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.
We guessed some integer number x. You are given a list of almost all its divisors. Almost all means that there are all divisors except 1 and x in the list.
Your task is to find the minimum possible integer x that can be the guessed number, or say that the input data is contradictory and it is impossible to find such number.
You have to answer t independent queries.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 25) — the number of queries. Then t queries follow.
The first line of the query contains one integer n (1 ≤ n ≤ 300) — the number of divisors in the list.
The second line of the query contains n integers d_1, d_2, ..., d_n (2 ≤ d_i ≤ 10^6), where d_i is the i-th divisor of the guessed number. It is guaranteed that all values d_i are distinct.
Output
For each query print the answer to it.
If the input data in the query is contradictory and it is impossible to find such number x that the given list of divisors is the list of almost all its divisors, print -1. Otherwise print the minimum possible x.
Example
Input
2
8
8 2 12 6 4 24 16 3
1
2
Output
48
4
Submitted Solution:
```
import math
for _ in range(int(input())):
n=int(input())
a=list(map(int,input().split()))
a.sort()
num=a[0]*a[-1]
b=[]
for i in range(2,int(math.sqrt(num))+1):
if num%i==0:
b.append(i)
if(i*i!=num):
b.append(num//i)
b.sort()
if(a==b):
print(num)
else :
print(-1)
``` | instruction | 0 | 38,723 | 22 | 77,446 |
Yes | output | 1 | 38,723 | 22 | 77,447 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We guessed some integer number x. You are given a list of almost all its divisors. Almost all means that there are all divisors except 1 and x in the list.
Your task is to find the minimum possible integer x that can be the guessed number, or say that the input data is contradictory and it is impossible to find such number.
You have to answer t independent queries.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 25) — the number of queries. Then t queries follow.
The first line of the query contains one integer n (1 ≤ n ≤ 300) — the number of divisors in the list.
The second line of the query contains n integers d_1, d_2, ..., d_n (2 ≤ d_i ≤ 10^6), where d_i is the i-th divisor of the guessed number. It is guaranteed that all values d_i are distinct.
Output
For each query print the answer to it.
If the input data in the query is contradictory and it is impossible to find such number x that the given list of divisors is the list of almost all its divisors, print -1. Otherwise print the minimum possible x.
Example
Input
2
8
8 2 12 6 4 24 16 3
1
2
Output
48
4
Submitted Solution:
```
I = input
for i in ' '*int(I()):
I()
a = {*map(int, I().split())}
s = min(a)*max(a)
d = set()
i = 2
while i*i <= s:
if s % i == 0:
d |= {i,s//i}
i += 1
if a != d:
s = -1
print(s)
``` | instruction | 0 | 38,724 | 22 | 77,448 |
Yes | output | 1 | 38,724 | 22 | 77,449 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We guessed some integer number x. You are given a list of almost all its divisors. Almost all means that there are all divisors except 1 and x in the list.
Your task is to find the minimum possible integer x that can be the guessed number, or say that the input data is contradictory and it is impossible to find such number.
You have to answer t independent queries.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 25) — the number of queries. Then t queries follow.
The first line of the query contains one integer n (1 ≤ n ≤ 300) — the number of divisors in the list.
The second line of the query contains n integers d_1, d_2, ..., d_n (2 ≤ d_i ≤ 10^6), where d_i is the i-th divisor of the guessed number. It is guaranteed that all values d_i are distinct.
Output
For each query print the answer to it.
If the input data in the query is contradictory and it is impossible to find such number x that the given list of divisors is the list of almost all its divisors, print -1. Otherwise print the minimum possible x.
Example
Input
2
8
8 2 12 6 4 24 16 3
1
2
Output
48
4
Submitted Solution:
```
t = int(input())
def make_divisors(n):
divisors = []
for i in range(1, int(n**0.5)+1):
if n % i == 0:
divisors.append(i)
if i != n // i:
divisors.append(n//i)
#divisors.sort(reverse=True)
return divisors
for _ in range(t):
n = int(input())
D = list(map(int, input().split()))
D.sort()
#print(D)
if len(D) == 1:
x = D[0]**2
else:
x = D[0]*D[-1]
D = set(D)
D.add(1)
D.add(x)
L = make_divisors(x)
if len(D) != len(L):
print(-1)
else:
flag = True
for i in L:
if i not in D:
flag = False
break
if flag:
print(x)
else:
print(-1)
``` | instruction | 0 | 38,725 | 22 | 77,450 |
Yes | output | 1 | 38,725 | 22 | 77,451 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We guessed some integer number x. You are given a list of almost all its divisors. Almost all means that there are all divisors except 1 and x in the list.
Your task is to find the minimum possible integer x that can be the guessed number, or say that the input data is contradictory and it is impossible to find such number.
You have to answer t independent queries.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 25) — the number of queries. Then t queries follow.
The first line of the query contains one integer n (1 ≤ n ≤ 300) — the number of divisors in the list.
The second line of the query contains n integers d_1, d_2, ..., d_n (2 ≤ d_i ≤ 10^6), where d_i is the i-th divisor of the guessed number. It is guaranteed that all values d_i are distinct.
Output
For each query print the answer to it.
If the input data in the query is contradictory and it is impossible to find such number x that the given list of divisors is the list of almost all its divisors, print -1. Otherwise print the minimum possible x.
Example
Input
2
8
8 2 12 6 4 24 16 3
1
2
Output
48
4
Submitted Solution:
```
if __name__ == '__main__':
t = int(input())
for i in range(t):
n = int(input())
ds = list(map(int, input().split()))
print(min(ds) * max(ds))
``` | instruction | 0 | 38,726 | 22 | 77,452 |
No | output | 1 | 38,726 | 22 | 77,453 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We guessed some integer number x. You are given a list of almost all its divisors. Almost all means that there are all divisors except 1 and x in the list.
Your task is to find the minimum possible integer x that can be the guessed number, or say that the input data is contradictory and it is impossible to find such number.
You have to answer t independent queries.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 25) — the number of queries. Then t queries follow.
The first line of the query contains one integer n (1 ≤ n ≤ 300) — the number of divisors in the list.
The second line of the query contains n integers d_1, d_2, ..., d_n (2 ≤ d_i ≤ 10^6), where d_i is the i-th divisor of the guessed number. It is guaranteed that all values d_i are distinct.
Output
For each query print the answer to it.
If the input data in the query is contradictory and it is impossible to find such number x that the given list of divisors is the list of almost all its divisors, print -1. Otherwise print the minimum possible x.
Example
Input
2
8
8 2 12 6 4 24 16 3
1
2
Output
48
4
Submitted Solution:
```
def almost(n, d):
d = sorted(d)
result = d[0] * d[-1]
div = []
for i in range(2, int(result**0.5) + 1):
if result % i == 0:
if result // i == i:
div.append(i)
else:
div.append(i)
div.append(result // i)
div = sorted(div)
return result if d != div else -1
if __name__ == '__main__':
t = int(input())
for i in range(t):
n = int(input())
d = [int(item) for item in input().split()]
print(almost(n, d))
``` | instruction | 0 | 38,727 | 22 | 77,454 |
No | output | 1 | 38,727 | 22 | 77,455 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We guessed some integer number x. You are given a list of almost all its divisors. Almost all means that there are all divisors except 1 and x in the list.
Your task is to find the minimum possible integer x that can be the guessed number, or say that the input data is contradictory and it is impossible to find such number.
You have to answer t independent queries.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 25) — the number of queries. Then t queries follow.
The first line of the query contains one integer n (1 ≤ n ≤ 300) — the number of divisors in the list.
The second line of the query contains n integers d_1, d_2, ..., d_n (2 ≤ d_i ≤ 10^6), where d_i is the i-th divisor of the guessed number. It is guaranteed that all values d_i are distinct.
Output
For each query print the answer to it.
If the input data in the query is contradictory and it is impossible to find such number x that the given list of divisors is the list of almost all its divisors, print -1. Otherwise print the minimum possible x.
Example
Input
2
8
8 2 12 6 4 24 16 3
1
2
Output
48
4
Submitted Solution:
```
t = int(input())
prime = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293, 307, 311, 313, 317, 331, 337, 347, 349, 353, 359, 367, 373, 379, 383, 389, 397, 401, 409, 419, 421, 431, 433, 439, 443, 449, 457, 461, 463, 467, 479, 487, 491, 499, 503, 509, 521, 523, 541, 547, 557, 563, 569, 571, 577, 587, 593, 599, 601, 607, 613, 617, 619, 631, 641, 643, 647, 653, 659, 661, 673, 677, 683, 691, 701, 709, 719, 727, 733, 739, 743, 751, 757, 761, 769, 773, 787, 797, 809, 811, 821, 823, 827, 829, 839, 853, 857, 859, 863, 877, 881, 883, 887, 907, 911, 919, 929, 937, 941, 947, 953, 967, 971, 977, 983, 991, 997, 1009]
for i in range(t):
n = int(input())
l = list(map(int,input().split()))
mini = min(l)
maksi = max(l)
a = mini * maksi
c = 0
for d in l:
if a % d != 0:
c = 1
break
if c == 1:
print(-1)
continue
# ile divisors
fac = []
i = 2
for p in prime:
if p > max(maksi, mini):
break
if maksi % p != 0 and mini % p != 0:
continue
chuj = 0
while True:
if maksi % p == 0:
maksi = maksi // p
chuj += 1
else:
break
while True:
if mini % p == 0:
mini = mini // p
chuj += 1
else:
break
if chuj > 0:
fac.append([p, chuj])
czek = 1
diwi = 1
for f in fac:
czek = czek * (f[0] ** f[1])
diwi = diwi * (f[1] + 1)
if czek != a:
diwi *= 2
print(diwi)
if diwi == 2:
diwi += 1
if diwi - 2 == n:
print(a)
else:
print(-1)
``` | instruction | 0 | 38,728 | 22 | 77,456 |
No | output | 1 | 38,728 | 22 | 77,457 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We guessed some integer number x. You are given a list of almost all its divisors. Almost all means that there are all divisors except 1 and x in the list.
Your task is to find the minimum possible integer x that can be the guessed number, or say that the input data is contradictory and it is impossible to find such number.
You have to answer t independent queries.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 25) — the number of queries. Then t queries follow.
The first line of the query contains one integer n (1 ≤ n ≤ 300) — the number of divisors in the list.
The second line of the query contains n integers d_1, d_2, ..., d_n (2 ≤ d_i ≤ 10^6), where d_i is the i-th divisor of the guessed number. It is guaranteed that all values d_i are distinct.
Output
For each query print the answer to it.
If the input data in the query is contradictory and it is impossible to find such number x that the given list of divisors is the list of almost all its divisors, print -1. Otherwise print the minimum possible x.
Example
Input
2
8
8 2 12 6 4 24 16 3
1
2
Output
48
4
Submitted Solution:
```
t=int(input())
for _ in range(0,t):
n=int(input())
d=list(map(int,input().split()))
d.sort()
ans=d[0]*d[-1]
if n==1:
print(ans)
else:
fl=0
fl1=0
if n%2==0:
for i in range(1,n//2):
r=d[i]*d[-1-i]
if ans!=r:
fl=1
print(-1)
break
else:
for i in range(1,n//2):
r=d[i]*d[-1-i]
if i==n//2-1:
if ans%d[i+1]:
fl1=1
print(-1)
if ans!=r and fl1==0:
fl=1
print(-1)
break
if fl==0:
print(ans)
``` | instruction | 0 | 38,729 | 22 | 77,458 |
No | output | 1 | 38,729 | 22 | 77,459 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given an array a, consisting of n integers, find:
$$$max_{1 ≤ i < j ≤ n} LCM(a_i,a_j),$$$
where LCM(x, y) is the smallest positive integer that is divisible by both x and y. For example, LCM(6, 8) = 24, LCM(4, 12) = 12, LCM(2, 3) = 6.
Input
The first line contains an integer n (2 ≤ n ≤ 10^5) — the number of elements in the array a.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^5) — the elements of the array a.
Output
Print one integer, the maximum value of the least common multiple of two elements in the array a.
Examples
Input
3
13 35 77
Output
1001
Input
6
1 2 4 8 16 32
Output
32
Submitted Solution:
```
def lcm(a,b):
m = a*b
while a != 0 and b != 0:
if a > b:
a %= b
else:
b %= a
return m // (a+b)
n=int(input())
b=[int (i) for i in input().split(' ')]
c=[]
for i in range(0,n-1):
c.append(lcm(b[n-1],b[i]))
print(max(c))
``` | instruction | 0 | 38,768 | 22 | 77,536 |
No | output | 1 | 38,768 | 22 | 77,537 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given an array a, consisting of n integers, find:
$$$max_{1 ≤ i < j ≤ n} LCM(a_i,a_j),$$$
where LCM(x, y) is the smallest positive integer that is divisible by both x and y. For example, LCM(6, 8) = 24, LCM(4, 12) = 12, LCM(2, 3) = 6.
Input
The first line contains an integer n (2 ≤ n ≤ 10^5) — the number of elements in the array a.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^5) — the elements of the array a.
Output
Print one integer, the maximum value of the least common multiple of two elements in the array a.
Examples
Input
3
13 35 77
Output
1001
Input
6
1 2 4 8 16 32
Output
32
Submitted Solution:
```
import math
n = int(input())
a = list(map(int, input().split()))
print((a[0]*a[n-1])//math.gcd(a[0], a[n-1]))
``` | instruction | 0 | 38,769 | 22 | 77,538 |
No | output | 1 | 38,769 | 22 | 77,539 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given an array a, consisting of n integers, find:
$$$max_{1 ≤ i < j ≤ n} LCM(a_i,a_j),$$$
where LCM(x, y) is the smallest positive integer that is divisible by both x and y. For example, LCM(6, 8) = 24, LCM(4, 12) = 12, LCM(2, 3) = 6.
Input
The first line contains an integer n (2 ≤ n ≤ 10^5) — the number of elements in the array a.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^5) — the elements of the array a.
Output
Print one integer, the maximum value of the least common multiple of two elements in the array a.
