
# p(n)は素数であるから、p(n)>10のとき9とp(n)および10とp(n)は互いに素でありx=0
n = int(input())
if n == 0:
# 1 mod 2
print(1)
elif n == 1:
# 11 mod 3
print(2)
elif n == 2:
# 111 mod 5
print(1)
else:
print(0)
``` | output | 1 | 74,142 | 22 | 148,285 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ikta loves fast programs. Recently, I'm trying to speed up the division program. However, it doesn't get much faster, so I thought it would be better to make it faster only for "common sense and typical" inputs. The problem Ikta is trying to solve is as follows.
For a given non-negative integer n, divide p (n) − 1-digit positive integer 11 ... 1 by p (n) in decimal notation. However, p (n) represents the smallest prime number larger than 22 {... 2} (n 2). Let p (0) = 2.
Your job is to complete the program faster than Ikta.
Input
The input is given in the following format.
n
Given the non-negative integer n of the input in question.
Constraints
Each variable being input satisfies the following constraints.
* 0 ≤ n <1000
Output
Print the solution to the problem on one line.
Examples
Input
0
Output
1
Input
1
Output
2
Input
2
Output
1
Submitted Solution:
```
n=int(input())
print([1,2,1,0][[n,3][n>3]])
``` | instruction | 0 | 74,143 | 22 | 148,286 |
Yes | output | 1 | 74,143 | 22 | 148,287 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ikta loves fast programs. Recently, I'm trying to speed up the division program. However, it doesn't get much faster, so I thought it would be better to make it faster only for "common sense and typical" inputs. The problem Ikta is trying to solve is as follows.
For a given non-negative integer n, divide p (n) − 1-digit positive integer 11 ... 1 by p (n) in decimal notation. However, p (n) represents the smallest prime number larger than 22 {... 2} (n 2). Let p (0) = 2.
Your job is to complete the program faster than Ikta.
Input
The input is given in the following format.
n
Given the non-negative integer n of the input in question.
Constraints
Each variable being input satisfies the following constraints.
* 0 ≤ n <1000
Output
Print the solution to the problem on one line.
Examples
Input
0
Output
1
Input
1
Output
2
Input
2
Output
1
Submitted Solution:
```
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time,copy,functools
sys.setrecursionlimit(10**7)
inf = 10**20
eps = 1.0 / 10**10
mod = 998244353
dd = [(0,-1),(1,0),(0,1),(-1,0)]
ddn = [(0,-1),(1,-1),(1,0),(1,1),(0,1),(-1,-1),(-1,0),(-1,1)]
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LI_(): return [int(x)-1 for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def S(): return input()
def pf(s): return print(s, flush=True)
def main():
rr = []
while True:
n = I()
r = 0
if n == 0 or n == 2:
r = 1
elif n == 1:
r = 2
rr.append(r)
break
return '\n'.join(map(str, rr))
print(main())
``` | instruction | 0 | 74,144 | 22 | 148,288 |
Yes | output | 1 | 74,144 | 22 | 148,289 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ikta loves fast programs. Recently, I'm trying to speed up the division program. However, it doesn't get much faster, so I thought it would be better to make it faster only for "common sense and typical" inputs. The problem Ikta is trying to solve is as follows.
For a given non-negative integer n, divide p (n) − 1-digit positive integer 11 ... 1 by p (n) in decimal notation. However, p (n) represents the smallest prime number larger than 22 {... 2} (n 2). Let p (0) = 2.
Your job is to complete the program faster than Ikta.
Input
The input is given in the following format.
n
Given the non-negative integer n of the input in question.
Constraints
Each variable being input satisfies the following constraints.
* 0 ≤ n <1000
Output
Print the solution to the problem on one line.
Examples
Input
0
Output
1
Input
1
Output
2
Input
2
Output
1
Submitted Solution:
```
n=int(input())
print(n?n>3?0:3-n:1)
``` | instruction | 0 | 74,145 | 22 | 148,290 |
No | output | 1 | 74,145 | 22 | 148,291 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Find out if it is possible to partition the first n positive integers into two non-empty disjoint sets S_1 and S_2 such that:
gcd(sum(S_1), sum(S_2)) > 1
Here sum(S) denotes the sum of all elements present in set S and gcd means the[greatest common divisor](https://en.wikipedia.org/wiki/Greatest_common_divisor).
Every integer number from 1 to n should be present in exactly one of S_1 or S_2.
Input
The only line of the input contains a single integer n (1 ≤ n ≤ 45 000)
Output
If such partition doesn't exist, print "No" (quotes for clarity).
Otherwise, print "Yes" (quotes for clarity), followed by two lines, describing S_1 and S_2 respectively.
Each set description starts with the set size, followed by the elements of the set in any order. Each set must be non-empty.
If there are multiple possible partitions — print any of them.
Examples
Input
1
Output
No
Input
3
Output
Yes
1 2
2 1 3
Note
In the first example, there is no way to partition a single number into two non-empty sets, hence the answer is "No".
In the second example, the sums of the sets are 2 and 4 respectively. The gcd(2, 4) = 2 > 1, hence that is one of the possible answers. | instruction | 0 | 74,181 | 22 | 148,362 |
Tags: constructive algorithms, math
Correct Solution:
```
# -*- coding: utf-8 -*-
"""
Created on Sat Aug 22 21:39:13 2020
@author: Dark Soul
"""
n=int(input(''))
if n<3:
print('No')
else:
s=n*(n+1)//2
sol=[]
for i in range(1,n+1):
sol.append(i)
if s%2==0:
print('Yes')
print(1,2)
print(n-1,*list(set(sol)-set([2])))
else:
for j in range(3,n+1):
if s%j==0:
print('Yes')
print(1,j)
print(n-1,*list(set(sol)-set([j])))
break
``` | output | 1 | 74,181 | 22 | 148,363 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Find out if it is possible to partition the first n positive integers into two non-empty disjoint sets S_1 and S_2 such that:
gcd(sum(S_1), sum(S_2)) > 1
Here sum(S) denotes the sum of all elements present in set S and gcd means the[greatest common divisor](https://en.wikipedia.org/wiki/Greatest_common_divisor).
