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.
Ayoub had an array a of integers of size n and this array had two interesting properties:
* All the integers in the array were between l and r (inclusive).
* The sum of all the elements was divisible by 3.
Unfortunately, Ayoub has lost his array, but he remembers the size of the array n and the numbers l and r, so he asked you to find the number of ways to restore the array.
Since the answer could be very large, print it modulo 10^9 + 7 (i.e. the remainder when dividing by 10^9 + 7). In case there are no satisfying arrays (Ayoub has a wrong memory), print 0.
Input
The first and only line contains three integers n, l and r (1 ≤ n ≤ 2 ⋅ 10^5 , 1 ≤ l ≤ r ≤ 10^9) — the size of the lost array and the range of numbers in the array.
Output
Print the remainder when dividing by 10^9 + 7 the number of ways to restore the array.
Examples
Input
2 1 3
Output
3
Input
3 2 2
Output
1
Input
9 9 99
Output
711426616
Note
In the first example, the possible arrays are : [1,2], [2,1], [3, 3].
In the second example, the only possible array is [2, 2, 2].
Submitted Solution:
```
n, l, r=map(int, input().split())
slf={}
slf[0]=0
slf[1]=0
slf[2]=0
for i in range(l, r+1):
slf[i%3]+=1
a=slf[0]
b=slf[1]
c=slf[2]
print(a, b, c)
m=[]
m.append(a)
m.append(a*a+b*c+c*b)
m.append(a*a*a+(a*b*c)*6+b*b*b+c*c*c)
for i in range(3, n):
m.append(m[i-1]*a+m[i-2]*(b*c+c*b)+m[i-3]*(c**3+b**3))
print(m[n-1]%(10**9+7))
``` | instruction | 0 | 66,488 | 22 | 132,976 |
No | output | 1 | 66,488 | 22 | 132,977 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ayoub had an array a of integers of size n and this array had two interesting properties:
* All the integers in the array were between l and r (inclusive).
* The sum of all the elements was divisible by 3.
Unfortunately, Ayoub has lost his array, but he remembers the size of the array n and the numbers l and r, so he asked you to find the number of ways to restore the array.
Since the answer could be very large, print it modulo 10^9 + 7 (i.e. the remainder when dividing by 10^9 + 7). In case there are no satisfying arrays (Ayoub has a wrong memory), print 0.
Input
The first and only line contains three integers n, l and r (1 ≤ n ≤ 2 ⋅ 10^5 , 1 ≤ l ≤ r ≤ 10^9) — the size of the lost array and the range of numbers in the array.
Output
Print the remainder when dividing by 10^9 + 7 the number of ways to restore the array.
Examples
Input
2 1 3
Output
3
Input
3 2 2
Output
1
Input
9 9 99
Output
711426616
Note
In the first example, the possible arrays are : [1,2], [2,1], [3, 3].
In the second example, the only possible array is [2, 2, 2].
Submitted Solution:
```
n, L, R = map(int, input().split())
q = 10 ** 9 + 7
u0 = [0] * n
u1 = [0] * n
u2 = [0] * n
k = (R - L + 1) // 3
c0 = k; c1 = k; c2 = k
if (R - L + 1) % 3 != 0:
if R % 3 == 0: c0 += 1
elif R % 3 == 1: c1 += 1
else: c2 += 1
if (R - L + 1) % 3 == 0:
if L % 3 == 0: c0 += 1
elif L % 3 == 1: c1 += 1
else: c2 += 1
u0[0] = c0; u1[0] = c1; u2[0] = c2
##print(u0)
##print(u1)
##print(u2)
for i in range(1, n):
u0[i] = (u0[i-1] * c0 % q + u1[i-1] * c2 % q + u2[i-1] * c1 % q) % q
u1[i] = (u0[i-1] * c1 % q + u1[i-1] * c0 % q + u2[i-1] * c2 % q) % q
u2[i] = (u0[i-1] * c2 % q + u1[i-1] * c1 % q + u2[i-1] * c0 % q) % q
## print(u0)
## print(u1)
## print(u2)
print(u0[n - 1])
``` | instruction | 0 | 66,489 | 22 | 132,978 |
No | output | 1 | 66,489 | 22 | 132,979 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ayoub had an array a of integers of size n and this array had two interesting properties:
* All the integers in the array were between l and r (inclusive).
* The sum of all the elements was divisible by 3.
Unfortunately, Ayoub has lost his array, but he remembers the size of the array n and the numbers l and r, so he asked you to find the number of ways to restore the array.
Since the answer could be very large, print it modulo 10^9 + 7 (i.e. the remainder when dividing by 10^9 + 7). In case there are no satisfying arrays (Ayoub has a wrong memory), print 0.