Examples
Input
3
13 35 77
Output
1001
Input
6
1 2 4 8 16 32
Output
32
Submitted Solution:
```
from fractions import gcd
_ = input()
a = list(map(int, input().split()))
a.sort(reverse = True)
a = a[:200]
L = len(a)
curr_max = 0
for i in range(L):
for j in range(i+1, L):
tmp = a[i]*a[j]//gcd(a[i], a[j])
if tmp > curr_max:
curr_max = tmp
print(curr_max)
``` | instruction | 0 | 38,770 | 22 | 77,540 |
No | output | 1 | 38,770 | 22 | 77,541 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given an array a, consisting of n integers, find:
$$$max_{1 ≤ i < j ≤ n} LCM(a_i,a_j),$$$
where LCM(x, y) is the smallest positive integer that is divisible by both x and y. For example, LCM(6, 8) = 24, LCM(4, 12) = 12, LCM(2, 3) = 6.
Input
The first line contains an integer n (2 ≤ n ≤ 10^5) — the number of elements in the array a.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^5) — the elements of the array a.
Output
Print one integer, the maximum value of the least common multiple of two elements in the array a.
Examples
Input
3
13 35 77
Output
1001
Input
6
1 2 4 8 16 32
Output
32
Submitted Solution:
```
from math import gcd
def main():
n = int(input())
a = [int(i) for i in input().split()]
a.sort(reverse = True)
k = 0
ma = a[0]
m = 0
for i,nm in enumerate(a):
num = nm*ma//gcd(nm, ma)
if num > m:
m = num
k = i
i = 1
while i < k:
ma = a[i]
for j,nm in enumerate(a[:k]):
num = nm*ma//gcd(nm, ma)
if num > m:
m = num
k = j
i += 1
print(m)
if __name__ == "__main__":
main()
``` | instruction | 0 | 38,771 | 22 | 77,542 |
No | output | 1 | 38,771 | 22 | 77,543 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two positive integer sequences a_1, …, a_n and b_1, …, b_m. For each j = 1, …, m find the greatest common divisor of a_1 + b_j, …, a_n + b_j.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 2 ⋅ 10^5).
The second line contains n integers a_1, …, a_n (1 ≤ a_i ≤ 10^{18}).
The third line contains m integers b_1, …, b_m (1 ≤ b_j ≤ 10^{18}).
Output
Print m integers. The j-th of them should be equal to GCD(a_1 + b_j, …, a_n + b_j).
Example
Input
4 4
1 25 121 169
1 2 7 23
Output
2 3 8 24 | instruction | 0 | 38,869 | 22 | 77,738 |
Tags: math, number theory
Correct Solution:
```
def gcd(a, b):
if b == 0:
return a
else:
return gcd(b, a%b)
def lgcd(a):
res = a[0]
for i in range(1, len(a)):
res = gcd(res, a[i])
return res
n, m = map(int, input().split())
a = sorted(list(map(int, input().split())))
b = list(map(int, input().split()))
if n == 1:
ans = [b[j] + a[0] for j in range(m)]
else:
A = lgcd([a[i + 1] - a[i] for i in range(n - 1)])
ans = [gcd(b[j] + a[0], A) for j in range(m)]
print(*ans)
``` | output | 1 | 38,869 | 22 | 77,739 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two positive integer sequences a_1, …, a_n and b_1, …, b_m. For each j = 1, …, m find the greatest common divisor of a_1 + b_j, …, a_n + b_j.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 2 ⋅ 10^5).
The second line contains n integers a_1, …, a_n (1 ≤ a_i ≤ 10^{18}).
The third line contains m integers b_1, …, b_m (1 ≤ b_j ≤ 10^{18}).
Output
Print m integers. The j-th of them should be equal to GCD(a_1 + b_j, …, a_n + b_j).
Example
Input
4 4
1 25 121 169
1 2 7 23
Output
2 3 8 24 | instruction | 0 | 38,870 | 22 | 77,740 |
Tags: math, number theory
Correct Solution:
```
n, m = map(int, input().split())
a = sorted(map(int, input().split()))
b = list(map(int, input().split()))
def gcd(x, y):
while x != 0 and y != 0:
if x > y:
x, y = y, x
x, y = y % x, x
return x + y
g = 0
for i in range(n - 1):
g = gcd(g, a[i + 1] - a[i])
for x in b:
print(gcd(g, x + a[0]), end=' ')
``` | output | 1 | 38,870 | 22 | 77,741 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two positive integer sequences a_1, …, a_n and b_1, …, b_m. For each j = 1, …, m find the greatest common divisor of a_1 + b_j, …, a_n + b_j.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 2 ⋅ 10^5).
The second line contains n integers a_1, …, a_n (1 ≤ a_i ≤ 10^{18}).
The third line contains m integers b_1, …, b_m (1 ≤ b_j ≤ 10^{18}).
Output
Print m integers. The j-th of them should be equal to GCD(a_1 + b_j, …, a_n + b_j).
Example
Input
4 4
1 25 121 169
1 2 7 23
Output
2 3 8 24 | instruction | 0 | 38,871 | 22 | 77,742 |
Tags: math, number theory
Correct Solution:
```
from sys import stdin
def gcd(a,b):
if b==0:
return(a)
return(gcd(b,a%b))
n,m=map(int,stdin.readline().split())
a=list(map(int,stdin.readline().split()))
b=list(map(int,stdin.readline().split()))
mini=min(a)
prevg=0
for i in a:
prevg=gcd(i-mini,prevg)
for i in b:
print(gcd(prevg,mini+i),end=" ")
``` | output | 1 | 38,871 | 22 | 77,743 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two positive integer sequences a_1, …, a_n and b_1, …, b_m. For each j = 1, …, m find the greatest common divisor of a_1 + b_j, …, a_n + b_j.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 2 ⋅ 10^5).
The second line contains n integers a_1, …, a_n (1 ≤ a_i ≤ 10^{18}).
The third line contains m integers b_1, …, b_m (1 ≤ b_j ≤ 10^{18}).
Output
Print m integers. The j-th of them should be equal to GCD(a_1 + b_j, …, a_n + b_j).
Example
Input
4 4
1 25 121 169
1 2 7 23
Output
2 3 8 24 | instruction | 0 | 38,872 | 22 | 77,744 |
Tags: math, number theory
Correct Solution:
```
from math import gcd
n,m=map(int,input().split());a=list(map(int,input().split()));b=list(map(int,input().split()));out=0
for i in range(1,n):out=gcd(out,abs(a[0]-a[i]))
print(*[gcd(out,a[0]+b[j]) for j in range(m)])
``` | output | 1 | 38,872 | 22 | 77,745 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two positive integer sequences a_1, …, a_n and b_1, …, b_m. For each j = 1, …, m find the greatest common divisor of a_1 + b_j, …, a_n + b_j.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 2 ⋅ 10^5).
The second line contains n integers a_1, …, a_n (1 ≤ a_i ≤ 10^{18}).
The third line contains m integers b_1, …, b_m (1 ≤ b_j ≤ 10^{18}).
Output
Print m integers. The j-th of them should be equal to GCD(a_1 + b_j, …, a_n + b_j).
Example
Input
4 4
1 25 121 169
1 2 7 23
Output
2 3 8 24 | instruction | 0 | 38,873 | 22 | 77,746 |
Tags: math, number theory
Correct Solution:
```
import sys
input = sys.stdin.readline
n,m=map(int,input().split())
A=list(map(int,input().split()))
B=list(map(int,input().split()))
from math import gcd
GCD=0
for i in range(1,n):
GCD=gcd(A[i]-A[i-1],GCD)
ANS=[]
for j in range(m):
ANS.append(gcd(GCD,A[0]+B[j]))
print(*ANS)
``` | output | 1 | 38,873 | 22 | 77,747 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two positive integer sequences a_1, …, a_n and b_1, …, b_m. For each j = 1, …, m find the greatest common divisor of a_1 + b_j, …, a_n + b_j.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 2 ⋅ 10^5).
The second line contains n integers a_1, …, a_n (1 ≤ a_i ≤ 10^{18}).
The third line contains m integers b_1, …, b_m (1 ≤ b_j ≤ 10^{18}).
Output
Print m integers. The j-th of them should be equal to GCD(a_1 + b_j, …, a_n + b_j).
Example
Input
4 4
1 25 121 169
1 2 7 23
Output
2 3 8 24 | instruction | 0 | 38,874 | 22 | 77,748 |
Tags: math, number theory
Correct Solution:
```
n,m = tuple(map(int,input().split()))
a = list(map(int,input().split()))
b = list(map(int,input().split()))
def gcd(a,b):
while b:
a,b = b,a%b
return a
temp = 0
for i in range(1,len(a)):
temp = gcd(temp,abs(a[i]-a[i-1]))
for i in range(len(b)):
print(gcd(b[i]+a[0],temp),end=' ')
``` | output | 1 | 38,874 | 22 | 77,749 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two positive integer sequences a_1, …, a_n and b_1, …, b_m. For each j = 1, …, m find the greatest common divisor of a_1 + b_j, …, a_n + b_j.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 2 ⋅ 10^5).
The second line contains n integers a_1, …, a_n (1 ≤ a_i ≤ 10^{18}).
The third line contains m integers b_1, …, b_m (1 ≤ b_j ≤ 10^{18}).
Output
Print m integers. The j-th of them should be equal to GCD(a_1 + b_j, …, a_n + b_j).
Example
Input
4 4
1 25 121 169
1 2 7 23
Output
2 3 8 24 | instruction | 0 | 38,875 | 22 | 77,750 |
Tags: math, number theory
Correct Solution:
```
n,m = map(int,input().split())
a = [ int(x) for x in input().split() ]
b = [ int(x) for x in input().split() ]
output = []
def gcd(a,b):
if b == 0:
return a
else:
return gcd(b,a%b)
res = 0
for index in range(1,n):
res = gcd(abs(a[index]-a[0]),res)
for index in range(0,m):
output += [gcd(res,a[0]+b[index])]
print(" ".join(map(str,output)))
``` | output | 1 | 38,875 | 22 | 77,751 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two positive integer sequences a_1, …, a_n and b_1, …, b_m. For each j = 1, …, m find the greatest common divisor of a_1 + b_j, …, a_n + b_j.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 2 ⋅ 10^5).
The second line contains n integers a_1, …, a_n (1 ≤ a_i ≤ 10^{18}).
The third line contains m integers b_1, …, b_m (1 ≤ b_j ≤ 10^{18}).
Output
Print m integers. The j-th of them should be equal to GCD(a_1 + b_j, …, a_n + b_j).
Example
Input
4 4
1 25 121 169
1 2 7 23
Output
2 3 8 24 | instruction | 0 | 38,876 | 22 | 77,752 |
Tags: math, number theory
Correct Solution:
```
from math import gcd
n, m = map(int, input().split())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
for i in b:
ans = a[0]+i
if(len(a)>=2):
ans = gcd(a[0]+i,a[1]+i)
if(len(a)>2):
ans = gcd(ans, a[(len(a)-1)//4]+i)
ans = gcd(ans, a[(len(a)-1)//2]+i)
ans = gcd(ans, a[((len(a)-1)*3)//4]+i)
ans = gcd(ans, a[(len(a)-1)]+i)
print(ans, end = " ")
``` | output | 1 | 38,876 | 22 | 77,753 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two positive integer sequences a_1, …, a_n and b_1, …, b_m. For each j = 1, …, m find the greatest common divisor of a_1 + b_j, …, a_n + b_j.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 2 ⋅ 10^5).
The second line contains n integers a_1, …, a_n (1 ≤ a_i ≤ 10^{18}).
The third line contains m integers b_1, …, b_m (1 ≤ b_j ≤ 10^{18}).
Output
Print m integers. The j-th of them should be equal to GCD(a_1 + b_j, …, a_n + b_j).
Example
Input
4 4
1 25 121 169
1 2 7 23
Output
2 3 8 24
Submitted Solution:
```
from math import gcd
n, m = map(int, input().split())
a = [int(x) for x in input().split()]
b = [int(x) for x in input().split()]
curGcd = None
for i in range(1, n):
if curGcd is None:
curGcd = a[i] - a[0]
else:
curGcd = gcd(curGcd, a[i] - a[0])
result = []
for val in b:
if curGcd is not None:
result.append(gcd(curGcd, a[0] + val))
else:
result.append(val + a[0])
print(*result)
``` | instruction | 0 | 38,877 | 22 | 77,754 |
Yes | output | 1 | 38,877 | 22 | 77,755 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two positive integer sequences a_1, …, a_n and b_1, …, b_m. For each j = 1, …, m find the greatest common divisor of a_1 + b_j, …, a_n + b_j.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 2 ⋅ 10^5).
The second line contains n integers a_1, …, a_n (1 ≤ a_i ≤ 10^{18}).
The third line contains m integers b_1, …, b_m (1 ≤ b_j ≤ 10^{18}).
Output
Print m integers. The j-th of them should be equal to GCD(a_1 + b_j, …, a_n + b_j).
Example
Input
4 4
1 25 121 169
1 2 7 23
Output
2 3 8 24
Submitted Solution:
```
from collections import defaultdict,deque
import sys
import bisect
import math
input=sys.stdin.readline
mod=1000000007
n,m=map(int,input().split())
a=[int(i) for i in input().split() if i!='\n']
b=[int(i) for i in input().split() if i!='\n']
ans=[]
if n==1:
for i in range(m):
ans.append(a[0]+b[i])
print(*ans)
else:
diff=abs(a[1]-a[0])
gcd=diff
for i in range(1,n-1):
gcd=math.gcd(gcd,abs(a[i]-a[i+1]))
for i in range(m):
ans.append(math.gcd(a[i]+b[i],gcd))
print(*ans)
``` | instruction | 0 | 38,878 | 22 | 77,756 |
Yes | output | 1 | 38,878 | 22 | 77,757 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two positive integer sequences a_1, …, a_n and b_1, …, b_m. For each j = 1, …, m find the greatest common divisor of a_1 + b_j, …, a_n + b_j.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 2 ⋅ 10^5).
The second line contains n integers a_1, …, a_n (1 ≤ a_i ≤ 10^{18}).
The third line contains m integers b_1, …, b_m (1 ≤ b_j ≤ 10^{18}).
Output
Print m integers. The j-th of them should be equal to GCD(a_1 + b_j, …, a_n + b_j).