Every integer number from 1 to n should be present in exactly one of S_1 or S_2.
Input
The only line of the input contains a single integer n (1 ≤ n ≤ 45 000)
Output
If such partition doesn't exist, print "No" (quotes for clarity).
Otherwise, print "Yes" (quotes for clarity), followed by two lines, describing S_1 and S_2 respectively.
Each set description starts with the set size, followed by the elements of the set in any order. Each set must be non-empty.
If there are multiple possible partitions — print any of them.
Examples
Input
1
Output
No
Input
3
Output
Yes
1 2
2 1 3
Note
In the first example, there is no way to partition a single number into two non-empty sets, hence the answer is "No".
In the second example, the sums of the sets are 2 and 4 respectively. The gcd(2, 4) = 2 > 1, hence that is one of the possible answers. | instruction | 0 | 74,182 | 22 | 148,364 |
Tags: constructive algorithms, math
Correct Solution:
```
import math as m
n = int(input())
l1 = []
l2 = []
for i in range(1,n+1):
if i%2==0:
l1.append(i)
else:
l2.append(i)
if m.gcd(sum(l1),sum(l2))<=1:
print("No")
else:
print("Yes")
print(len(l1),*l1)
print(len(l2),*l2)
``` | output | 1 | 74,182 | 22 | 148,365 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Find out if it is possible to partition the first n positive integers into two non-empty disjoint sets S_1 and S_2 such that:
gcd(sum(S_1), sum(S_2)) > 1
Here sum(S) denotes the sum of all elements present in set S and gcd means the[greatest common divisor](https://en.wikipedia.org/wiki/Greatest_common_divisor).
Every integer number from 1 to n should be present in exactly one of S_1 or S_2.
Input
The only line of the input contains a single integer n (1 ≤ n ≤ 45 000)
Output
If such partition doesn't exist, print "No" (quotes for clarity).
Otherwise, print "Yes" (quotes for clarity), followed by two lines, describing S_1 and S_2 respectively.
Each set description starts with the set size, followed by the elements of the set in any order. Each set must be non-empty.
If there are multiple possible partitions — print any of them.
Examples
Input
1
Output
No
Input
3
Output
Yes
1 2
2 1 3
Note
In the first example, there is no way to partition a single number into two non-empty sets, hence the answer is "No".
In the second example, the sums of the sets are 2 and 4 respectively. The gcd(2, 4) = 2 > 1, hence that is one of the possible answers. | instruction | 0 | 74,183 | 22 | 148,366 |
Tags: constructive algorithms, math
Correct Solution:
```
n=int(input())
if n<3:print("No")
else:print("Yes",1,n,n-1,*range(1,n))
``` | output | 1 | 74,183 | 22 | 148,367 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Find out if it is possible to partition the first n positive integers into two non-empty disjoint sets S_1 and S_2 such that:
gcd(sum(S_1), sum(S_2)) > 1
Here sum(S) denotes the sum of all elements present in set S and gcd means the[greatest common divisor](https://en.wikipedia.org/wiki/Greatest_common_divisor).
Every integer number from 1 to n should be present in exactly one of S_1 or S_2.
Input
The only line of the input contains a single integer n (1 ≤ n ≤ 45 000)
Output
If such partition doesn't exist, print "No" (quotes for clarity).
Otherwise, print "Yes" (quotes for clarity), followed by two lines, describing S_1 and S_2 respectively.
Each set description starts with the set size, followed by the elements of the set in any order. Each set must be non-empty.
If there are multiple possible partitions — print any of them.
Examples
Input
1
Output
No
Input
3
Output
Yes
1 2
2 1 3
Note
In the first example, there is no way to partition a single number into two non-empty sets, hence the answer is "No".
In the second example, the sums of the sets are 2 and 4 respectively. The gcd(2, 4) = 2 > 1, hence that is one of the possible answers. | instruction | 0 | 74,184 | 22 | 148,368 |
Tags: constructive algorithms, math
Correct Solution:
```
import io
import sys
from math import gcd
def solve(n):
return [list(range(1, n+1, 2)), list(range(2, n+1, 2))]
inp = sys.stdin
n = int(inp.readline().strip())
if n <= 2:
print("No")
else:
print("Yes")
ans = solve(n)
print(len(ans[0]), " ".join([str(x) for x in ans[0]]))
print(len(ans[1]), " ".join([str(x) for x in ans[1]]))
``` | output | 1 | 74,184 | 22 | 148,369 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Find out if it is possible to partition the first n positive integers into two non-empty disjoint sets S_1 and S_2 such that:
gcd(sum(S_1), sum(S_2)) > 1
Here sum(S) denotes the sum of all elements present in set S and gcd means the[greatest common divisor](https://en.wikipedia.org/wiki/Greatest_common_divisor).
Every integer number from 1 to n should be present in exactly one of S_1 or S_2.
Input
The only line of the input contains a single integer n (1 ≤ n ≤ 45 000)
Output
If such partition doesn't exist, print "No" (quotes for clarity).
Otherwise, print "Yes" (quotes for clarity), followed by two lines, describing S_1 and S_2 respectively.
Each set description starts with the set size, followed by the elements of the set in any order. Each set must be non-empty.
If there are multiple possible partitions — print any of them.
Examples
Input
1
Output
No
Input
3
Output
Yes
1 2
2 1 3
Note
In the first example, there is no way to partition a single number into two non-empty sets, hence the answer is "No".
In the second example, the sums of the sets are 2 and 4 respectively. The gcd(2, 4) = 2 > 1, hence that is one of the possible answers. | instruction | 0 | 74,185 | 22 | 148,370 |
Tags: constructive algorithms, math
Correct Solution:
```
n=int(input())
if n ==1 or n == 2:
print('No')
else:
print('Yes')
evenlist = [str(2*i) for i in range(1,int((n+2)/2))]
oddlist = [str(2*i+1) for i in range(0,int((n+1)/2))]
print(len(evenlist),' '.join(evenlist))
print(len(oddlist), ' '.join(oddlist))
``` | output | 1 | 74,185 | 22 | 148,371 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Find out if it is possible to partition the first n positive integers into two non-empty disjoint sets S_1 and S_2 such that:
gcd(sum(S_1), sum(S_2)) > 1
Here sum(S) denotes the sum of all elements present in set S and gcd means the[greatest common divisor](https://en.wikipedia.org/wiki/Greatest_common_divisor).