Input
The first and only line contains three integers n, l and r (1 ≤ n ≤ 2 ⋅ 10^5 , 1 ≤ l ≤ r ≤ 10^9) — the size of the lost array and the range of numbers in the array.
Output
Print the remainder when dividing by 10^9 + 7 the number of ways to restore the array.
Examples
Input
2 1 3
Output
3
Input
3 2 2
Output
1
Input
9 9 99
Output
711426616
Note
In the first example, the possible arrays are : [1,2], [2,1], [3, 3].
In the second example, the only possible array is [2, 2, 2].
Submitted Solution:
```
from math import ceil
n,a,b=list(map(int,input().split()))
d=[[0]*n for i in range(3)]
s=b-a+1
mod=10**9+7
t=s%3
x,y,z=[ceil(s/3)]*t+[s//3]*(3-t)
d[0][0],d[1][0],d[2][0]=x,y,z
for i in range(1,n):
d[0][i]=(d[0][i-1]*x+(d[1][i-1]*z)+(d[2][i-1]*y))%mod
d[1][i]=((d[0][i-1]*y)+(d[1][i-1]*x)+(d[2][i-1]*z))%mod
d[2][i]=((d[0][i-1]*z)+(d[1][i-1]*y)+(d[2][i-1]*x))%mod
print(d[0][n-1]%mod)
``` | instruction | 0 | 66,490 | 22 | 132,980 |
No | output | 1 | 66,490 | 22 | 132,981 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ayoub had an array a of integers of size n and this array had two interesting properties:
* All the integers in the array were between l and r (inclusive).
* The sum of all the elements was divisible by 3.
Unfortunately, Ayoub has lost his array, but he remembers the size of the array n and the numbers l and r, so he asked you to find the number of ways to restore the array.
Since the answer could be very large, print it modulo 10^9 + 7 (i.e. the remainder when dividing by 10^9 + 7). In case there are no satisfying arrays (Ayoub has a wrong memory), print 0.
Input
The first and only line contains three integers n, l and r (1 ≤ n ≤ 2 ⋅ 10^5 , 1 ≤ l ≤ r ≤ 10^9) — the size of the lost array and the range of numbers in the array.
Output
Print the remainder when dividing by 10^9 + 7 the number of ways to restore the array.
Examples
Input
2 1 3
Output
3
Input
3 2 2
Output
1
Input
9 9 99
Output
711426616
Note
In the first example, the possible arrays are : [1,2], [2,1], [3, 3].
In the second example, the only possible array is [2, 2, 2].
Submitted Solution:
```
######### ## ## ## #### ##### ## # ## # ##
# # # # # # # # # # # # # # # # # # #
# # # # ### # # # # # # # # # # # #
# ##### # # # # ### # # # # # # # # #####
# # # # # # # # # # # # # # # # # #
######### # # # # ##### # ##### # ## # ## # #
"""
PPPPPPP RRRRRRR OOOO VV VV EEEEEEEEEE
PPPPPPPP RRRRRRRR OOOOOO VV VV EE
PPPPPPPPP RRRRRRRRR OOOOOOOO VV VV EE
PPPPPPPP RRRRRRRR OOOOOOOO VV VV EEEEEE
PPPPPPP RRRRRRR OOOOOOOO VV VV EEEEEEE
PP RRRR OOOOOOOO VV VV EEEEEE
PP RR RR OOOOOOOO VV VV EE
PP RR RR OOOOOO VV VV EE
PP RR RR OOOO VVVV EEEEEEEEEE
"""
"""
Perfection is achieved not when there is nothing more to add, but rather when there is nothing more to take away.