Example
Input
4 4
1 25 121 169
1 2 7 23
Output
2 3 8 24
Submitted Solution:
```
import sys
input = lambda: sys.stdin.readline().rstrip("\r\n")
from math import sqrt,ceil,gcd
n,m=map(int,input().split())
a=list(map(int,input().split()))
b=list(map(int,input().split()))
t=a[0]
for i in range(n):
a[i]-=t
ans=0
#print(a)
ck=0
for i in a:
ck=gcd(ck,i)
#print(ck)
for i in b:
print(gcd(t+i,ck),end=' ')
print()
``` | instruction | 0 | 38,879 | 22 | 77,758 |
Yes | output | 1 | 38,879 | 22 | 77,759 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two positive integer sequences a_1, …, a_n and b_1, …, b_m. For each j = 1, …, m find the greatest common divisor of a_1 + b_j, …, a_n + b_j.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 2 ⋅ 10^5).
The second line contains n integers a_1, …, a_n (1 ≤ a_i ≤ 10^{18}).
The third line contains m integers b_1, …, b_m (1 ≤ b_j ≤ 10^{18}).
Output
Print m integers. The j-th of them should be equal to GCD(a_1 + b_j, …, a_n + b_j).
Example
Input
4 4
1 25 121 169
1 2 7 23
Output
2 3 8 24
Submitted Solution:
```
def gcd(x,y):
if y== 0:
return x
else:
return gcd(y,x%y)
res=0
n,m = map(int,input().split())
a = list(map(int,input().split()))
b = list(map(int,input().split()))
for i in range(1,n):
res = gcd(abs(a[i]-a[0]),res)
for j in range(m):
print(gcd(b[j]+a[0],res),end=" ")
``` | instruction | 0 | 38,880 | 22 | 77,760 |
Yes | output | 1 | 38,880 | 22 | 77,761 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two positive integer sequences a_1, …, a_n and b_1, …, b_m. For each j = 1, …, m find the greatest common divisor of a_1 + b_j, …, a_n + b_j.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 2 ⋅ 10^5).
The second line contains n integers a_1, …, a_n (1 ≤ a_i ≤ 10^{18}).
The third line contains m integers b_1, …, b_m (1 ≤ b_j ≤ 10^{18}).
Output
Print m integers. The j-th of them should be equal to GCD(a_1 + b_j, …, a_n + b_j).
Example
Input
4 4
1 25 121 169
1 2 7 23
Output
2 3 8 24
Submitted Solution:
```
def gcd(a, b):
if b == 0:
return a
else:
return gcd(b, a % b)
def lgcd(a):
res = a[0]
for i in range(1, len(a)):
res = gcd(res, a[i])
return res
n, m = map(int, input().split())
a = sorted(list(map(int, input().split())))
b = list(map(int, input().split()))
if n == 1:
A = a[0]
else:
A = lgcd([a[i + 1] - a[i] for i in range(n - 1)])
ans = [gcd(b[j] + a[0], A) for j in range(m)]
print(*ans)
``` | instruction | 0 | 38,881 | 22 | 77,762 |
No | output | 1 | 38,881 | 22 | 77,763 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two positive integer sequences a_1, …, a_n and b_1, …, b_m. For each j = 1, …, m find the greatest common divisor of a_1 + b_j, …, a_n + b_j.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 2 ⋅ 10^5).
The second line contains n integers a_1, …, a_n (1 ≤ a_i ≤ 10^{18}).
The third line contains m integers b_1, …, b_m (1 ≤ b_j ≤ 10^{18}).
Output
Print m integers. The j-th of them should be equal to GCD(a_1 + b_j, …, a_n + b_j).
Example
Input
4 4
1 25 121 169
1 2 7 23
Output
2 3 8 24
Submitted Solution:
```
from math import gcd
n, m = map(int, input().split())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
for i in b:
ans = b[0]+i
if(len(a)>=2):
ans = gcd(a[0]+i,a[1]+i)
if(len(a)>2):
for r in range(2, len(a)):
ans = gcd(ans, a[r]+i)
print(ans, end = " ")
``` | instruction | 0 | 38,882 | 22 | 77,764 |
No | output | 1 | 38,882 | 22 | 77,765 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two positive integer sequences a_1, …, a_n and b_1, …, b_m. For each j = 1, …, m find the greatest common divisor of a_1 + b_j, …, a_n + b_j.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 2 ⋅ 10^5).
The second line contains n integers a_1, …, a_n (1 ≤ a_i ≤ 10^{18}).
The third line contains m integers b_1, …, b_m (1 ≤ b_j ≤ 10^{18}).
Output
Print m integers. The j-th of them should be equal to GCD(a_1 + b_j, …, a_n + b_j).
Example
Input
4 4
1 25 121 169
1 2 7 23
Output
2 3 8 24
Submitted Solution:
```
n, m = list(map(int, input().split()))
alst = list(map(int, input().split()))
blst = list(map(int, input().split()))
alst = set(alst)
alst = list(alst)
alst = sorted(alst)
# Now find the least difference between the numbers
mn = 10**18
for i in range(0, n-1):
mn = min(mn, alst[i+1] - alst[i])
# Now everyone will have gcd smaller than mn
def gcd(a, b):
if b == 0:
return a
else:
return gcd(b, a%b)
for x in blst:
gc = gcd(x + alst[0], mn)
for y in alst:
gc = gcd(x + y, mn)
print(gc, end=' ')
print()
``` | instruction | 0 | 38,883 | 22 | 77,766 |
No | output | 1 | 38,883 | 22 | 77,767 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two positive integer sequences a_1, …, a_n and b_1, …, b_m. For each j = 1, …, m find the greatest common divisor of a_1 + b_j, …, a_n + b_j.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 2 ⋅ 10^5).
The second line contains n integers a_1, …, a_n (1 ≤ a_i ≤ 10^{18}).
The third line contains m integers b_1, …, b_m (1 ≤ b_j ≤ 10^{18}).
Output
Print m integers. The j-th of them should be equal to GCD(a_1 + b_j, …, a_n + b_j).
Example
Input
4 4
1 25 121 169
1 2 7 23
Output
2 3 8 24
Submitted Solution:
```
n,m = map(int, input().split())
a=[int(i) for i in input().split()]
b=[int(i) for i in input().split()]
a.sort()
def gcd(a,b):
if b==0:
return a
else:
return gcd(b,a%b)
tmp = 0
for i in range(1,len(a)):
tmp = gcd(tmp,a[i]-a[i-1])
print(tmp)
ret = []
for i in range(len(b)):
ret.append(gcd(tmp, a[-1]+b[i]))
print(*ret)
``` | instruction | 0 | 38,884 | 22 | 77,768 |
No | output | 1 | 38,884 | 22 | 77,769 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is an interactive problem. In the output section below you will see the information about flushing the output.
Bear Limak thinks of some hidden number — an integer from interval [2, 100]. Your task is to say if the hidden number is prime or composite.
Integer x > 1 is called prime if it has exactly two distinct divisors, 1 and x. If integer x > 1 is not prime, it's called composite.
You can ask up to 20 queries about divisors of the hidden number. In each query you should print an integer from interval [2, 100]. The system will answer "yes" if your integer is a divisor of the hidden number. Otherwise, the answer will be "no".
For example, if the hidden number is 14 then the system will answer "yes" only if you print 2, 7 or 14.
When you are done asking queries, print "prime" or "composite" and terminate your program.
You will get the Wrong Answer verdict if you ask more than 20 queries, or if you print an integer not from the range [2, 100]. Also, you will get the Wrong Answer verdict if the printed answer isn't correct.
You will get the Idleness Limit Exceeded verdict if you don't print anything (but you should) or if you forget about flushing the output (more info below).
Input
After each query you should read one string from the input. It will be "yes" if the printed integer is a divisor of the hidden number, and "no" otherwise.
Output
Up to 20 times you can ask a query — print an integer from interval [2, 100] in one line. You have to both print the end-of-line character and flush the output. After flushing you should read a response from the input.
In any moment you can print the answer "prime" or "composite" (without the quotes). After that, flush the output and terminate your program.
To flush you can use (just after printing an integer and end-of-line):
* fflush(stdout) in C++;
* System.out.flush() in Java;
* stdout.flush() in Python;
* flush(output) in Pascal;
* See the documentation for other languages.
Hacking. To hack someone, as the input you should print the hidden number — one integer from the interval [2, 100]. Of course, his/her solution won't be able to read the hidden number from the input.
Examples
Input
yes
no
yes
Output
2
80
5
composite
Input
no
yes
no
no
no
Output
58
59
78
78
2
prime
Note
The hidden number in the first query is 30. In a table below you can see a better form of the provided example of the communication process.
<image>
The hidden number is divisible by both 2 and 5. Thus, it must be composite. Note that it isn't necessary to know the exact value of the hidden number. In this test, the hidden number is 30.
<image>
59 is a divisor of the hidden number. In the interval [2, 100] there is only one number with this divisor. The hidden number must be 59, which is prime. Note that the answer is known even after the second query and you could print it then and terminate. Though, it isn't forbidden to ask unnecessary queries (unless you exceed the limit of 20 queries). | instruction | 0 | 39,151 | 22 | 78,302 |
Tags: constructive algorithms, interactive, math
Correct Solution:
```
import sys
from functools import lru_cache, cmp_to_key
from heapq import merge, heapify, heappop, heappush
from math import ceil, floor, gcd, fabs, factorial, fmod, sqrt, inf
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
from time import perf_counter
from fractions import Fraction
# sys.setrecursionlimit(pow(10, 6))
# sys.stdin = open("input.txt", "r")
# sys.stdout = open("output.txt", "w")
mod = 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)]
def prime(number):
if number == 2:
return True
if number & 1:
i = 2
while i * i <= number:
if number % i == 0:
return False
i += 1
return True
return False
arr = []
for i in range(2, 101):
if prime(i):
arr.append(i)
if i * i <= 100:
arr.append(i*i)
arr.sort()
low, high = 0, 24
c = 0
for i in range(20):
out(arr[i])
sys.stdout.flush()
verdict = data()
if verdict.lower() == 'yes':
c += 1
if c == 2:
out("composite")
exit()
out("prime")
``` | output | 1 | 39,151 | 22 | 78,303 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is an interactive problem. In the output section below you will see the information about flushing the output.
Bear Limak thinks of some hidden number — an integer from interval [2, 100]. Your task is to say if the hidden number is prime or composite.
Integer x > 1 is called prime if it has exactly two distinct divisors, 1 and x. If integer x > 1 is not prime, it's called composite.
You can ask up to 20 queries about divisors of the hidden number. In each query you should print an integer from interval [2, 100]. The system will answer "yes" if your integer is a divisor of the hidden number. Otherwise, the answer will be "no".
For example, if the hidden number is 14 then the system will answer "yes" only if you print 2, 7 or 14.
When you are done asking queries, print "prime" or "composite" and terminate your program.
You will get the Wrong Answer verdict if you ask more than 20 queries, or if you print an integer not from the range [2, 100]. Also, you will get the Wrong Answer verdict if the printed answer isn't correct.
You will get the Idleness Limit Exceeded verdict if you don't print anything (but you should) or if you forget about flushing the output (more info below).
Input
After each query you should read one string from the input. It will be "yes" if the printed integer is a divisor of the hidden number, and "no" otherwise.
Output
Up to 20 times you can ask a query — print an integer from interval [2, 100] in one line. You have to both print the end-of-line character and flush the output. After flushing you should read a response from the input.
In any moment you can print the answer "prime" or "composite" (without the quotes). After that, flush the output and terminate your program.
To flush you can use (just after printing an integer and end-of-line):
* fflush(stdout) in C++;
* System.out.flush() in Java;
* stdout.flush() in Python;
* flush(output) in Pascal;
* See the documentation for other languages.
Hacking. To hack someone, as the input you should print the hidden number — one integer from the interval [2, 100]. Of course, his/her solution won't be able to read the hidden number from the input.
Examples
Input
yes
no
yes
Output
2
80
5
composite
Input
no
yes
no
no
no
Output
58
59
78
78
2
prime
Note
The hidden number in the first query is 30. In a table below you can see a better form of the provided example of the communication process.
<image>
The hidden number is divisible by both 2 and 5. Thus, it must be composite. Note that it isn't necessary to know the exact value of the hidden number. In this test, the hidden number is 30.
<image>
59 is a divisor of the hidden number. In the interval [2, 100] there is only one number with this divisor. The hidden number must be 59, which is prime. Note that the answer is known even after the second query and you could print it then and terminate. Though, it isn't forbidden to ask unnecessary queries (unless you exceed the limit of 20 queries). | instruction | 0 | 39,152 | 22 | 78,304 |
Tags: constructive algorithms, interactive, math
Correct Solution:
```
#------------------------template--------------------------#
import os
import sys
from math import *
from collections import *
# from fractions import *
# from heapq import*
from bisect import *
from io import BytesIO, IOBase
def vsInput():
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
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")
ALPHA='abcdefghijklmnopqrstuvwxyz'
M=10**9+7
EPS=1e-6
def Ceil(a,b): return a//b+int(a%b>0)
def value():return tuple(map(int,input().split()))
def array():return [int(i) for i in input().split()]
def Int():return int(input())
def Str():return input()
def arrayS():return [i for i in input().split()]
#-------------------------code---------------------------#
# vsInput()
def primeN(n):
prime = [True for i in range(n+1)]
prime[0]=False
prime[1]=False
p=2
while(p*p<=n):
if(prime[p]):
for i in range(p*p,n+1,p):
prime[i]=False
p+=1
return [p for p in range(n+1) if(prime[p])]
primes=primeN(50)
# print(primes)
have={}
have[2]=[37, 41, 43, 47]
have[3]=[23, 29, 31]
have[5]=[17, 19]
have[7]=[11, 13]
found=0
for i in [2,3,5,7]:
print(i,flush=True)
res=input()
if(res=='yes'):found+=1
for p in have[i]:
print(p,flush=True)
res=input()
if(res=='yes'):found+=1
print(i*i,flush=True)
res=input()
if(res=='yes'):found+=1
if(found>1):
print("composite",flush=True)
exit()
print("prime",flush=True)
exit()
``` | output | 1 | 39,152 | 22 | 78,305 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is an interactive problem. In the output section below you will see the information about flushing the output.
Bear Limak thinks of some hidden number — an integer from interval [2, 100]. Your task is to say if the hidden number is prime or composite.