Every integer number from 1 to n should be present in exactly one of S_1 or S_2.
Input
The only line of the input contains a single integer n (1 ≤ n ≤ 45 000)
Output
If such partition doesn't exist, print "No" (quotes for clarity).
Otherwise, print "Yes" (quotes for clarity), followed by two lines, describing S_1 and S_2 respectively.
Each set description starts with the set size, followed by the elements of the set in any order. Each set must be non-empty.
If there are multiple possible partitions — print any of them.
Examples
Input
1
Output
No
Input
3
Output
Yes
1 2
2 1 3
Note
In the first example, there is no way to partition a single number into two non-empty sets, hence the answer is "No".
In the second example, the sums of the sets are 2 and 4 respectively. The gcd(2, 4) = 2 > 1, hence that is one of the possible answers. | instruction | 0 | 74,186 | 22 | 148,372 |
Tags: constructive algorithms, math
Correct Solution:
```
from math import gcd
n=int(input());odd=[];even=[];print
for i in range(1,n+1):
if i%2==1:odd.append(i)
else:even.append(i)
if len(odd)>0 and len(even)>0 and gcd(sum(odd),sum(even))>1:
print('Yes');print(len(odd),*odd);print(len(even),*even)
else:print('No')
``` | output | 1 | 74,186 | 22 | 148,373 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Find out if it is possible to partition the first n positive integers into two non-empty disjoint sets S_1 and S_2 such that:
gcd(sum(S_1), sum(S_2)) > 1
Here sum(S) denotes the sum of all elements present in set S and gcd means the[greatest common divisor](https://en.wikipedia.org/wiki/Greatest_common_divisor).
Every integer number from 1 to n should be present in exactly one of S_1 or S_2.
Input
The only line of the input contains a single integer n (1 ≤ n ≤ 45 000)
Output
If such partition doesn't exist, print "No" (quotes for clarity).
Otherwise, print "Yes" (quotes for clarity), followed by two lines, describing S_1 and S_2 respectively.
Each set description starts with the set size, followed by the elements of the set in any order. Each set must be non-empty.
If there are multiple possible partitions — print any of them.
Examples
Input
1
Output
No
Input
3
Output
Yes
1 2
2 1 3
Note
In the first example, there is no way to partition a single number into two non-empty sets, hence the answer is "No".
In the second example, the sums of the sets are 2 and 4 respectively. The gcd(2, 4) = 2 > 1, hence that is one of the possible answers. | instruction | 0 | 74,187 | 22 | 148,374 |
Tags: constructive algorithms, math
Correct Solution:
```
# cook your dish here
from math import gcd
def evenodd(integer):
a = []
b = []
for i in range(1,integer+1):
if i%2 is 0:
a.append(i)
else:
b.append(i)
sa = sum(a)
sb = sum(b)
# print(sum(a))
# print(sum(b))
if gcd(sa,sb) > 1:
print("Yes")
a = [len(a)] + a
b = [len(b)] + b
print(*a)
print(*b)
else:
print("No")
def main():
k = int(input())
evenodd(k)
if __name__ == "__main__":
main()
``` | output | 1 | 74,187 | 22 | 148,375 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Find out if it is possible to partition the first n positive integers into two non-empty disjoint sets S_1 and S_2 such that:
gcd(sum(S_1), sum(S_2)) > 1
Here sum(S) denotes the sum of all elements present in set S and gcd means the[greatest common divisor](https://en.wikipedia.org/wiki/Greatest_common_divisor).
Every integer number from 1 to n should be present in exactly one of S_1 or S_2.
Input
The only line of the input contains a single integer n (1 ≤ n ≤ 45 000)
Output
If such partition doesn't exist, print "No" (quotes for clarity).
Otherwise, print "Yes" (quotes for clarity), followed by two lines, describing S_1 and S_2 respectively.
Each set description starts with the set size, followed by the elements of the set in any order. Each set must be non-empty.
If there are multiple possible partitions — print any of them.
Examples
Input
1
Output
No
Input
3
Output
Yes
1 2
2 1 3
Note
In the first example, there is no way to partition a single number into two non-empty sets, hence the answer is "No".
In the second example, the sums of the sets are 2 and 4 respectively. The gcd(2, 4) = 2 > 1, hence that is one of the possible answers. | instruction | 0 | 74,188 | 22 | 148,376 |
Tags: constructive algorithms, math
Correct Solution:
```
# print("Input n")
n = int(input())
# Special cases
if n==1 or n==2:
print("No")
# Which of n/2 and (n+1)/2 is an int is correct
elif n%2 == 0:
print("Yes")
print("1 " + str(n//2))
print(str(n-1), end = " ")
for i in range(1, n+1):
if i != n//2:
print(i, end = " ")
else:
print("Yes")
print("1 " + str((n+1)//2))
print(str(n-1), end = " ")
for i in range(1, n+1):
if i != (n+1)//2:
print(i, end = " ")
``` | output | 1 | 74,188 | 22 | 148,377 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Find out if it is possible to partition the first n positive integers into two non-empty disjoint sets S_1 and S_2 such that:
gcd(sum(S_1), sum(S_2)) > 1
Here sum(S) denotes the sum of all elements present in set S and gcd means the[greatest common divisor](https://en.wikipedia.org/wiki/Greatest_common_divisor).
Every integer number from 1 to n should be present in exactly one of S_1 or S_2.
Input
The only line of the input contains a single integer n (1 ≤ n ≤ 45 000)
Output
If such partition doesn't exist, print "No" (quotes for clarity).
Otherwise, print "Yes" (quotes for clarity), followed by two lines, describing S_1 and S_2 respectively.
Each set description starts with the set size, followed by the elements of the set in any order. Each set must be non-empty.
If there are multiple possible partitions — print any of them.
Examples
Input
1
Output
No
Input
3
Output
Yes
1 2
2 1 3
Note
In the first example, there is no way to partition a single number into two non-empty sets, hence the answer is "No".
In the second example, the sums of the sets are 2 and 4 respectively. The gcd(2, 4) = 2 > 1, hence that is one of the possible answers.