"""
import sys
input = sys.stdin.readline
read = lambda: map(int, input().split())
# from bisect import bisect_left as lower_bound;
# from bisect import bisect_right as upper_bound;
# from math import ceil, factorial;
def ceil(x):
if x != int(x):
x = int(x) + 1
return x
def factorial(x, m):
val = 1
while x>0:
val = (val * x) % m
x -= 1
return val
def fact(x):
val = 1
while x > 0:
val *= x
x -= 1
return val
# swap_array function
def swaparr(arr, a,b):
temp = arr[a];
arr[a] = arr[b];
arr[b] = temp;
## gcd function
def gcd(a,b):
if b == 0:
return a;
return gcd(b, a % b);
## lcm function
def lcm(a, b):
return (a * b) // gcd(a, b)
## nCr function efficient using Binomial Cofficient
def nCr(n, k):
if k > n:
return 0
if(k > n - k):
k = n - k
res = 1
for i in range(k):
res = res * (n - i)
res = res / (i + 1)
return int(res)
## upper bound function code -- such that e in a[:i] e < x;
def upper_bound(a, x, lo=0, hi = None):
if hi == None:
hi = len(a);
while lo < hi:
mid = (lo+hi)//2;
if a[mid] < x:
lo = mid+1;
else:
hi = mid;
return lo;
## prime factorization
def primefs(n):
## if n == 1 ## calculating primes
primes = {}
while(n%2 == 0 and n > 0):
primes[2] = primes.get(2, 0) + 1
n = n//2
for i in range(3, int(n**0.5)+2, 2):
while(n%i == 0 and n > 0):
primes[i] = primes.get(i, 0) + 1
n = n//i
if n > 2:
primes[n] = primes.get(n, 0) + 1
## prime factoriazation of n is stored in dictionary
## primes and can be accesed. O(sqrt n)
return primes
## MODULAR EXPONENTIATION FUNCTION
def power(x, y, p):
res = 1
x = x % p
if (x == 0) :
return 0
while (y > 0) :
if ((y & 1) == 1) :
res = (res * x) % p
y = y >> 1
x = (x * x) % p
return res
## DISJOINT SET UNINON FUNCTIONS
def swap(a,b):
temp = a
a = b
b = temp
return a,b;
# find function with path compression included (recursive)
# def find(x, link):
# if link[x] == x:
# return x
# link[x] = find(link[x], link);
# return link[x];
# find function with path compression (ITERATIVE)
def find(x, link):
p = x;
while( p != link[p]):
p = link[p];
while( x != p):
nex = link[x];
link[x] = p;
x = nex;
return p;
# the union function which makes union(x,y)
# of two nodes x and y
def union(x, y, link, size):
x = find(x, link)
y = find(y, link)
if size[x] < size[y]:
x,y = swap(x,y)
if x != y:
size[x] += size[y]
link[y] = x
## returns an array of boolean if primes or not USING SIEVE OF ERATOSTHANES
def sieve(n):
prime = [True for i in range(n+1)]
prime[0], prime[1] = False, False
p = 2
while (p * p <= n):
if (prime[p] == True):
for i in range(p * p, n+1, p):
prime[i] = False
p += 1
return prime
# Euler's Toitent Function phi
def phi(n) :
result = n
p = 2
while(p * p<= n) :
if (n % p == 0) :
while (n % p == 0) :
n = n // p
result = result * (1.0 - (1.0 / (float) (p)))
p = p + 1
if (n > 1) :
result = result * (1.0 - (1.0 / (float)(n)))
return (int)(result)
def is_prime(n):
if n == 0:
return False
if n == 1:
return True
for i in range(2, int(n ** (1 / 2)) + 1):
if not n % i:
return False
return True
#### PRIME FACTORIZATION IN O(log n) using Sieve ####
MAXN = int(1e5 + 5)
def spf_sieve():
spf[1] = 1;
for i in range(2, MAXN):
spf[i] = i;
for i in range(4, MAXN, 2):
spf[i] = 2;
for i in range(3, ceil(MAXN ** 0.5), 2):
if spf[i] == i:
for j in range(i*i, MAXN, i):
if spf[j] == j:
spf[j] = i;
## function for storing smallest prime factors (spf) in the array
################## un-comment below 2 lines when using factorization #################
spf = [0 for i in range(MAXN)]
# spf_sieve();
def factoriazation(x):
res = []
for i in range(2, int(x ** 0.5) + 1):
while x % i == 0:
res.append(i)
x //= i
if x != 1:
res.append(x)
return res
## this function is useful for multiple queries only, o/w use
## primefs function above. complexity O(log n)
def factors(n):
res = []
for i in range(1, int(n ** 0.5) + 1):
if n % i == 0:
res.append(i)
res.append(n // i)
return list(set(res))
## taking integer array input
def int_array():
return list(map(int, input().strip().split()));
def float_array():
return list(map(float, input().strip().split()));
## taking string array input
def str_array():
return input().strip().split();
#defining a couple constants
MOD = int(1e9)+7;
CMOD = 998244353;
INF = float('inf'); NINF = -float('inf');
################### ---------------- TEMPLATE ENDS HERE ---------------- ###################
from itertools import permutations
import math
from bisect import bisect_left
def solve():
n, l, r = read()
mod0 = r // 3 - math.ceil(l / 3) + 1
mod1 = (r - 1) // 3 - math.ceil((l - 1) / 3) + 1
mod2 = (r - 2) // 3 - math.ceil((l - 2) / 3) + 1
dp = [mod0, mod1, mod2]
for i in range(n - 1):
d = []
d.append((dp[0] * mod0 + dp[1] * mod1 + dp[2] * mod2) % MOD)
d.append((dp[0] * mod1 + dp[1] * mod0 + dp[2] * mod2) % MOD)
d.append((dp[0] * mod2 + dp[1] * mod1 + dp[2] * mod0) % MOD)
dp = d
print(dp[0] % MOD)
if __name__ == '__main__':
for _ in range(1):
solve()
# fin_time = datetime.now()
# print("Execution time (for loop): ", (fin_time-init_time))
``` | instruction | 0 | 66,491 | 22 | 132,982 |
No | output | 1 | 66,491 | 22 | 132,983 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an integer n and an integer k.