Integer x > 1 is called prime if it has exactly two distinct divisors, 1 and x. If integer x > 1 is not prime, it's called composite.
You can ask up to 20 queries about divisors of the hidden number. In each query you should print an integer from interval [2, 100]. The system will answer "yes" if your integer is a divisor of the hidden number. Otherwise, the answer will be "no".
For example, if the hidden number is 14 then the system will answer "yes" only if you print 2, 7 or 14.
When you are done asking queries, print "prime" or "composite" and terminate your program.
You will get the Wrong Answer verdict if you ask more than 20 queries, or if you print an integer not from the range [2, 100]. Also, you will get the Wrong Answer verdict if the printed answer isn't correct.
You will get the Idleness Limit Exceeded verdict if you don't print anything (but you should) or if you forget about flushing the output (more info below).
Input
After each query you should read one string from the input. It will be "yes" if the printed integer is a divisor of the hidden number, and "no" otherwise.
Output
Up to 20 times you can ask a query — print an integer from interval [2, 100] in one line. You have to both print the end-of-line character and flush the output. After flushing you should read a response from the input.
In any moment you can print the answer "prime" or "composite" (without the quotes). After that, flush the output and terminate your program.
To flush you can use (just after printing an integer and end-of-line):
* fflush(stdout) in C++;
* System.out.flush() in Java;
* stdout.flush() in Python;
* flush(output) in Pascal;
* See the documentation for other languages.
Hacking. To hack someone, as the input you should print the hidden number — one integer from the interval [2, 100]. Of course, his/her solution won't be able to read the hidden number from the input.
Examples
Input
yes
no
yes
Output
2
80
5
composite
Input
no
yes
no
no
no
Output
58
59
78
78
2
prime
Note
The hidden number in the first query is 30. In a table below you can see a better form of the provided example of the communication process.
<image>
The hidden number is divisible by both 2 and 5. Thus, it must be composite. Note that it isn't necessary to know the exact value of the hidden number. In this test, the hidden number is 30.
<image>
59 is a divisor of the hidden number. In the interval [2, 100] there is only one number with this divisor. The hidden number must be 59, which is prime. Note that the answer is known even after the second query and you could print it then and terminate. Though, it isn't forbidden to ask unnecessary queries (unless you exceed the limit of 20 queries). | instruction | 0 | 39,153 | 22 | 78,306 |
Tags: constructive algorithms, interactive, math
Correct Solution:
```
from sys import stdout, stdin
PS = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97]
# PSL50 = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47] ## len == 15
# PSU50 = [53, 59, 61, 67, 71, 73, 79, 83, 89, 97] ## len == 10
def ask_and_yes( num):
print(num)
stdout.flush()
## get cpu answer
line = stdin.readline().strip()
return line == "yes"
def solve():
# yes_cnt = 0
first_p = -1
for p in PS:
if p >= 53:
print("prime")
return
else:
if ask_and_yes( p):
first_p = p
break
## rewind back
quo = int(100/first_p)
## deal with edge cases
if first_p >= quo:
print( "prime")
return
## get the first feasible prime
for i in range( len( PS)-1, -1, -1):
if PS[ i] < quo:
break
## start questioning again
for j in range( i, -1, -1):
if PS[ j] == first_p:
break ## fixed
else:
if ask_and_yes( PS[ j]):
print( "composite")
return
## one last time ## fixed
if ask_and_yes( first_p*first_p):
print( "composite")
else:
print( "prime")
solve()
``` | output | 1 | 39,153 | 22 | 78,307 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is an interactive problem. In the output section below you will see the information about flushing the output.
Bear Limak thinks of some hidden number — an integer from interval [2, 100]. Your task is to say if the hidden number is prime or composite.
Integer x > 1 is called prime if it has exactly two distinct divisors, 1 and x. If integer x > 1 is not prime, it's called composite.
You can ask up to 20 queries about divisors of the hidden number. In each query you should print an integer from interval [2, 100]. The system will answer "yes" if your integer is a divisor of the hidden number. Otherwise, the answer will be "no".
For example, if the hidden number is 14 then the system will answer "yes" only if you print 2, 7 or 14.
When you are done asking queries, print "prime" or "composite" and terminate your program.
You will get the Wrong Answer verdict if you ask more than 20 queries, or if you print an integer not from the range [2, 100]. Also, you will get the Wrong Answer verdict if the printed answer isn't correct.
You will get the Idleness Limit Exceeded verdict if you don't print anything (but you should) or if you forget about flushing the output (more info below).
Input
After each query you should read one string from the input. It will be "yes" if the printed integer is a divisor of the hidden number, and "no" otherwise.
Output
Up to 20 times you can ask a query — print an integer from interval [2, 100] in one line. You have to both print the end-of-line character and flush the output. After flushing you should read a response from the input.
In any moment you can print the answer "prime" or "composite" (without the quotes). After that, flush the output and terminate your program.
To flush you can use (just after printing an integer and end-of-line):
* fflush(stdout) in C++;
* System.out.flush() in Java;
* stdout.flush() in Python;
* flush(output) in Pascal;
* See the documentation for other languages.
Hacking. To hack someone, as the input you should print the hidden number — one integer from the interval [2, 100]. Of course, his/her solution won't be able to read the hidden number from the input.
Examples
Input
yes
no
yes
Output
2
80
5
composite
Input
no
yes
no
no
no
Output
58
59
78
78
2
prime
Note
The hidden number in the first query is 30. In a table below you can see a better form of the provided example of the communication process.
<image>
The hidden number is divisible by both 2 and 5. Thus, it must be composite. Note that it isn't necessary to know the exact value of the hidden number. In this test, the hidden number is 30.
<image>
59 is a divisor of the hidden number. In the interval [2, 100] there is only one number with this divisor. The hidden number must be 59, which is prime. Note that the answer is known even after the second query and you could print it then and terminate. Though, it isn't forbidden to ask unnecessary queries (unless you exceed the limit of 20 queries). | instruction | 0 | 39,154 | 22 | 78,308 |
Tags: constructive algorithms, interactive, math
Correct Solution:
```
import sys
def is_prime(n, primes):
for p in primes:
if n % p == 0:
return False
else:
return True
def main():
primes = [2, 3, 5, 7]
for i in range(primes[-1] + 2, 49, 2):
if is_prime(i, primes):
primes.append(i)
ndivs = 0
last_divisor = -1
for p in primes[:4]:
print(p)
sys.stdout.flush()
ans = sys.stdin.readline().strip() == "no"
ndivs += not ans
if not ans:
last_divisor = p
if ndivs >= 2:
break
if ndivs > 1:
print("composite")
elif ndivs == 0:
print("prime")
else:
# ndivs == 1
print(last_divisor * last_divisor)
sys.stdout.flush()
ans = sys.stdin.readline().strip() == "no"
if not ans:
print("composite")
else:
for p in primes[4:]:
print(p)
sys.stdout.flush()
ans = sys.stdin.readline().strip() == "no"
if not ans:
print("composite")
break
else:
print("prime")
sys.stdout.flush()
if __name__ == "__main__":
main()
``` | output | 1 | 39,154 | 22 | 78,309 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is an interactive problem. In the output section below you will see the information about flushing the output.
Bear Limak thinks of some hidden number — an integer from interval [2, 100]. Your task is to say if the hidden number is prime or composite.
Integer x > 1 is called prime if it has exactly two distinct divisors, 1 and x. If integer x > 1 is not prime, it's called composite.
You can ask up to 20 queries about divisors of the hidden number. In each query you should print an integer from interval [2, 100]. The system will answer "yes" if your integer is a divisor of the hidden number. Otherwise, the answer will be "no".
For example, if the hidden number is 14 then the system will answer "yes" only if you print 2, 7 or 14.
When you are done asking queries, print "prime" or "composite" and terminate your program.
You will get the Wrong Answer verdict if you ask more than 20 queries, or if you print an integer not from the range [2, 100]. Also, you will get the Wrong Answer verdict if the printed answer isn't correct.
You will get the Idleness Limit Exceeded verdict if you don't print anything (but you should) or if you forget about flushing the output (more info below).
Input
After each query you should read one string from the input. It will be "yes" if the printed integer is a divisor of the hidden number, and "no" otherwise.
Output
Up to 20 times you can ask a query — print an integer from interval [2, 100] in one line. You have to both print the end-of-line character and flush the output. After flushing you should read a response from the input.
In any moment you can print the answer "prime" or "composite" (without the quotes). After that, flush the output and terminate your program.
To flush you can use (just after printing an integer and end-of-line):
* fflush(stdout) in C++;
* System.out.flush() in Java;
* stdout.flush() in Python;
* flush(output) in Pascal;
* See the documentation for other languages.
Hacking. To hack someone, as the input you should print the hidden number — one integer from the interval [2, 100]. Of course, his/her solution won't be able to read the hidden number from the input.
Examples
Input
yes
no
yes
Output
2
80
5
composite
Input
no
yes
no
no
no
Output
58
59
78
78
2
prime
Note
The hidden number in the first query is 30. In a table below you can see a better form of the provided example of the communication process.
<image>
The hidden number is divisible by both 2 and 5. Thus, it must be composite. Note that it isn't necessary to know the exact value of the hidden number. In this test, the hidden number is 30.
<image>
59 is a divisor of the hidden number. In the interval [2, 100] there is only one number with this divisor. The hidden number must be 59, which is prime. Note that the answer is known even after the second query and you could print it then and terminate. Though, it isn't forbidden to ask unnecessary queries (unless you exceed the limit of 20 queries). | instruction | 0 | 39,155 | 22 | 78,310 |
Tags: constructive algorithms, interactive, math
Correct Solution:
```
from sys import stdin, stdout
dividers = ['4', '9', '25', '49', '2', '3', '5', '7', '11', '13', '17', '19', '23', '29', '31', '37', '41', '43', '47']
cnt = 0
for v in dividers:
stdout.write(v + '\n')
stdout.flush()
if stdin.readline().rstrip() == 'yes':
cnt += 1
if cnt > 1:
stdout.write('composite')
else:
stdout.write('prime')
stdout.flush()
``` | output | 1 | 39,155 | 22 | 78,311 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is an interactive problem. In the output section below you will see the information about flushing the output.
Bear Limak thinks of some hidden number — an integer from interval [2, 100]. Your task is to say if the hidden number is prime or composite.
Integer x > 1 is called prime if it has exactly two distinct divisors, 1 and x. If integer x > 1 is not prime, it's called composite.
You can ask up to 20 queries about divisors of the hidden number. In each query you should print an integer from interval [2, 100]. The system will answer "yes" if your integer is a divisor of the hidden number. Otherwise, the answer will be "no".
For example, if the hidden number is 14 then the system will answer "yes" only if you print 2, 7 or 14.
When you are done asking queries, print "prime" or "composite" and terminate your program.
You will get the Wrong Answer verdict if you ask more than 20 queries, or if you print an integer not from the range [2, 100]. Also, you will get the Wrong Answer verdict if the printed answer isn't correct.
You will get the Idleness Limit Exceeded verdict if you don't print anything (but you should) or if you forget about flushing the output (more info below).
Input
After each query you should read one string from the input. It will be "yes" if the printed integer is a divisor of the hidden number, and "no" otherwise.
Output
Up to 20 times you can ask a query — print an integer from interval [2, 100] in one line. You have to both print the end-of-line character and flush the output. After flushing you should read a response from the input.
In any moment you can print the answer "prime" or "composite" (without the quotes). After that, flush the output and terminate your program.
To flush you can use (just after printing an integer and end-of-line):
* fflush(stdout) in C++;
* System.out.flush() in Java;
* stdout.flush() in Python;
* flush(output) in Pascal;
* See the documentation for other languages.
Hacking. To hack someone, as the input you should print the hidden number — one integer from the interval [2, 100]. Of course, his/her solution won't be able to read the hidden number from the input.
Examples
Input
yes
no
yes
Output
2
80
5
composite
Input
no
yes
no
no
no
Output
58
59
78
78
2
prime
Note
The hidden number in the first query is 30. In a table below you can see a better form of the provided example of the communication process.
<image>
The hidden number is divisible by both 2 and 5. Thus, it must be composite. Note that it isn't necessary to know the exact value of the hidden number. In this test, the hidden number is 30.
<image>
59 is a divisor of the hidden number. In the interval [2, 100] there is only one number with this divisor. The hidden number must be 59, which is prime. Note that the answer is known even after the second query and you could print it then and terminate. Though, it isn't forbidden to ask unnecessary queries (unless you exceed the limit of 20 queries). | instruction | 0 | 39,156 | 22 | 78,312 |
Tags: constructive algorithms, interactive, math
Correct Solution:
```
#------------------------template--------------------------#
import os
import sys
from math import *
from collections import *
# from fractions import *
# from heapq import*
from bisect import *
from io import BytesIO, IOBase
def vsInput():
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
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")
ALPHA='abcdefghijklmnopqrstuvwxyz'
M=10**9+7
EPS=1e-6
def Ceil(a,b): return a//b+int(a%b>0)
def value():return tuple(map(int,input().split()))
def array():return [int(i) for i in input().split()]
def Int():return int(input())
def Str():return input()
def arrayS():return [i for i in input().split()]
#-------------------------code---------------------------#
# vsInput()
def primeN(n):
prime = [True for i in range(n+1)]
prime[0]=False
prime[1]=False
p=2
while(p*p<=n):
if(prime[p]):
for i in range(p*p,n+1,p):
prime[i]=False
p+=1
return [p for p in range(n+1) if(prime[p])]
primes=primeN(50)
found=0
for i in primes:
print(i,flush=True)
res=input()
if(res=='yes'):found+=1
if(i>7):continue
print(i*i,flush=True)
res=input()
if(res=='yes'):found+=1
if(found>1): print("composite",flush=True)
else: print("prime",flush=True)
exit()
``` | output | 1 | 39,156 | 22 | 78,313 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is an interactive problem. In the output section below you will see the information about flushing the output.
Bear Limak thinks of some hidden number — an integer from interval [2, 100]. Your task is to say if the hidden number is prime or composite.
Integer x > 1 is called prime if it has exactly two distinct divisors, 1 and x. If integer x > 1 is not prime, it's called composite.
You can ask up to 20 queries about divisors of the hidden number. In each query you should print an integer from interval [2, 100]. The system will answer "yes" if your integer is a divisor of the hidden number. Otherwise, the answer will be "no".