Submitted Solution:
```
# ✪ H4WK3yE乡
# Mayank Chaudhary
# ABES EC , Ghaziabad
# ///==========Libraries, Constants and Functions=============///
import sys
from bisect import bisect_left,bisect_right
from collections import deque,Counter
from math import gcd,sqrt,factorial,ceil,log10
from itertools import permutations
inf = float("inf")
mod = 1000000007
mini=1000000007
def fact(n): # <------ To calculate factorial of n under modulo m
if n==0:
return 1
p=1;d=(10**9)+7
for i in range(1,n+1):
p=p*i
p=p%d
return p
def ncr(n,r): # < ------ To calculate nCr mod p value using Fermat Little under modulo m
d=10**9+7
num=fact(n)
den=(fact(r)*fact(n-r))%d
den=pow(den,d-2,d)
return (num*den)%d
def sieve(n): # <----- sieve of eratosthenes for prime no.
prime=[True for i in range(n+1)]
lst=[0]*(n+1)
p=2
while p*p<=n:
if prime[p]:
lst[p]=p
for i in range(p*p,n+1,p):
if prime[i]==True:
prime[i]=False
lst[i]=p
p=p+1
for i in range(1,n+1):
if lst[i]==0:
lst[i]=i
mylist=[0]*(n+1)
for i in range(2,n+1):
mylist[lst[i]]=mylist[lst[i]]+1
return mylist
def binary(number): # <----- calculate the no. of 1's in binary representation of number
result=0
while number:
result=result+1
number=number&(number-1)
return result
def calculate_factors(n): #<---- most efficient method to calculate no. of factors of number
hh = [1] * (n + 1);
p = 2;
while((p * p) < n):
if (hh[p] == 1):
for i in range((p * 2), n, p):
hh[i] = 0;
p += 1;
total = 1;
for p in range(2, n + 1):
if (hh[p] == 1):
count = 0;
if (n % p == 0):
while (n % p == 0):
n = int(n / p);
count += 1;
total *= (count + 1);
return total;
def prime_factors(n): #<------------ to find prime factors of a no.
i = 2
factors = set()
while i * i <= n:
if n % i:
i += 1
else:
factors.add(n//i)
n=n//i
factors.add(i)
if n > 1:
factors.add(n)
return (factors)
def isPrime(n): #<-----------check whether a no. is prime or not
if n==2 or n==3: return True
if n%2==0 or n<2: return False
for i in range(3,int(n**0.5)+1,2): # only odd numbers
if n%i==0:
return False
return True
def get_array(): return list(map(int, sys.stdin.readline().strip().split()))
def get_ints(): return map(int, sys.stdin.readline().strip().split())
def input(): return sys.stdin.readline().strip()
# ///==========MAIN=============///
n=int(input())
if n==1 or n==2:
print("No")
else:
print("Yes")
print(1,n)
Arr=[n-1]
for i in range(1,n):
Arr.append(i)
print(*Arr,sep=' ')
``` | instruction | 0 | 74,189 | 22 | 148,378 |
Yes | output | 1 | 74,189 | 22 | 148,379 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Find out if it is possible to partition the first n positive integers into two non-empty disjoint sets S_1 and S_2 such that:
gcd(sum(S_1), sum(S_2)) > 1
Here sum(S) denotes the sum of all elements present in set S and gcd means the[greatest common divisor](https://en.wikipedia.org/wiki/Greatest_common_divisor).
Every integer number from 1 to n should be present in exactly one of S_1 or S_2.
Input
The only line of the input contains a single integer n (1 ≤ n ≤ 45 000)
Output
If such partition doesn't exist, print "No" (quotes for clarity).
Otherwise, print "Yes" (quotes for clarity), followed by two lines, describing S_1 and S_2 respectively.
Each set description starts with the set size, followed by the elements of the set in any order. Each set must be non-empty.
If there are multiple possible partitions — print any of them.
Examples
Input
1
Output
No
Input
3
Output
Yes
1 2
2 1 3
Note
In the first example, there is no way to partition a single number into two non-empty sets, hence the answer is "No".
In the second example, the sums of the sets are 2 and 4 respectively. The gcd(2, 4) = 2 > 1, hence that is one of the possible answers.
Submitted Solution:
```
n=int(input())
x=35
#import math
#print(math.gcd (35,5))
if (n<=2) :
exit (print("No"))
t=(n+1)//2
#if (t%2==0) or n%2==0:
print("Yes")
print((n+1)//2,end=" ")
for i in range(1,n+1,2) :
print(i,end=" ")
print("")
print(n//2,end=" ")
for i in range(2,n+1,2) :
print(i,end=" ")
``` | instruction | 0 | 74,190 | 22 | 148,380 |
Yes | output | 1 | 74,190 | 22 | 148,381 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Find out if it is possible to partition the first n positive integers into two non-empty disjoint sets S_1 and S_2 such that:
gcd(sum(S_1), sum(S_2)) > 1
Here sum(S) denotes the sum of all elements present in set S and gcd means the[greatest common divisor](https://en.wikipedia.org/wiki/Greatest_common_divisor).
Every integer number from 1 to n should be present in exactly one of S_1 or S_2.
Input
The only line of the input contains a single integer n (1 ≤ n ≤ 45 000)
Output
If such partition doesn't exist, print "No" (quotes for clarity).
Otherwise, print "Yes" (quotes for clarity), followed by two lines, describing S_1 and S_2 respectively.
Each set description starts with the set size, followed by the elements of the set in any order. Each set must be non-empty.
If there are multiple possible partitions — print any of them.
Examples
Input
1
Output
No
Input
3
Output
Yes
1 2
2 1 3
Note
In the first example, there is no way to partition a single number into two non-empty sets, hence the answer is "No".
In the second example, the sums of the sets are 2 and 4 respectively. The gcd(2, 4) = 2 > 1, hence that is one of the possible answers.