For example, if the hidden number is 14 then the system will answer "yes" only if you print 2, 7 or 14.
When you are done asking queries, print "prime" or "composite" and terminate your program.
You will get the Wrong Answer verdict if you ask more than 20 queries, or if you print an integer not from the range [2, 100]. Also, you will get the Wrong Answer verdict if the printed answer isn't correct.
You will get the Idleness Limit Exceeded verdict if you don't print anything (but you should) or if you forget about flushing the output (more info below).
Input
After each query you should read one string from the input. It will be "yes" if the printed integer is a divisor of the hidden number, and "no" otherwise.
Output
Up to 20 times you can ask a query — print an integer from interval [2, 100] in one line. You have to both print the end-of-line character and flush the output. After flushing you should read a response from the input.
In any moment you can print the answer "prime" or "composite" (without the quotes). After that, flush the output and terminate your program.
To flush you can use (just after printing an integer and end-of-line):
* fflush(stdout) in C++;
* System.out.flush() in Java;
* stdout.flush() in Python;
* flush(output) in Pascal;
* See the documentation for other languages.
Hacking. To hack someone, as the input you should print the hidden number — one integer from the interval [2, 100]. Of course, his/her solution won't be able to read the hidden number from the input.
Examples
Input
yes
no
yes
Output
2
80
5
composite
Input
no
yes
no
no
no
Output
58
59
78
78
2
prime
Note
The hidden number in the first query is 30. In a table below you can see a better form of the provided example of the communication process.
<image>
The hidden number is divisible by both 2 and 5. Thus, it must be composite. Note that it isn't necessary to know the exact value of the hidden number. In this test, the hidden number is 30.
<image>
59 is a divisor of the hidden number. In the interval [2, 100] there is only one number with this divisor. The hidden number must be 59, which is prime. Note that the answer is known even after the second query and you could print it then and terminate. Though, it isn't forbidden to ask unnecessary queries (unless you exceed the limit of 20 queries). | instruction | 0 | 39,157 | 22 | 78,314 |
Tags: constructive algorithms, interactive, math
Correct Solution:
```
"""
Author : thekushalghosh
Team : CodeDiggers
"""
import sys,math,cmath,time
start_time = time.time()
################# ---- USER DEFINED INPUT FUNCTIONS ---- #################
def inp():
return(int(input()))
def inlt():
return(list(map(int,input().split())))
def insr():
s = input()
return(s[:len(s) - 1])
def invr():
return(map(int,input().split()))
##########################################################################
################# ---- THE ACTUAL CODE STARTS BELOW ---- #################
def solve():
a = []
for i in range(2,50):
qw = 1
for j in range(2,math.ceil(i ** 0.5)):
if i % j == 0:
qw = 0
break
if qw == 1:
a.append(i)
qw = "prime"
c = 0
for i in range(len(a)):
print(a[i])
sys.stdout.flush()
s = insr()
if s == "yes":
c = c + 1
if c >= 2:
qw = "composite"
break
print(qw)
################## ---- THE ACTUAL CODE ENDS ABOVE ---- ##################
##########################################################################
ONLINE_JUDGE = __debug__
if not ONLINE_JUDGE:
sys.stdin = open('input.txt','r')
sys.stdout = open('output.txt','w')
else:
input = sys.stdin.readline
t = 1
for tt in range(t):
solve()
if not ONLINE_JUDGE:
print("Time Elapsed:",time.time() - start_time,"seconds")
sys.stdout.close()
``` | output | 1 | 39,157 | 22 | 78,315 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is an interactive problem. In the output section below you will see the information about flushing the output.
Bear Limak thinks of some hidden number — an integer from interval [2, 100]. Your task is to say if the hidden number is prime or composite.
Integer x > 1 is called prime if it has exactly two distinct divisors, 1 and x. If integer x > 1 is not prime, it's called composite.
You can ask up to 20 queries about divisors of the hidden number. In each query you should print an integer from interval [2, 100]. The system will answer "yes" if your integer is a divisor of the hidden number. Otherwise, the answer will be "no".
For example, if the hidden number is 14 then the system will answer "yes" only if you print 2, 7 or 14.
When you are done asking queries, print "prime" or "composite" and terminate your program.
You will get the Wrong Answer verdict if you ask more than 20 queries, or if you print an integer not from the range [2, 100]. Also, you will get the Wrong Answer verdict if the printed answer isn't correct.
You will get the Idleness Limit Exceeded verdict if you don't print anything (but you should) or if you forget about flushing the output (more info below).
Input
After each query you should read one string from the input. It will be "yes" if the printed integer is a divisor of the hidden number, and "no" otherwise.
Output
Up to 20 times you can ask a query — print an integer from interval [2, 100] in one line. You have to both print the end-of-line character and flush the output. After flushing you should read a response from the input.
In any moment you can print the answer "prime" or "composite" (without the quotes). After that, flush the output and terminate your program.
To flush you can use (just after printing an integer and end-of-line):
* fflush(stdout) in C++;
* System.out.flush() in Java;
* stdout.flush() in Python;
* flush(output) in Pascal;
* See the documentation for other languages.
Hacking. To hack someone, as the input you should print the hidden number — one integer from the interval [2, 100]. Of course, his/her solution won't be able to read the hidden number from the input.
Examples
Input
yes
no
yes
Output
2
80
5
composite
Input
no
yes
no
no
no
Output
58
59
78
78
2
prime
Note
The hidden number in the first query is 30. In a table below you can see a better form of the provided example of the communication process.
<image>
The hidden number is divisible by both 2 and 5. Thus, it must be composite. Note that it isn't necessary to know the exact value of the hidden number. In this test, the hidden number is 30.
<image>
59 is a divisor of the hidden number. In the interval [2, 100] there is only one number with this divisor. The hidden number must be 59, which is prime. Note that the answer is known even after the second query and you could print it then and terminate. Though, it isn't forbidden to ask unnecessary queries (unless you exceed the limit of 20 queries). | instruction | 0 | 39,158 | 22 | 78,316 |
Tags: constructive algorithms, interactive, math
Correct Solution:
```
# from __future__ import division, print_function
import sys
import os
from io import BytesIO, IOBase
# def inp(): return sys.stdin.readline().rstrip("\r\n") #for fast input
# n = int(os.read(0,100))
# if(n==1):
# os.write(1,b'! 1')
# sys.stdout.flush()
# else:
# per = [0]*n
# # print('? 2 1')
# # sys.stdout.flush()
# a1 = query(1,0)
# # print('? 1 2')
# # sys.stdout.flush()
# a2 = int(query(0,1))
# if(a2>a1):
# max = 2
# per[0] = a2
# else:
# max = 1
# per[1] = a1
# # print(per)
# for i in range(3,n+1):
# # print('? '+str(i)+" "+str(max))
# # sys.stdout.flush()
# a1 = int(query(i-1,max-1))
# # print('? '+str(max)+' '+str(i))
# # sys.stdout.flush()
# a2 = int(query(max-1,i-1))
# if(a1>a2):
# per[i-1] = a1
# else:
# per[max-1] = a2
# max = i
# # print(per)
# per[max-1] = n
# # print('! ',end="")
# # print(' '.join(str(per[i]) for i in range(len(per))))
# # sys.stdout.flush()
# os.write(1, b'! ')
# os.write(1, b' '.join(str(x).encode('ascii') for x in per))
def isPrime(n) :
if (n <= 1) : return False
if (n <= 3) : return True
if (n % 2 == 0 or n % 3 == 0) : return False
i = 5
while(i * i <= n) :
if (n % i == 0 or n % (i + 2) == 0) :
return False
i = i + 6
return True
def query(a):
# print(1)
os.write(1,b"%d\n" % a)
return str(os.read(0,100))
def solve(case):
cnt = 0
primes = [2,3,5,7]
div = []
for i in range(len(primes)):
# print(primes[i])
# sys.stdout.flush()
# ans = inp()
ans = query(primes[i])
# print(ans)
if('yes' in ans):
div.append(primes[i])
cnt+=1
if(cnt == 0):
print('prime')
else:
if(len(div) == 1):
i = div[0]
x = 2
cnt = 0
while((i**x)<101):
# print(i**x)
# sys.stdout.flush()
if('yes' in query(i**x)):
cnt+=1
break
x+=1
new = []
for i in range(10,50):
if(isPrime(i)):
new.append(i)
for i in new:
if(div[0]*i>100):
break
# print(i)
# sys.stdout.flush()
if('yes' in query(i)):
cnt+=1
if(cnt == 0):
print('prime')
else:
print('composite')
else:
print('composite')
solve(1)
``` | output | 1 | 39,158 | 22 | 78,317 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is an interactive problem. In the output section below you will see the information about flushing the output.
Bear Limak thinks of some hidden number — an integer from interval [2, 100]. Your task is to say if the hidden number is prime or composite.
Integer x > 1 is called prime if it has exactly two distinct divisors, 1 and x. If integer x > 1 is not prime, it's called composite.
You can ask up to 20 queries about divisors of the hidden number. In each query you should print an integer from interval [2, 100]. The system will answer "yes" if your integer is a divisor of the hidden number. Otherwise, the answer will be "no".
For example, if the hidden number is 14 then the system will answer "yes" only if you print 2, 7 or 14.
When you are done asking queries, print "prime" or "composite" and terminate your program.
You will get the Wrong Answer verdict if you ask more than 20 queries, or if you print an integer not from the range [2, 100]. Also, you will get the Wrong Answer verdict if the printed answer isn't correct.
You will get the Idleness Limit Exceeded verdict if you don't print anything (but you should) or if you forget about flushing the output (more info below).
Input
After each query you should read one string from the input. It will be "yes" if the printed integer is a divisor of the hidden number, and "no" otherwise.
Output
Up to 20 times you can ask a query — print an integer from interval [2, 100] in one line. You have to both print the end-of-line character and flush the output. After flushing you should read a response from the input.
In any moment you can print the answer "prime" or "composite" (without the quotes). After that, flush the output and terminate your program.
To flush you can use (just after printing an integer and end-of-line):
* fflush(stdout) in C++;
* System.out.flush() in Java;
* stdout.flush() in Python;
* flush(output) in Pascal;
* See the documentation for other languages.
Hacking. To hack someone, as the input you should print the hidden number — one integer from the interval [2, 100]. Of course, his/her solution won't be able to read the hidden number from the input.
Examples
Input
yes
no
yes
Output
2
80
5
composite
Input
no
yes
no
no
no
Output
58
59
78
78
2
prime
Note
The hidden number in the first query is 30. In a table below you can see a better form of the provided example of the communication process.
<image>
The hidden number is divisible by both 2 and 5. Thus, it must be composite. Note that it isn't necessary to know the exact value of the hidden number. In this test, the hidden number is 30.
<image>
59 is a divisor of the hidden number. In the interval [2, 100] there is only one number with this divisor. The hidden number must be 59, which is prime. Note that the answer is known even after the second query and you could print it then and terminate. Though, it isn't forbidden to ask unnecessary queries (unless you exceed the limit of 20 queries).
Submitted Solution:
```
"""
Satwik_Tiwari ;) .
6th Sept , 2020 - Sunday
"""
#===============================================================================================
#importing some useful libraries.
from __future__ import division, print_function
from fractions import Fraction
import sys
import os
from io import BytesIO, IOBase
from itertools import *
import bisect
from heapq import *
from math import *
from copy import *
from collections import deque
from collections import Counter as counter # Counter(list) return a dict with {key: count}
from itertools import combinations as comb # if a = [1,2,3] then print(list(comb(a,2))) -----> [(1, 2), (1, 3), (2, 3)]
from itertools import permutations as permutate
from bisect import bisect_left as bl
#If the element is already present in the list,
# the left most position where element has to be inserted is returned.
from bisect import bisect_right as br
from bisect import bisect
#If the element is already present in the list,
# the right most position where element has to be inserted is returned
#==============================================================================================
#fast I/O 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")
def print(*args, **kwargs):
"""Prints the values to a stream, or to sys.stdout by default."""
sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
at_start = True
for x in args:
if not at_start:
file.write(sep)
file.write(str(x))
at_start = False
file.write(kwargs.pop("end", "\n"))
if kwargs.pop("flush", False):
file.flush()
if sys.version_info[0] < 3:
sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
# inp = lambda: sys.stdin.readline().rstrip("\r\n")
#===============================================================================================
### START ITERATE RECURSION ###
from types import GeneratorType
def iterative(f, stack=[]):
def wrapped_func(*args, **kwargs):
if stack: return f(*args, **kwargs)
to = f(*args, **kwargs)
while True:
if type(to) is GeneratorType:
stack.append(to)
to = next(to)
continue
stack.pop()
if not stack: break
to = stack[-1].send(to)
return to
return wrapped_func
#### END ITERATE RECURSION ####
#===============================================================================================
#some shortcuts
mod = 1000000007
def inp(): return sys.stdin.readline().rstrip("\r\n") #for fast input
def out(var): sys.stdout.write(str(var)) #for fast output, always take string
def lis(): return list(map(int, inp().split()))
def stringlis(): return list(map(str, inp().split()))
def sep(): return map(int, inp().split())
def strsep(): return map(str, inp().split())
# def graph(vertex): return [[] for i in range(0,vertex+1)]
def zerolist(n): return [0]*n
def nextline(): out("\n") #as stdout.write always print sring.
def testcase(t):
for pp in range(t):
solve(pp)
def printlist(a) :
for p in range(0,len(a)):
out(str(a[p]) + ' ')
def google(p):
print('Case #'+str(p)+': ',end='')
def lcm(a,b): return (a*b)//gcd(a,b)
def power(x, y, p) :
res = 1 # Initialize result
x = x % p # Update x if it is more , than or equal to p
if (x == 0) :
return 0
while (y > 0) :
if ((y & 1) == 1) : # If y is odd, multiply, x with result
res = (res * x) % p
y = y >> 1 # y = y/2
x = (x * x) % p
return res
def ncr(n,r): return factorial(n)//(factorial(r)*factorial(max(n-r,1)))
def isPrime(n) :
if (n <= 1) : return False
if (n <= 3) : return True
if (n % 2 == 0 or n % 3 == 0) : return False
i = 5
while(i * i <= n) :
if (n % i == 0 or n % (i + 2) == 0) :
return False
i = i + 6
return True
#===============================================================================================
# code here ;))
def solve(case):
cnt = 0
primes = [2,3,5,7]
div = []
for i in range(len(primes)):
print(primes[i])
sys.stdout.flush()
ans = inp()
if(ans == 'yes'):
div.append(primes[i])
cnt+=1
if(cnt == 0):
print('prime')
else:
if(len(div) == 1):
i = div[0]
x = 2
cnt = 0
while((i**x)<101):
print(i**x)
sys.stdout.flush()
if(inp() == 'yes'):
cnt+=1
break
x+=1
new = []
for i in range(10,50):
if(isPrime(i)):
new.append(i)
for i in new:
if(div[0]*i>100):
break
print(i)
sys.stdout.flush()
if(inp() == 'yes'):
cnt+=1
if(cnt == 0):
print('prime')
else:
print('composite')
else:
print('composite')
testcase(1)
# testcase(int(inp()))
``` | instruction | 0 | 39,159 | 22 | 78,318 |
Yes | output | 1 | 39,159 | 22 | 78,319 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is an interactive problem. In the output section below you will see the information about flushing the output.