Submitted Solution:
```
n = int(input())
a=[]
b=[]
check=False
if n%3==0:
check=True
for i in range(1,n+1):
if i%3==0:
b.append(i)
else:
a.append(i)
elif (n%2==1 and (n//2)%2==1) or (n%2==0 and (n//2)%2==0):
check=True
for i in range(1,n+1):
if i%2==0:
a.append(i)
else:
b.append(i)
elif n%2==1 and n>1:
check=True
a.append(n)
for i in range(1,n):
b.append(i)
elif n%2==0 and (n//2)%2==1 and n!=2:
check=True
a.append(n//2)
for i in range(1,n+1):
b.append(i)
b.remove(n//2)
if check:
print("Yes")
print(len(a),end=' ')
for i in a:
print(i,end=' ')
print()
print(len(b),end=' ')
for i in b:
print(i,end=' ')
print()
else:
print("No")
``` | instruction | 0 | 74,191 | 22 | 148,382 |
Yes | output | 1 | 74,191 | 22 | 148,383 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Find out if it is possible to partition the first n positive integers into two non-empty disjoint sets S_1 and S_2 such that:
gcd(sum(S_1), sum(S_2)) > 1
Here sum(S) denotes the sum of all elements present in set S and gcd means the[greatest common divisor](https://en.wikipedia.org/wiki/Greatest_common_divisor).
Every integer number from 1 to n should be present in exactly one of S_1 or S_2.
Input
The only line of the input contains a single integer n (1 ≤ n ≤ 45 000)
Output
If such partition doesn't exist, print "No" (quotes for clarity).
Otherwise, print "Yes" (quotes for clarity), followed by two lines, describing S_1 and S_2 respectively.
Each set description starts with the set size, followed by the elements of the set in any order. Each set must be non-empty.
If there are multiple possible partitions — print any of them.
Examples
Input
1
Output
No
Input
3
Output
Yes
1 2
2 1 3
Note
In the first example, there is no way to partition a single number into two non-empty sets, hence the answer is "No".
In the second example, the sums of the sets are 2 and 4 respectively. The gcd(2, 4) = 2 > 1, hence that is one of the possible answers.
Submitted Solution:
```
# cook your dish here
#import sys
#sys.setrecursionlimit(10**9)
ll=lambda:map(int,input().split())
t=lambda:int(input())
ss=lambda:input()
lx=lambda x:map(int,input().split(x))
yy=lambda:print("Yes")
nn=lambda:print("No")
from math import log10 ,log2,ceil,factorial as fac,gcd,inf,sqrt,log
#from itertools import combinations_with_replacement as cs
#from functools import reduce
from bisect import bisect_right as br,bisect_left as bl
from collections import Counter
#from math import inf
mod=10**9+7
#for _ in range(t()):
def f():
n=t()
s=(n*(n+1))//2 -1
s1=1
l=list(range(2,n+1))
p=[1]
while len(l)-1>0:
x=l.pop(0)
p.append(x)
s1+=x
s-=x
if gcd(s,s1)>1:
yy()
print(len(p),*p)
print(len(l),*l)
return
nn()
f()
'''
baca
bac
1 2 3
baaccca
abbaccccaba
'''
``` | instruction | 0 | 74,192 | 22 | 148,384 |
Yes | output | 1 | 74,192 | 22 | 148,385 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Find out if it is possible to partition the first n positive integers into two non-empty disjoint sets S_1 and S_2 such that:
gcd(sum(S_1), sum(S_2)) > 1
Here sum(S) denotes the sum of all elements present in set S and gcd means the[greatest common divisor](https://en.wikipedia.org/wiki/Greatest_common_divisor).
Every integer number from 1 to n should be present in exactly one of S_1 or S_2.
Input
The only line of the input contains a single integer n (1 ≤ n ≤ 45 000)
Output
If such partition doesn't exist, print "No" (quotes for clarity).
Otherwise, print "Yes" (quotes for clarity), followed by two lines, describing S_1 and S_2 respectively.
Each set description starts with the set size, followed by the elements of the set in any order. Each set must be non-empty.
If there are multiple possible partitions — print any of them.
Examples
Input
1
Output
No
Input
3
Output
Yes
1 2
2 1 3
Note
In the first example, there is no way to partition a single number into two non-empty sets, hence the answer is "No".
In the second example, the sums of the sets are 2 and 4 respectively. The gcd(2, 4) = 2 > 1, hence that is one of the possible answers.
Submitted Solution:
```
n=int(input())
if n==1 or n==2:
print('No')
raise SystemExit
print ('Yes')
if n%2==0:
print (1,int(n/2))
print(2,(' '.join( [str(i) for i in range(1,n+1) if i!=n/2])))
else:
print (1,int((n-1)/2+1))
print (2,(' '.join([str(i) for i in range(1,n+1) if i!=(n-1)/2+1])))
``` | instruction | 0 | 74,193 | 22 | 148,386 |
No | output | 1 | 74,193 | 22 | 148,387 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Find out if it is possible to partition the first n positive integers into two non-empty disjoint sets S_1 and S_2 such that:
gcd(sum(S_1), sum(S_2)) > 1
Here sum(S) denotes the sum of all elements present in set S and gcd means the[greatest common divisor](https://en.wikipedia.org/wiki/Greatest_common_divisor).
Every integer number from 1 to n should be present in exactly one of S_1 or S_2.
Input
The only line of the input contains a single integer n (1 ≤ n ≤ 45 000)
Output
If such partition doesn't exist, print "No" (quotes for clarity).
Otherwise, print "Yes" (quotes for clarity), followed by two lines, describing S_1 and S_2 respectively.
Each set description starts with the set size, followed by the elements of the set in any order. Each set must be non-empty.
If there are multiple possible partitions — print any of them.
Examples
Input
1
Output
No
Input
3
Output
Yes
1 2
2 1 3
Note
In the first example, there is no way to partition a single number into two non-empty sets, hence the answer is "No".
In the second example, the sums of the sets are 2 and 4 respectively. The gcd(2, 4) = 2 > 1, hence that is one of the possible answers.
Submitted Solution:
```
n = int(input())
if n==1:
print("NO")
else:
print("YES")
print("1")
print(*[i for i in range(1,n+1())])
``` | instruction | 0 | 74,194 | 22 | 148,388 |
No | output | 1 | 74,194 | 22 | 148,389 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Find out if it is possible to partition the first n positive integers into two non-empty disjoint sets S_1 and S_2 such that:
gcd(sum(S_1), sum(S_2)) > 1
Here sum(S) denotes the sum of all elements present in set S and gcd means the[greatest common divisor](https://en.wikipedia.org/wiki/Greatest_common_divisor).