Bear Limak thinks of some hidden number — an integer from interval [2, 100]. Your task is to say if the hidden number is prime or composite.
Integer x > 1 is called prime if it has exactly two distinct divisors, 1 and x. If integer x > 1 is not prime, it's called composite.
You can ask up to 20 queries about divisors of the hidden number. In each query you should print an integer from interval [2, 100]. The system will answer "yes" if your integer is a divisor of the hidden number. Otherwise, the answer will be "no".
For example, if the hidden number is 14 then the system will answer "yes" only if you print 2, 7 or 14.
When you are done asking queries, print "prime" or "composite" and terminate your program.
You will get the Wrong Answer verdict if you ask more than 20 queries, or if you print an integer not from the range [2, 100]. Also, you will get the Wrong Answer verdict if the printed answer isn't correct.
You will get the Idleness Limit Exceeded verdict if you don't print anything (but you should) or if you forget about flushing the output (more info below).
Input
After each query you should read one string from the input. It will be "yes" if the printed integer is a divisor of the hidden number, and "no" otherwise.
Output
Up to 20 times you can ask a query — print an integer from interval [2, 100] in one line. You have to both print the end-of-line character and flush the output. After flushing you should read a response from the input.
In any moment you can print the answer "prime" or "composite" (without the quotes). After that, flush the output and terminate your program.
To flush you can use (just after printing an integer and end-of-line):
* fflush(stdout) in C++;
* System.out.flush() in Java;
* stdout.flush() in Python;
* flush(output) in Pascal;
* See the documentation for other languages.
Hacking. To hack someone, as the input you should print the hidden number — one integer from the interval [2, 100]. Of course, his/her solution won't be able to read the hidden number from the input.
Examples
Input
yes
no
yes
Output
2
80
5
composite
Input
no
yes
no
no
no
Output
58
59
78
78
2
prime
Note
The hidden number in the first query is 30. In a table below you can see a better form of the provided example of the communication process.
<image>
The hidden number is divisible by both 2 and 5. Thus, it must be composite. Note that it isn't necessary to know the exact value of the hidden number. In this test, the hidden number is 30.
<image>
59 is a divisor of the hidden number. In the interval [2, 100] there is only one number with this divisor. The hidden number must be 59, which is prime. Note that the answer is known even after the second query and you could print it then and terminate. Though, it isn't forbidden to ask unnecessary queries (unless you exceed the limit of 20 queries).
Submitted Solution:
```
#2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97
import sys,math
primes=[2,3,5,7]
an=[11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97]
ctr=0
used=0
for i in primes:
print(i)
sys.stdout.flush()
ans=input()
if ans=='yes':
ctr+=1
used=i
if ctr>=2:
print('composite')
sys.stdout.flush()
sys.exit(0)
elif ctr==0:
print('prime')
sys.stdout.flush()
sys.exit(0)
for i in an:
if used*i>100:
break
print(i)
sys.stdout.flush()
ans=input()
if ans=='yes':
print('composite')
sys.stdout.flush()
sys.exit(0)
print('prime')
sys.stdout.flush()
``` | instruction | 0 | 39,160 | 22 | 78,320 |
No | output | 1 | 39,160 | 22 | 78,321 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is an interactive problem. In the output section below you will see the information about flushing the output.
Bear Limak thinks of some hidden number — an integer from interval [2, 100]. Your task is to say if the hidden number is prime or composite.
Integer x > 1 is called prime if it has exactly two distinct divisors, 1 and x. If integer x > 1 is not prime, it's called composite.
You can ask up to 20 queries about divisors of the hidden number. In each query you should print an integer from interval [2, 100]. The system will answer "yes" if your integer is a divisor of the hidden number. Otherwise, the answer will be "no".
For example, if the hidden number is 14 then the system will answer "yes" only if you print 2, 7 or 14.
When you are done asking queries, print "prime" or "composite" and terminate your program.
You will get the Wrong Answer verdict if you ask more than 20 queries, or if you print an integer not from the range [2, 100]. Also, you will get the Wrong Answer verdict if the printed answer isn't correct.
You will get the Idleness Limit Exceeded verdict if you don't print anything (but you should) or if you forget about flushing the output (more info below).
Input
After each query you should read one string from the input. It will be "yes" if the printed integer is a divisor of the hidden number, and "no" otherwise.
Output
Up to 20 times you can ask a query — print an integer from interval [2, 100] in one line. You have to both print the end-of-line character and flush the output. After flushing you should read a response from the input.
In any moment you can print the answer "prime" or "composite" (without the quotes). After that, flush the output and terminate your program.
To flush you can use (just after printing an integer and end-of-line):
* fflush(stdout) in C++;
* System.out.flush() in Java;
* stdout.flush() in Python;
* flush(output) in Pascal;
* See the documentation for other languages.
Hacking. To hack someone, as the input you should print the hidden number — one integer from the interval [2, 100]. Of course, his/her solution won't be able to read the hidden number from the input.
Examples
Input
yes
no
yes
Output
2
80
5
composite
Input
no
yes
no
no
no
Output
58
59
78
78
2
prime
Note
The hidden number in the first query is 30. In a table below you can see a better form of the provided example of the communication process.
<image>
The hidden number is divisible by both 2 and 5. Thus, it must be composite. Note that it isn't necessary to know the exact value of the hidden number. In this test, the hidden number is 30.
<image>
59 is a divisor of the hidden number. In the interval [2, 100] there is only one number with this divisor. The hidden number must be 59, which is prime. Note that the answer is known even after the second query and you could print it then and terminate. Though, it isn't forbidden to ask unnecessary queries (unless you exceed the limit of 20 queries).
Submitted Solution:
```
for x in range(2, 11, 1):
print(str(x))
divisible = input()
if divisible == "yes":
print("composite")
exit()
print("prime")
``` | instruction | 0 | 39,161 | 22 | 78,322 |
No | output | 1 | 39,161 | 22 | 78,323 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is an interactive problem. In the output section below you will see the information about flushing the output.
Bear Limak thinks of some hidden number — an integer from interval [2, 100]. Your task is to say if the hidden number is prime or composite.
Integer x > 1 is called prime if it has exactly two distinct divisors, 1 and x. If integer x > 1 is not prime, it's called composite.
You can ask up to 20 queries about divisors of the hidden number. In each query you should print an integer from interval [2, 100]. The system will answer "yes" if your integer is a divisor of the hidden number. Otherwise, the answer will be "no".
For example, if the hidden number is 14 then the system will answer "yes" only if you print 2, 7 or 14.
When you are done asking queries, print "prime" or "composite" and terminate your program.
You will get the Wrong Answer verdict if you ask more than 20 queries, or if you print an integer not from the range [2, 100]. Also, you will get the Wrong Answer verdict if the printed answer isn't correct.
You will get the Idleness Limit Exceeded verdict if you don't print anything (but you should) or if you forget about flushing the output (more info below).
Input
After each query you should read one string from the input. It will be "yes" if the printed integer is a divisor of the hidden number, and "no" otherwise.
Output
Up to 20 times you can ask a query — print an integer from interval [2, 100] in one line. You have to both print the end-of-line character and flush the output. After flushing you should read a response from the input.
In any moment you can print the answer "prime" or "composite" (without the quotes). After that, flush the output and terminate your program.
To flush you can use (just after printing an integer and end-of-line):
* fflush(stdout) in C++;
* System.out.flush() in Java;
* stdout.flush() in Python;
* flush(output) in Pascal;
* See the documentation for other languages.
Hacking. To hack someone, as the input you should print the hidden number — one integer from the interval [2, 100]. Of course, his/her solution won't be able to read the hidden number from the input.
Examples
Input
yes
no
yes
Output
2
80
5
composite
Input
no
yes
no
no
no
Output
58
59
78
78
2
prime
Note
The hidden number in the first query is 30. In a table below you can see a better form of the provided example of the communication process.
<image>
The hidden number is divisible by both 2 and 5. Thus, it must be composite. Note that it isn't necessary to know the exact value of the hidden number. In this test, the hidden number is 30.
<image>
59 is a divisor of the hidden number. In the interval [2, 100] there is only one number with this divisor. The hidden number must be 59, which is prime. Note that the answer is known even after the second query and you could print it then and terminate. Though, it isn't forbidden to ask unnecessary queries (unless you exceed the limit of 20 queries).
Submitted Solution:
```
import sys
primes = [2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97]
print(2)
ans = "prime"
state = False
reply = input()
if (reply == "yes"):
for i in range(1,15):
print(primes[i])
reply = input()
if (reply == "yes"):
ans = "composite"
break
if (ans != "prime"):
print(ans)
sys.stdout.flush()
else:
print(3)
reply = input()
if (reply == "yes"):
for i in range(2,12):
print(primes[i])
reply = input()
if (reply == "yes"):
ans = "composite"
break
if (ans != "prime"):
print(ans)
sys.stdout.flush()
else:
print(5)
reply = input()
if (reply == "yes"):
for i in range(3,9):
print(primes[i])
reply = input()
if (reply == "yes"):
ans = "composite"
break
if (ans != "prime"):
print(ans)
sys.stdout.flush()
else:
print(7)
reply = input()
if (reply == "yes"):
for i in range(4,7):
print(primes[i])
reply = input()
if (reply == "yes"):
ans = "composite"
break
if (ans != "prime"):
print(ans)
sys.stdout.flush()
if (ans == "prime"):
print(ans)
sys.stdout.flush()
``` | instruction | 0 | 39,162 | 22 | 78,324 |
No | output | 1 | 39,162 | 22 | 78,325 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is an interactive problem. In the output section below you will see the information about flushing the output.
Bear Limak thinks of some hidden number — an integer from interval [2, 100]. Your task is to say if the hidden number is prime or composite.
Integer x > 1 is called prime if it has exactly two distinct divisors, 1 and x. If integer x > 1 is not prime, it's called composite.
You can ask up to 20 queries about divisors of the hidden number. In each query you should print an integer from interval [2, 100]. The system will answer "yes" if your integer is a divisor of the hidden number. Otherwise, the answer will be "no".
For example, if the hidden number is 14 then the system will answer "yes" only if you print 2, 7 or 14.
When you are done asking queries, print "prime" or "composite" and terminate your program.
You will get the Wrong Answer verdict if you ask more than 20 queries, or if you print an integer not from the range [2, 100]. Also, you will get the Wrong Answer verdict if the printed answer isn't correct.
You will get the Idleness Limit Exceeded verdict if you don't print anything (but you should) or if you forget about flushing the output (more info below).
Input
After each query you should read one string from the input. It will be "yes" if the printed integer is a divisor of the hidden number, and "no" otherwise.
Output
Up to 20 times you can ask a query — print an integer from interval [2, 100] in one line. You have to both print the end-of-line character and flush the output. After flushing you should read a response from the input.
In any moment you can print the answer "prime" or "composite" (without the quotes). After that, flush the output and terminate your program.
To flush you can use (just after printing an integer and end-of-line):
* fflush(stdout) in C++;
* System.out.flush() in Java;
* stdout.flush() in Python;
* flush(output) in Pascal;
* See the documentation for other languages.
Hacking. To hack someone, as the input you should print the hidden number — one integer from the interval [2, 100]. Of course, his/her solution won't be able to read the hidden number from the input.
Examples
Input
yes
no
yes
Output
2
80
5
composite
Input
no
yes
no
no
no
Output
58
59
78
78
2
prime
Note
The hidden number in the first query is 30. In a table below you can see a better form of the provided example of the communication process.
<image>
The hidden number is divisible by both 2 and 5. Thus, it must be composite. Note that it isn't necessary to know the exact value of the hidden number. In this test, the hidden number is 30.
<image>
59 is a divisor of the hidden number. In the interval [2, 100] there is only one number with this divisor. The hidden number must be 59, which is prime. Note that the answer is known even after the second query and you could print it then and terminate. Though, it isn't forbidden to ask unnecessary queries (unless you exceed the limit of 20 queries).
Submitted Solution:
```
def isPrime(x):
if x < 2 :
return False
tmp = 2
while True :
if tmp*tmp > x :
return True
if x % tmp ==0 :
return False ;
tmp = tmp + 1
if __name__=='__main__':
prime = []
for i in range(54) :
if isPrime(i) :
prime.append(i)
count = 0
for x in prime :
#print(x)
print(x)
s = input()
if s == 'yes' :
count = count + 1
else :
pass
if count > 1 :
print('composite')
else :
print('prime')
#stdout.flush()
``` | instruction | 0 | 39,163 | 22 | 78,326 |
No | output | 1 | 39,163 | 22 | 78,327 |
Provide a correct Python 3 solution for this coding contest problem.
You are given an integer N.
Construct any one N-by-N matrix a that satisfies the conditions below. It can be proved that a solution always exists under the constraints of this problem.
* 1 \leq a_{i,j} \leq 10^{15}
* a_{i,j} are pairwise distinct integers.
* There exists a positive integer m such that the following holds: Let x and y be two elements of the matrix that are vertically or horizontally adjacent. Then, {\rm max}(x,y) {\rm mod} {\rm min}(x,y) is always m.
Constraints
* 2 \leq N \leq 500
Input
Input is given from Standard Input in the following format:
N
Output
Print your solution in the following format:
a_{1,1} ... a_{1,N}
:
a_{N,1} ... a_{N,N}
Output
Print your solution in the following format:
a_{1,1} ... a_{1,N}
:
a_{N,1} ... a_{N,N}
Example
Input
2
Output
4 7
23 10 | instruction | 0 | 39,447 | 22 | 78,894 |
"Correct Solution:
```
import sys
#sys.stdin=open("data.txt")
input=sys.stdin.readline
mii=lambda:map(int,input().split())
# get primes
lim=10000
sieve=[1]*lim
prime=[]
for i in range(2,lim):
if not sieve[i]: continue
prime.append(i)
for j in range(i*i,lim,i):
sieve[j]=0
prime.sort()
prime=[i for i in prime if i>500]
prime=prime[:500]
li=[]
while prime:
li.append(prime.pop(0))
li.append(prime.pop(-1))
powers=li
# print matrix
n=int(input())
out=[[0 for j in range(n)] for i in range(n)]
s=set()
for i in range(n):
for j in range(n):
if (i+j)%2==0:
# put small number
out[i][j]=powers[(i+j)//2]*(i+1)
if out[i][j] in s:
print("why",i,j)
s.add(out[i][j])
for i in range(n):
for j in range(n):
if (i+j)%2==1:
# put big number
big=i+1
if i>0: big*=i
if i<n-1: big*=i+2
big*=powers[(i+j)//2]
big*=powers[(i+j+1)//2]
k=1
while big*k+1 in s: k+=1
out[i][j]=big*k+1
s.add(out[i][j])
# checking
if 0:
for i in range(n-1):
for j in range(n):
x=out[i][j]
y=out[i+1][j]
if x<y: x,y=y,x
while x%y!=1: 1
x=out[j][i]
y=out[j][i+1]
if x<y: x,y=y,x
while x%y!=1: 1
s=set()
for i in out:
for j in i:
s.add(j)
while len(s)!=n**2: 1
while max(s)>10**15: 1
while min(s)<1: 1
for li in out: print(" ".join(map(str,li)))
``` | output | 1 | 39,447 | 22 | 78,895 |
Provide a correct Python 3 solution for this coding contest problem.
You are given an integer N.
Construct any one N-by-N matrix a that satisfies the conditions below. It can be proved that a solution always exists under the constraints of this problem.
* 1 \leq a_{i,j} \leq 10^{15}
* a_{i,j} are pairwise distinct integers.
* There exists a positive integer m such that the following holds: Let x and y be two elements of the matrix that are vertically or horizontally adjacent. Then, {\rm max}(x,y) {\rm mod} {\rm min}(x,y) is always m.
Constraints
* 2 \leq N \leq 500
Input
Input is given from Standard Input in the following format:
N
Output
Print your solution in the following format:
a_{1,1} ... a_{1,N}
:
a_{N,1} ... a_{N,N}
Output
Print your solution in the following format:
a_{1,1} ... a_{1,N}
:
a_{N,1} ... a_{N,N}
Example
Input
2
Output
4 7
23 10 | instruction | 0 | 39,449 | 22 | 78,898 |
"Correct Solution:
```
N = int(input())
if N == 2:
print(4, 7)
print(23, 10)
exit()
a= [[0] * N for _ in range(N)]
def sieve(n):
'''
:param n:
:return: n以下の素数のリストを返す
エラトステネスの篩→O(n log log n)
'''
prime = []
is_prime = [True] * (n + 1) #is_prime[i] = Trueならiは素数
is_prime[0] = False
is_prime[1] = False
for i in range(2, n+1):
if is_prime[i]:
prime.append(i)
for j in range(2 * i, n + 1, i):
is_prime[j] = False
return prime
P = sieve(8000) #len(P) > 1000
def p1(k):
return P[(k // 2) + 1]
def p2(k):
if k >= 0:
return P[(k // 2) + N + 1]
else:
return P[k // 2]
for i in range(N):
for j in range(N):
if (i + j) % 2 == 0:
a[i][j] = p1(i + j) * p2(i - j)
else:
a[i][j] = p1(i + j + 1) * p1(i + j - 1) * p2(i - j + 1) * p2(i - j - 1) + 1
for i in range(N):
print(' '.join(map(str, a[i])))
``` | output | 1 | 39,449 | 22 | 78,899 |
Provide a correct Python 3 solution for this coding contest problem.
You are given an integer N.
Construct any one N-by-N matrix a that satisfies the conditions below. It can be proved that a solution always exists under the constraints of this problem.
* 1 \leq a_{i,j} \leq 10^{15}
* a_{i,j} are pairwise distinct integers.
* There exists a positive integer m such that the following holds: Let x and y be two elements of the matrix that are vertically or horizontally adjacent. Then, {\rm max}(x,y) {\rm mod} {\rm min}(x,y) is always m.
Constraints
* 2 \leq N \leq 500
Input
Input is given from Standard Input in the following format:
N
Output
Print your solution in the following format:
a_{1,1} ... a_{1,N}
:
a_{N,1} ... a_{N,N}
Output
Print your solution in the following format:
a_{1,1} ... a_{1,N}
:
a_{N,1} ... a_{N,N}
Example
Input
2
Output
4 7
23 10 | instruction | 0 | 39,450 | 22 | 78,900 |
"Correct Solution:
```
from math import gcd
Primes = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293, 307, 311, 313, 317, 331, 337, 347, 349, 353, 359, 367, 373, 379, 383, 389, 397, 401, 409, 419, 421, 431, 433, 439, 443, 449, 457, 461, 463, 467, 479, 487, 491, 499, 503, 509, 521, 523, 541, 547, 557, 563, 569, 571, 577, 587, 593, 599, 601, 607, 613, 617, 619, 631, 641, 643, 647, 653, 659, 661, 673, 677, 683, 691, 701, 709, 719, 727, 733, 739, 743, 751, 757, 761, 769, 773, 787, 797, 809, 811, 821, 823, 827, 829, 839, 853, 857, 859, 863, 877, 881, 883, 887, 907, 911, 919, 929, 937, 941, 947, 953, 967, 971, 977, 983, 991, 997, 1009, 1013, 1019, 1021, 1031, 1033, 1039, 1049, 1051, 1061, 1063, 1069, 1087, 1091, 1093, 1097, 1103, 1109, 1117, 1123, 1129, 1151, 1153, 1163, 1171, 1181, 1187, 1193, 1201, 1213, 1217, 1223, 1229, 1231, 1237, 1249, 1259, 1277, 1279, 1283, 1289, 1291, 1297, 1301, 1303, 1307, 1319, 1321, 1327, 1361, 1367, 1373, 1381, 1399, 1409, 1423, 1427, 1429, 1433, 1439, 1447, 1451, 1453, 1459, 1471, 1481, 1483, 1487, 1489, 1493, 1499, 1511, 1523, 1531, 1543, 1549, 1553, 1559, 1567, 1571, 1579, 1583, 1597, 1601, 1607, 1609, 1613, 1619, 1621, 1627, 1637, 1657, 1663, 1667, 1669, 1693, 1697, 1699, 1709, 1721, 1723, 1733, 1741, 1747, 1753, 1759, 1777, 1783, 1787, 1789, 1801, 1811, 1823, 1831, 1847, 1861, 1867, 1871, 1873, 1877, 1879, 1889, 1901, 1907, 1913, 1931, 1933, 1949, 1951, 1973, 1979, 1987, 1993, 1997, 1999, 2003, 2011, 2017, 2027, 2029, 2039, 2053, 2063, 2069, 2081, 2083, 2087, 2089, 2099, 2111, 2113, 2129, 2131, 2137, 2141, 2143, 2153, 2161, 2179, 2203, 2207, 2213, 2221, 2237, 2239, 2243, 2251, 2267, 2269, 2273, 2281, 2287, 2293, 2297, 2309, 2311, 2333, 2339, 2341, 2347, 2351, 2357, 2371, 2377, 2381, 2383, 2389, 2393, 2399, 2411, 2417, 2423, 2437, 2441, 2447, 2459, 2467, 2473, 2477, 2503, 2521, 2531, 2539, 2543, 2549, 2551, 2557, 2579, 2591, 2593, 2609, 2617, 2621, 2633, 2647, 2657, 2659, 2663, 2671, 2677, 2683, 2687, 2689, 2693, 2699, 2707, 2711, 2713, 2719, 2729, 2731, 2741, 2749, 2753, 2767, 2777, 2789, 2791, 2797, 2801, 2803, 2819, 2833, 2837, 2843, 2851, 2857, 2861, 2879, 2887, 2897, 2903, 2909, 2917, 2927, 2939, 2953, 2957, 2963, 2969, 2971, 2999, 3001, 3011, 3019, 3023, 3037, 3041, 3049, 3061, 3067, 3079, 3083, 3089, 3109, 3119, 3121, 3137, 3163, 3167, 3169, 3181, 3187, 3191, 3203, 3209, 3217, 3221, 3229, 3251, 3253, 3257, 3259, 3271, 3299, 3301, 3307, 3313, 3319, 3323, 3329, 3331, 3343, 3347, 3359, 3361, 3371, 3373, 3389, 3391, 3407, 3413, 3433, 3449, 3457, 3461, 3463, 3467, 3469, 3491, 3499, 3511, 3517, 3527, 3529, 3533, 3539, 3541, 3547, 3557, 3559, 3571, 3581, 3583, 3593, 3607, 3613, 3617, 3623, 3631, 3637, 3643, 3659, 3671, 3673, 3677, 3691, 3697, 3701, 3709, 3719, 3727, 3733, 3739, 3761, 3767, 3769, 3779, 3793, 3797, 3803, 3821, 3823, 3833, 3847, 3851, 3853, 3863, 3877, 3881, 3889, 3907, 3911, 3917, 3919, 3923, 3929, 3931, 3943, 3947, 3967, 3989, 4001, 4003, 4007, 4013, 4019, 4021, 4027, 4049, 4051, 4057, 4073, 4079, 4091, 4093, 4099, 4111, 4127, 4129, 4133, 4139, 4153, 4157, 4159, 4177, 4201, 4211, 4217, 4219, 4229, 4231, 4241, 4243, 4253, 4259, 4261, 4271, 4273, 4283, 4289, 4297, 4327, 4337, 4339, 4349, 4357, 4363, 4373, 4391, 4397, 4409, 4421, 4423, 4441, 4447, 4451, 4457, 4463, 4481, 4483, 4493, 4507, 4513, 4517, 4519, 4523, 4547, 4549, 4561, 4567, 4583, 4591, 4597, 4603, 4621, 4637, 4639, 4643, 4649, 4651, 4657, 4663, 4673, 4679, 4691, 4703, 4721, 4723, 4729, 4733, 4751, 4759, 4783, 4787, 4789, 4793, 4799, 4801, 4813, 4817, 4831, 4861, 4871, 4877, 4889, 4903, 4909, 4919, 4931, 4933, 4937, 4943, 4951, 4957, 4967, 4969, 4973, 4987, 4993, 4999, 5003, 5009, 5011, 5021, 5023, 5039, 5051, 5059, 5077, 5081, 5087, 5099, 5101, 5107, 5113, 5119, 5147, 5153, 5167, 5171, 5179, 5189, 5197, 5209, 5227, 5231, 5233, 5237, 5261, 5273, 5279, 5281, 5297, 5303, 5309, 5323, 5333, 5347, 5351, 5381, 5387, 5393, 5399, 5407, 5413, 5417, 5419, 5431, 5437, 5441, 5443, 5449, 5471, 5477, 5479, 5483, 5501, 5503, 5507, 5519, 5521, 5527, 5531, 5557, 5563, 5569, 5573, 5581, 5591, 5623, 5639, 5641, 5647, 5651, 5653, 5657, 5659, 5669, 5683, 5689, 5693, 5701, 5711, 5717, 5737, 5741, 5743, 5749, 5779, 5783, 5791, 5801, 5807, 5813, 5821, 5827, 5839, 5843, 5849, 5851, 5857, 5861, 5867, 5869, 5879, 5881, 5897, 5903, 5923, 5927, 5939, 5953, 5981, 5987, 6007, 6011, 6029, 6037, 6043, 6047, 6053, 6067, 6073, 6079, 6089, 6091, 6101, 6113, 6121, 6131, 6133, 6143, 6151, 6163, 6173, 6197, 6199, 6203, 6211, 6217, 6221, 6229, 6247, 6257, 6263, 6269, 6271, 6277, 6287, 6299, 6301, 6311, 6317, 6323, 6329, 6337, 6343, 6353, 6359, 6361, 6367, 6373, 6379, 6389, 6397, 6421, 6427, 6449, 6451, 6469, 6473, 6481, 6491, 6521, 6529, 6547, 6551, 6553, 6563, 6569, 6571, 6577, 6581, 6599, 6607, 6619, 6637, 6653, 6659, 6661, 6673, 6679, 6689, 6691, 6701, 6703, 6709, 6719, 6733, 6737, 6761, 6763, 6779, 6781, 6791, 6793, 6803, 6823, 6827, 6829, 6833, 6841, 6857, 6863, 6869, 6871, 6883, 6899, 6907, 6911, 6917, 6947, 6949, 6959, 6961, 6967, 6971, 6977, 6983, 6991, 6997, 7001, 7013, 7019, 7027, 7039, 7043, 7057, 7069, 7079, 7103, 7109, 7121, 7127, 7129, 7151, 7159, 7177, 7187, 7193, 7207, 7211, 7213, 7219, 7229, 7237, 7243, 7247, 7253, 7283, 7297, 7307, 7309, 7321, 7331, 7333, 7349, 7351, 7369, 7393, 7411, 7417, 7433, 7451, 7457, 7459, 7477, 7481, 7487, 7489, 7499, 7507, 7517, 7523, 7529, 7537, 7541, 7547, 7549, 7559, 7561, 7573, 7577, 7583, 7589, 7591, 7603, 7607, 7621, 7639, 7643, 7649, 7669, 7673, 7681, 7687, 7691, 7699, 7703, 7717, 7723, 7727, 7741, 7753, 7757, 7759, 7789, 7793, 7817, 7823, 7829, 7841, 7853, 7867, 7873, 7877, 7879, 7883, 7901, 7907, 7919, 7927, 7933, 7937, 7949, 7951, 7963, 7993, 8009, 8011, 8017, 8039, 8053, 8059, 8069, 8081, 8087, 8089, 8093, 8101, 8111, 8117, 8123, 8147, 8161, 8167, 8171, 8179, 8191, 8209, 8219, 8221, 8231, 8233, 8237, 8243, 8263, 8269, 8273, 8287, 8291, 8293, 8297, 8311, 8317, 8329, 8353, 8363, 8369, 8377, 8387, 8389, 8419, 8423, 8429, 8431, 8443, 8447, 8461, 8467, 8501, 8513, 8521, 8527, 8537, 8539, 8543, 8563, 8573, 8581, 8597, 8599, 8609, 8623, 8627, 8629, 8641, 8647, 8663, 8669, 8677, 8681, 8689, 8693, 8699, 8707, 8713, 8719, 8731, 8737, 8741, 8747, 8753, 8761, 8779, 8783, 8803, 8807, 8819, 8821, 8831, 8837, 8839, 8849, 8861, 8863, 8867, 8887, 8893, 8923, 8929, 8933, 8941, 8951, 8963, 8969, 8971, 8999, 9001, 9007, 9011, 9013, 9029, 9041, 9043, 9049, 9059, 9067, 9091, 9103, 9109, 9127, 9133, 9137, 9151, 9157, 9161, 9173, 9181, 9187, 9199, 9203, 9209, 9221, 9227, 9239, 9241, 9257, 9277, 9281, 9283, 9293, 9311, 9319, 9323, 9337, 9341, 9343, 9349, 9371, 9377, 9391, 9397, 9403, 9413, 9419, 9421, 9431, 9433, 9437, 9439, 9461, 9463, 9467, 9473, 9479, 9491, 9497, 9511, 9521, 9533, 9539, 9547, 9551, 9587, 9601, 9613, 9619, 9623, 9629, 9631, 9643, 9649, 9661, 9677, 9679, 9689, 9697, 9719, 9721, 9733, 9739, 9743, 9749, 9767, 9769, 9781, 9787, 9791, 9803, 9811, 9817, 9829, 9833, 9839, 9851, 9857, 9859, 9871, 9883, 9887, 9901, 9907, 9923, 9929, 9931, 9941, 9949, 9967, 9973]
N = int(input())
a = []
for i in range(N):
a.append([])
for j in range(N):
if (i + j) % 2 == 0:
a[i].append(Primes[250 + (i - j) // 2] * Primes[500 + (i + j) // 2])
else:
a[i].append(0)
for i in range(N):
for j in range(N):
if (i + j) % 2 != 0:
mul = 1
if i > 0:
mul = mul * a[i-1][j] // gcd(mul,a[i-1][j])
if j > 0:
mul = mul * a[i][j-1] // gcd(mul,a[i][j-1])
if i < N-1:
mul = mul * a[i+1][j] // gcd(mul,a[i+1][j])
if j < N - 1:
mul = mul * a[i][j+1] // gcd(mul,a[i][j+1])
mul += 1
a[i][j] = mul
if N == 2:
print("4 7\n23 10")
else:
for row in a:
print(" ".join(map(str,row)))
``` | output | 1 | 39,450 | 22 | 78,901 |
Provide a correct Python 3 solution for this coding contest problem.