Every integer number from 1 to n should be present in exactly one of S_1 or S_2.
Input
The only line of the input contains a single integer n (1 ≤ n ≤ 45 000)
Output
If such partition doesn't exist, print "No" (quotes for clarity).
Otherwise, print "Yes" (quotes for clarity), followed by two lines, describing S_1 and S_2 respectively.
Each set description starts with the set size, followed by the elements of the set in any order. Each set must be non-empty.
If there are multiple possible partitions — print any of them.
Examples
Input
1
Output
No
Input
3
Output
Yes
1 2
2 1 3
Note
In the first example, there is no way to partition a single number into two non-empty sets, hence the answer is "No".
In the second example, the sums of the sets are 2 and 4 respectively. The gcd(2, 4) = 2 > 1, hence that is one of the possible answers.
Submitted Solution:
```
n = int(input())
if n <= 2:
print("No")
else:
print("Yes")
if n % 2 == 0:
k = n / 2
else:
k = (n + 1)/ 2
print("1" , int(k))
print(str(n - 1), end="")
k = int(k)
for i in range(1, n + 1):
if i == k:
continue
print(" " + str (i), end="")
print("\n")
``` | instruction | 0 | 74,195 | 22 | 148,390 |
No | output | 1 | 74,195 | 22 | 148,391 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Find out if it is possible to partition the first n positive integers into two non-empty disjoint sets S_1 and S_2 such that:
gcd(sum(S_1), sum(S_2)) > 1
Here sum(S) denotes the sum of all elements present in set S and gcd means the[greatest common divisor](https://en.wikipedia.org/wiki/Greatest_common_divisor).
Every integer number from 1 to n should be present in exactly one of S_1 or S_2.
Input
The only line of the input contains a single integer n (1 ≤ n ≤ 45 000)
Output
If such partition doesn't exist, print "No" (quotes for clarity).
Otherwise, print "Yes" (quotes for clarity), followed by two lines, describing S_1 and S_2 respectively.
Each set description starts with the set size, followed by the elements of the set in any order. Each set must be non-empty.
If there are multiple possible partitions — print any of them.
Examples
Input
1
Output
No
Input
3
Output
Yes
1 2
2 1 3
Note
In the first example, there is no way to partition a single number into two non-empty sets, hence the answer is "No".
In the second example, the sums of the sets are 2 and 4 respectively. The gcd(2, 4) = 2 > 1, hence that is one of the possible answers.
Submitted Solution:
```
def ok(n):
if n<=2:
print('No')
else:
for _ in range(1):
print('Yes')
print(1,n-1)
print(n-1,n-2,n)
ok(int(input()))
``` | instruction | 0 | 74,196 | 22 | 148,392 |
No | output | 1 | 74,196 | 22 | 148,393 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Petya loves looking for numbers' divisors. One day Petya came across the following problem:
You are given n queries in the form "xi yi". For each query Petya should count how many divisors of number xi divide none of the numbers xi - yi, xi - yi + 1, ..., xi - 1. Help him.
Input
The first line contains an integer n (1 ≤ n ≤ 105). Each of the following n lines contain two space-separated integers xi and yi (1 ≤ xi ≤ 105, 0 ≤ yi ≤ i - 1, where i is the query's ordinal number; the numeration starts with 1).
If yi = 0 for the query, then the answer to the query will be the number of divisors of the number xi. In this case you do not need to take the previous numbers x into consideration.
Output
For each query print the answer on a single line: the number of positive integers k such that <image>
Examples
Input
6
4 0
3 1
5 2
6 2
18 4
10000 3
Output
3
1
1
2
2
22
Note
Let's write out the divisors that give answers for the first 5 queries:
1) 1, 2, 4
2) 3
3) 5
4) 2, 6
5) 9, 18 | instruction | 0 | 74,219 | 22 | 148,438 |
Tags: binary search, data structures, number theory
Correct Solution:
```
import sys,collections as cc
input = sys.stdin.readline
I = lambda : list(map(int,input().split()))
def div(b):
an=[]
for i in range(1,int(b**0.5)+1):
if b%i==0:
an.append(i)
if i!=b//i:
an.append(b//i)
return an
n,=I()
ar=[]
vis=[-1]*(10**5+2)
for i in range(n):
a,b=I()
an=0
dv=div(a)
#print(dv,vis[:20])
for j in dv:
if vis[j]<i-b:
an+=1
vis[j]=i
print(an)
``` | output | 1 | 74,219 | 22 | 148,439 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Petya loves looking for numbers' divisors. One day Petya came across the following problem:
You are given n queries in the form "xi yi". For each query Petya should count how many divisors of number xi divide none of the numbers xi - yi, xi - yi + 1, ..., xi - 1. Help him.
Input
The first line contains an integer n (1 ≤ n ≤ 105). Each of the following n lines contain two space-separated integers xi and yi (1 ≤ xi ≤ 105, 0 ≤ yi ≤ i - 1, where i is the query's ordinal number; the numeration starts with 1).
If yi = 0 for the query, then the answer to the query will be the number of divisors of the number xi. In this case you do not need to take the previous numbers x into consideration.
Output
For each query print the answer on a single line: the number of positive integers k such that <image>
Examples
Input
6
4 0
3 1
5 2
6 2
18 4
10000 3
Output
3
1
1
2
2
22
Note
Let's write out the divisors that give answers for the first 5 queries:
1) 1, 2, 4
2) 3
3) 5
4) 2, 6
5) 9, 18 | instruction | 0 | 74,220 | 22 | 148,440 |
Tags: binary search, data structures, number theory
Correct Solution:
```
import sys
import collections as cc
input=sys.stdin.buffer.readline
I=lambda:list(map(int,input().split()))
prev=cc.defaultdict(int)
for tc in range(int(input())):
x,y=I()
div=set()
for i in range(1,int(x**0.5)+1):
if x%i==0:
div.add(i)
div.add(x//i)
ans=0
now=tc+1
for i in div:
if now-prev[i]>y:
ans+=1
prev[i]=now
print(ans)
``` | output | 1 | 74,220 | 22 | 148,441 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Petya loves looking for numbers' divisors. One day Petya came across the following problem:
You are given n queries in the form "xi yi". For each query Petya should count how many divisors of number xi divide none of the numbers xi - yi, xi - yi + 1, ..., xi - 1. Help him.
Input
The first line contains an integer n (1 ≤ n ≤ 105). Each of the following n lines contain two space-separated integers xi and yi (1 ≤ xi ≤ 105, 0 ≤ yi ≤ i - 1, where i is the query's ordinal number; the numeration starts with 1).
If yi = 0 for the query, then the answer to the query will be the number of divisors of the number xi. In this case you do not need to take the previous numbers x into consideration.
Output
For each query print the answer on a single line: the number of positive integers k such that <image>
Examples
Input
6
4 0
3 1
5 2
6 2
18 4
10000 3
Output
3
1
1
2
2
22
Note
Let's write out the divisors that give answers for the first 5 queries:
1) 1, 2, 4
2) 3
3) 5
4) 2, 6
5) 9, 18 | instruction | 0 | 74,221 | 22 | 148,442 |
Tags: binary search, data structures, number theory
Correct Solution:
```
from collections import defaultdict
mp=defaultdict(lambda :0)
x=int(input())
for k in range(1,x+1):
a,b=map(int,input().split())
res=0
for n in range(1,int(a**.5)+1):
if a%n==0:
if mp[n]==0:
res+=1
elif mp[n]<k-b:
res+=1
mp[n]=k
if n*n!=a:
v=a//n
if mp[v]==0:
res+=1
elif mp[v]<k-b:
res+=1
mp[v]=k
print(res)
``` | output | 1 | 74,221 | 22 | 148,443 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Petya loves looking for numbers' divisors. One day Petya came across the following problem:
You are given n queries in the form "xi yi". For each query Petya should count how many divisors of number xi divide none of the numbers xi - yi, xi - yi + 1, ..., xi - 1. Help him.
Input
The first line contains an integer n (1 ≤ n ≤ 105). Each of the following n lines contain two space-separated integers xi and yi (1 ≤ xi ≤ 105, 0 ≤ yi ≤ i - 1, where i is the query's ordinal number; the numeration starts with 1).
If yi = 0 for the query, then the answer to the query will be the number of divisors of the number xi. In this case you do not need to take the previous numbers x into consideration.
Output
For each query print the answer on a single line: the number of positive integers k such that <image>
Examples
Input
6
4 0
3 1
5 2
6 2
18 4
10000 3
Output
3
1
1
2
2
22
Note
Let's write out the divisors that give answers for the first 5 queries:
1) 1, 2, 4
2) 3
3) 5
4) 2, 6
5) 9, 18 | instruction | 0 | 74,222 | 22 | 148,444 |
Tags: binary search, data structures, number theory
Correct Solution:
```
import sys
import collections as cc
input=sys.stdin.readline
I=lambda:list(map(int,input().split()))
prev=cc.defaultdict(int)
for tc in range(int(input())):
x,y=I()
div=set()
for i in range(1,int(x**0.5)+1):
if x%i==0:
div.add(i)
div.add(x//i)
ans=0
now=tc+1
for i in div:
if now-prev[i]>y:
ans+=1
prev[i]=now
print(ans)
``` | output | 1 | 74,222 | 22 | 148,445 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Greatest common divisor GCD(a, b) of two positive integers a and b is equal to the biggest integer d such that both integers a and b are divisible by d. There are many efficient algorithms to find greatest common divisor GCD(a, b), for example, Euclid algorithm.
Formally, find the biggest integer d, such that all integers a, a + 1, a + 2, ..., b are divisible by d. To make the problem even more complicated we allow a and b to be up to googol, 10100 — such number do not fit even in 64-bit integer type!
Input
The only line of the input contains two integers a and b (1 ≤ a ≤ b ≤ 10100).
Output
Output one integer — greatest common divisor of all integers from a to b inclusive.
Examples
Input
1 2
Output
1
Input
61803398874989484820458683436563811772030917980576 61803398874989484820458683436563811772030917980576
Output
61803398874989484820458683436563811772030917980576 | instruction | 0 | 74,604 | 22 | 149,208 |
Tags: math, number theory
Correct Solution:
```
s , s1 = map(str , input().split())
print(s if s == s1 else 1)
``` | output | 1 | 74,604 | 22 | 149,209 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Greatest common divisor GCD(a, b) of two positive integers a and b is equal to the biggest integer d such that both integers a and b are divisible by d. There are many efficient algorithms to find greatest common divisor GCD(a, b), for example, Euclid algorithm.
Formally, find the biggest integer d, such that all integers a, a + 1, a + 2, ..., b are divisible by d. To make the problem even more complicated we allow a and b to be up to googol, 10100 — such number do not fit even in 64-bit integer type!
Input
The only line of the input contains two integers a and b (1 ≤ a ≤ b ≤ 10100).
Output
Output one integer — greatest common divisor of all integers from a to b inclusive.
Examples
Input
1 2
Output
1
Input
61803398874989484820458683436563811772030917980576 61803398874989484820458683436563811772030917980576
Output
61803398874989484820458683436563811772030917980576 | instruction | 0 | 74,605 | 22 | 149,210 |
Tags: math, number theory
Correct Solution:
```
num,num2 = map(str,input().split())
if num == num2:
print(num)
else:
print(1)
``` | output | 1 | 74,605 | 22 | 149,211 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Greatest common divisor GCD(a, b) of two positive integers a and b is equal to the biggest integer d such that both integers a and b are divisible by d. There are many efficient algorithms to find greatest common divisor GCD(a, b), for example, Euclid algorithm.
Formally, find the biggest integer d, such that all integers a, a + 1, a + 2, ..., b are divisible by d. To make the problem even more complicated we allow a and b to be up to googol, 10100 — such number do not fit even in 64-bit integer type!
Input
The only line of the input contains two integers a and b (1 ≤ a ≤ b ≤ 10100).
Output
Output one integer — greatest common divisor of all integers from a to b inclusive.