You are given an integer N.
Construct any one N-by-N matrix a that satisfies the conditions below. It can be proved that a solution always exists under the constraints of this problem.
* 1 \leq a_{i,j} \leq 10^{15}
* a_{i,j} are pairwise distinct integers.
* There exists a positive integer m such that the following holds: Let x and y be two elements of the matrix that are vertically or horizontally adjacent. Then, {\rm max}(x,y) {\rm mod} {\rm min}(x,y) is always m.
Constraints
* 2 \leq N \leq 500
Input
Input is given from Standard Input in the following format:
N
Output
Print your solution in the following format:
a_{1,1} ... a_{1,N}
:
a_{N,1} ... a_{N,N}
Output
Print your solution in the following format:
a_{1,1} ... a_{1,N}
:
a_{N,1} ... a_{N,N}
Example
Input
2
Output
4 7
23 10 | instruction | 0 | 39,451 | 22 | 78,902 |
"Correct Solution:
```
#!/usr/bin/python3
import random, math, fractions
def lcm(a, b):
return (a // fractions.gcd(a, b)) * b
def is_prime(n):
i = 2
while i * i <= n:
if n % i == 0:
return False
i += 1
return True
def primes(n):
lst = []
for i in range(2, n + 1):
if is_prime(i):
lst.append(i)
return lst
n = int(input())
a = [[None for j in range(n)] for i in range(n)]
lst = [(i,j) for i in range(n) for j in range(n - 1, -1, -1)]
used = set()
pr = primes(10 ** 4)
off = 0
while pr[off] < n + 3:
off += 1
for (i, j) in lst:
if (i + j) % 2 == 0:
A = (i + j) // 2
B = n + (i - j) // 2
a[i][j] = (A + 1) * (pr[off + B])
if a[i][j] in used:
print(101, a[i][j])
raise ValueError()
used.add(a[i][j])
for (i, j) in lst:
if (i + j) % 2 == 0:
continue
val = 1
for (ip, jp) in [(i - 1, j), (i + 1, j), (i, j - 1), (i, j + 1)]:
if 0 <= ip < n and 0 <= jp < n and a[ip][jp] != None:
val = lcm(val, a[ip][jp])
if val >= 5 * 10 ** 14:
print("ERR", val)
raise ValueError()
res = val + 1
while res in used:
k = random.randint(1, (10 ** 15 - 1) // val)
res = k * val + 1
a[i][j] = res
used.add(res)
for i in range(n):
print(" ".join(map(str, a[i])))
``` | output | 1 | 39,451 | 22 | 78,903 |
Provide a correct Python 3 solution for this coding contest problem.
You are given an integer N.
Construct any one N-by-N matrix a that satisfies the conditions below. It can be proved that a solution always exists under the constraints of this problem.
* 1 \leq a_{i,j} \leq 10^{15}
* a_{i,j} are pairwise distinct integers.
* There exists a positive integer m such that the following holds: Let x and y be two elements of the matrix that are vertically or horizontally adjacent. Then, {\rm max}(x,y) {\rm mod} {\rm min}(x,y) is always m.
Constraints
* 2 \leq N \leq 500
Input
Input is given from Standard Input in the following format:
N
Output
Print your solution in the following format:
a_{1,1} ... a_{1,N}
:
a_{N,1} ... a_{N,N}
Output
Print your solution in the following format:
a_{1,1} ... a_{1,N}
:
a_{N,1} ... a_{N,N}
Example
Input
2
Output
4 7
23 10 | instruction | 0 | 39,452 | 22 | 78,904 |
"Correct Solution:
```
from fractions import gcd
from itertools import chain
def eratosthenes_generator():
yield 2
n = 3
h = {}
while True:
m = n
if n in h:
b = h[n]
m += 2 * b
while m in h:
m += 2 * b
h[m] = b
del h[n]
else:
m += 2 * n
while m in h:
m += 2 * n
h[m] = n
yield n
n += 2
def solve(n):
if n == 2:
return [[2, 3], [5, 4]]
ans = [[0] * n for _ in range(n)]
eg = eratosthenes_generator()
primes = [next(eg) for _ in range(2 * n)]
primes = list(chain.from_iterable(zip(primes, primes[::-1])))
ps = primes[:n]
qs = primes[n:2 * n]
m = (n - 1) // 2
for i in range(n):
for j in range(i % 2, n, 2):
l1 = ps[(i + j) // 2 - m]
l2 = qs[(i - j) // 2]
ans[i][j] = l1 * l2
for i in range(n):
for j in range((i % 2) ^ 1, n, 2):
surroundings = []
if i > 0:
surroundings.append(ans[i - 1][j])
if i < n - 1:
surroundings.append(ans[i + 1][j])
if j > 0:
surroundings.append(ans[i][j - 1])
if j < n - 1:
surroundings.append(ans[i][j + 1])
lcm = surroundings[0]
for s in surroundings[1:]:
lcm = lcm * s // gcd(lcm, s)
ans[i][j] = lcm + 1
return ans
def check(ans):
for i in range(n):
for j in range(n):
x = ans[i][j]
assert x <= 1e15, x
if i > 0:
y = ans[i - 1][j]
assert max(x, y) % min(x, y) == 1
if i < n - 1:
y = ans[i + 1][j]
assert max(x, y) % min(x, y) == 1
if j > 0:
y = ans[i][j - 1]
assert max(x, y) % min(x, y) == 1
if j < n - 1:
y = ans[i][j + 1]
assert max(x, y) % min(x, y) == 1
print('OK')
n = int(input())
ans = solve(n)
# check(ans)
print('\n'.join(' '.join(map(str, row)) for row in ans))
``` | output | 1 | 39,452 | 22 | 78,905 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp knows that if the sum of the digits of a number is divisible by 3, then the number itself is divisible by 3. He assumes that the numbers, the sum of the digits of which is divisible by 4, are also somewhat interesting. Thus, he considers a positive integer n interesting if its sum of digits is divisible by 4.
Help Polycarp find the nearest larger or equal interesting number for the given number a. That is, find the interesting number n such that n ≥ a and n is minimal.
Input
The only line in the input contains an integer a (1 ≤ a ≤ 1000).
Output
Print the nearest greater or equal interesting number for the given number a. In other words, print the interesting number n such that n ≥ a and n is minimal.
Examples
Input
432
Output
435
Input
99
Output
103
Input
237
Output
237
Input
42
Output
44 | instruction | 0 | 39,672 | 22 | 79,344 |
Tags: implementation
Correct Solution:
```
a = int(input())
def sum(n):
s = 0
while n > 0:
s += n % 10
n //= 10
return s
while (sum(a) % 4 != 0):
a += 1
print(a)
``` | output | 1 | 39,672 | 22 | 79,345 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp knows that if the sum of the digits of a number is divisible by 3, then the number itself is divisible by 3. He assumes that the numbers, the sum of the digits of which is divisible by 4, are also somewhat interesting. Thus, he considers a positive integer n interesting if its sum of digits is divisible by 4.
Help Polycarp find the nearest larger or equal interesting number for the given number a. That is, find the interesting number n such that n ≥ a and n is minimal.
Input
The only line in the input contains an integer a (1 ≤ a ≤ 1000).
Output
Print the nearest greater or equal interesting number for the given number a. In other words, print the interesting number n such that n ≥ a and n is minimal.
Examples
Input
432
Output
435
Input
99
Output
103
Input
237
Output
237
Input
42
Output
44 | instruction | 0 | 39,674 | 22 | 79,348 |
Tags: implementation
Correct Solution:
```
a=int(input())
for i in range(a,10000000000000000) :
sum=0
for j in range (len(str(a))):
sum=sum+int(str(a)[j])
if sum%4==0 :
print(a)
break
else :
a=a+1
``` | output | 1 | 39,674 | 22 | 79,349 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp knows that if the sum of the digits of a number is divisible by 3, then the number itself is divisible by 3. He assumes that the numbers, the sum of the digits of which is divisible by 4, are also somewhat interesting. Thus, he considers a positive integer n interesting if its sum of digits is divisible by 4.
Help Polycarp find the nearest larger or equal interesting number for the given number a. That is, find the interesting number n such that n ≥ a and n is minimal.
Input
The only line in the input contains an integer a (1 ≤ a ≤ 1000).
Output
Print the nearest greater or equal interesting number for the given number a. In other words, print the interesting number n such that n ≥ a and n is minimal.
Examples
Input
432
Output
435
Input
99
Output
103
Input
237
Output
237
Input
42
Output
44 | instruction | 0 | 39,675 | 22 | 79,350 |
Tags: implementation
Correct Solution:
```
'''test, p = map(int, input().split())
for _ in range(test):
l, r = map(int, input().split())
ans = 0
if r%2 == 1:
ans += (pow(3, r//2+1, p)-1)//2
else:
ans += (pow(3, r//2, p)-1)//2
ans -= (pow(3, l//2, p)-1)//2
'''
a = int(input())
while True:
temp = a; s=0
while temp != 0:
s += temp%10
temp //= 10
if s%4 == 0:
break
a += 1
print(a)
``` | output | 1 | 39,675 | 22 | 79,351 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp knows that if the sum of the digits of a number is divisible by 3, then the number itself is divisible by 3. He assumes that the numbers, the sum of the digits of which is divisible by 4, are also somewhat interesting. Thus, he considers a positive integer n interesting if its sum of digits is divisible by 4.
Help Polycarp find the nearest larger or equal interesting number for the given number a. That is, find the interesting number n such that n ≥ a and n is minimal.
Input
The only line in the input contains an integer a (1 ≤ a ≤ 1000).
Output
Print the nearest greater or equal interesting number for the given number a. In other words, print the interesting number n such that n ≥ a and n is minimal.
Examples
Input
432
Output
435
Input
99
Output
103
Input
237
Output
237
Input
42
Output
44 | instruction | 0 | 39,677 | 22 | 79,354 |
Tags: implementation
Correct Solution:
```
a=int(input())
ma=a
for i in range(a,a+1000):
x=list(str(i))
x=list(map(int,x))
if sum(x)%4==0:
ma=i
break
print(ma)
``` | output | 1 | 39,677 | 22 | 79,355 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp knows that if the sum of the digits of a number is divisible by 3, then the number itself is divisible by 3. He assumes that the numbers, the sum of the digits of which is divisible by 4, are also somewhat interesting. Thus, he considers a positive integer n interesting if its sum of digits is divisible by 4.
Help Polycarp find the nearest larger or equal interesting number for the given number a. That is, find the interesting number n such that n ≥ a and n is minimal.
Input
The only line in the input contains an integer a (1 ≤ a ≤ 1000).
Output
Print the nearest greater or equal interesting number for the given number a. In other words, print the interesting number n such that n ≥ a and n is minimal.
Examples
Input
432
Output
435
Input
99
Output
103
Input
237
Output
237
Input
42
Output
44 | instruction | 0 | 39,678 | 22 | 79,356 |
Tags: implementation
Correct Solution:
```
def sum_of_digits(n):
sum = 0
while n > 0:
sum += n % 10
n //= 10
return sum
n = int(input())
while sum_of_digits(n) % 4 != 0 :
n += 1
print(n)
``` | output | 1 | 39,678 | 22 | 79,357 |
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