Examples
Input
1 2
Output
1
Input
61803398874989484820458683436563811772030917980576 61803398874989484820458683436563811772030917980576
Output
61803398874989484820458683436563811772030917980576 | instruction | 0 | 74,606 | 22 | 149,212 |
Tags: math, number theory
Correct Solution:
```
a, b = [int(x) for x in input().split()]
print(1 if a != b else a)
``` | output | 1 | 74,606 | 22 | 149,213 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Greatest common divisor GCD(a, b) of two positive integers a and b is equal to the biggest integer d such that both integers a and b are divisible by d. There are many efficient algorithms to find greatest common divisor GCD(a, b), for example, Euclid algorithm.
Formally, find the biggest integer d, such that all integers a, a + 1, a + 2, ..., b are divisible by d. To make the problem even more complicated we allow a and b to be up to googol, 10100 — such number do not fit even in 64-bit integer type!
Input
The only line of the input contains two integers a and b (1 ≤ a ≤ b ≤ 10100).
Output
Output one integer — greatest common divisor of all integers from a to b inclusive.
Examples
Input
1 2
Output
1
Input
61803398874989484820458683436563811772030917980576 61803398874989484820458683436563811772030917980576
Output
61803398874989484820458683436563811772030917980576 | instruction | 0 | 74,607 | 22 | 149,214 |
Tags: math, number theory
Correct Solution:
```
import math
n = input()
n = n.split()
s = eval(n[0])
n = eval(n[1])
if n == s:
print(n)
else:
print(1)
``` | output | 1 | 74,607 | 22 | 149,215 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Greatest common divisor GCD(a, b) of two positive integers a and b is equal to the biggest integer d such that both integers a and b are divisible by d. There are many efficient algorithms to find greatest common divisor GCD(a, b), for example, Euclid algorithm.
Formally, find the biggest integer d, such that all integers a, a + 1, a + 2, ..., b are divisible by d. To make the problem even more complicated we allow a and b to be up to googol, 10100 — such number do not fit even in 64-bit integer type!
Input
The only line of the input contains two integers a and b (1 ≤ a ≤ b ≤ 10100).
Output
Output one integer — greatest common divisor of all integers from a to b inclusive.
Examples
Input
1 2
Output
1
Input
61803398874989484820458683436563811772030917980576 61803398874989484820458683436563811772030917980576
Output
61803398874989484820458683436563811772030917980576 | instruction | 0 | 74,608 | 22 | 149,216 |
Tags: math, number theory
Correct Solution:
```
m,g=map(int,input().split())
if m==g:
print(m)
else:
print("1")
``` | output | 1 | 74,608 | 22 | 149,217 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Greatest common divisor GCD(a, b) of two positive integers a and b is equal to the biggest integer d such that both integers a and b are divisible by d. There are many efficient algorithms to find greatest common divisor GCD(a, b), for example, Euclid algorithm.
Formally, find the biggest integer d, such that all integers a, a + 1, a + 2, ..., b are divisible by d. To make the problem even more complicated we allow a and b to be up to googol, 10100 — such number do not fit even in 64-bit integer type!
Input
The only line of the input contains two integers a and b (1 ≤ a ≤ b ≤ 10100).
Output
Output one integer — greatest common divisor of all integers from a to b inclusive.
Examples
Input
1 2
Output
1
Input
61803398874989484820458683436563811772030917980576 61803398874989484820458683436563811772030917980576
Output
61803398874989484820458683436563811772030917980576 | instruction | 0 | 74,609 | 22 | 149,218 |
Tags: math, number theory
Correct Solution:
```
x,y=map(int,input().split())
print(x if x==y else 1)
``` | output | 1 | 74,609 | 22 | 149,219 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Greatest common divisor GCD(a, b) of two positive integers a and b is equal to the biggest integer d such that both integers a and b are divisible by d. There are many efficient algorithms to find greatest common divisor GCD(a, b), for example, Euclid algorithm.
Formally, find the biggest integer d, such that all integers a, a + 1, a + 2, ..., b are divisible by d. To make the problem even more complicated we allow a and b to be up to googol, 10100 — such number do not fit even in 64-bit integer type!
Input
The only line of the input contains two integers a and b (1 ≤ a ≤ b ≤ 10100).
Output
Output one integer — greatest common divisor of all integers from a to b inclusive.
Examples
Input
1 2
Output
1
Input
61803398874989484820458683436563811772030917980576 61803398874989484820458683436563811772030917980576
Output
61803398874989484820458683436563811772030917980576 | instruction | 0 | 74,610 | 22 | 149,220 |
Tags: math, number theory
Correct Solution:
```
def my_gcd(x, y):
temp = x % y
if temp == 0:
return y
return my_gcd(y, temp)
m, n = map(int, input().split())
if m == n:
print(m)
else:
if m < n:
s = m
l = n
else:
s = n
l = m
gcd = 0
for i in range(s, l):
for j in range(i+1, l+1):
gcd = my_gcd(i, j)
if gcd == 1:
break
if gcd == 1:
break
print(gcd)
``` | output | 1 | 74,610 | 22 | 149,221 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Greatest common divisor GCD(a, b) of two positive integers a and b is equal to the biggest integer d such that both integers a and b are divisible by d. There are many efficient algorithms to find greatest common divisor GCD(a, b), for example, Euclid algorithm.
Formally, find the biggest integer d, such that all integers a, a + 1, a + 2, ..., b are divisible by d. To make the problem even more complicated we allow a and b to be up to googol, 10100 — such number do not fit even in 64-bit integer type!
Input
The only line of the input contains two integers a and b (1 ≤ a ≤ b ≤ 10100).
Output
Output one integer — greatest common divisor of all integers from a to b inclusive.
Examples
Input
1 2
Output
1
Input
61803398874989484820458683436563811772030917980576 61803398874989484820458683436563811772030917980576
Output
61803398874989484820458683436563811772030917980576 | instruction | 0 | 74,611 | 22 | 149,222 |
Tags: math, number theory
Correct Solution:
```
#print("muri")
a, b = map(int, input().split())
if a==b:print(a)
else: print("1")
``` | output | 1 | 74,611 | 22 | 149,223 |
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