In one step you can do one of the following moves:
* decrease n by 1;
* divide n by k if n is divisible by k.
For example, if n = 27 and k = 3 you can do the following steps: 27 → 26 → 25 → 24 → 8 → 7 → 6 → 2 → 1 → 0.
You are asked to calculate the minimum number of steps to reach 0 from n.
Input
The first line contains one integer t (1 ≤ t ≤ 100) — the number of queries.
The only line of each query contains two integers n and k (1 ≤ n ≤ 10^{18}, 2 ≤ k ≤ 10^{18}).
Output
For each query print the minimum number of steps to reach 0 from n in single line.
Example
Input
2
59 3
1000000000000000000 10
Output
8
19
Note
Steps for the first test case are: 59 → 58 → 57 → 19 → 18 → 6 → 2 → 1 → 0.
In the second test case you have to divide n by k 18 times and then decrease n by 1. | instruction | 0 | 66,513 | 22 | 133,026 |
Tags: implementation, math
Correct Solution:
```
t = int(input())
for _ in range(t):
n, k = map(int, input().split())
cnt = 0
while n != 0:
if n % k == 0:
n //= k
cnt += 1
else:
cnt += n % k
n = n - n % k
print(cnt)
``` | output | 1 | 66,513 | 22 | 133,027 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an integer n and an integer k.
In one step you can do one of the following moves:
* decrease n by 1;
* divide n by k if n is divisible by k.
For example, if n = 27 and k = 3 you can do the following steps: 27 → 26 → 25 → 24 → 8 → 7 → 6 → 2 → 1 → 0.
You are asked to calculate the minimum number of steps to reach 0 from n.
Input
The first line contains one integer t (1 ≤ t ≤ 100) — the number of queries.
The only line of each query contains two integers n and k (1 ≤ n ≤ 10^{18}, 2 ≤ k ≤ 10^{18}).
Output
For each query print the minimum number of steps to reach 0 from n in single line.
Example
Input
2
59 3
1000000000000000000 10
Output
8
19
Note
Steps for the first test case are: 59 → 58 → 57 → 19 → 18 → 6 → 2 → 1 → 0.
In the second test case you have to divide n by k 18 times and then decrease n by 1. | instruction | 0 | 66,514 | 22 | 133,028 |
Tags: implementation, math
Correct Solution:
```
q=int(input())
for i in range(q):
t=list(map(int,input().split()))
n=t[0]
k=t[1]
t=int(n%k)
kq=0
while(n!=0):
if(n<k):
kq=kq+n
n=0
else:
if(t==0):
kq=kq+1
n=int(n//k)
else:
kq=kq+t+1
n=int((n-t)//k)
t=int(n%k)
#print(n,kq)
print(kq)
``` | output | 1 | 66,514 | 22 | 133,029 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an integer n and an integer k.
In one step you can do one of the following moves:
* decrease n by 1;
* divide n by k if n is divisible by k.
For example, if n = 27 and k = 3 you can do the following steps: 27 → 26 → 25 → 24 → 8 → 7 → 6 → 2 → 1 → 0.
You are asked to calculate the minimum number of steps to reach 0 from n.
Input
The first line contains one integer t (1 ≤ t ≤ 100) — the number of queries.
The only line of each query contains two integers n and k (1 ≤ n ≤ 10^{18}, 2 ≤ k ≤ 10^{18}).
Output
For each query print the minimum number of steps to reach 0 from n in single line.
Example
Input
2
59 3
1000000000000000000 10
Output
8
19
Note
Steps for the first test case are: 59 → 58 → 57 → 19 → 18 → 6 → 2 → 1 → 0.
In the second test case you have to divide n by k 18 times and then decrease n by 1. | instruction | 0 | 66,515 | 22 | 133,030 |
Tags: implementation, math
Correct Solution:
```
n = int(input())
for i in range(n):
a,b = map(int,input().split())
t = 0
while a>0:
if a%b==0:
a=a//b
t+=1
else:
t+=a%b
a= a - a%b
print(t)
``` | output | 1 | 66,515 | 22 | 133,031 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an integer n and an integer k.
In one step you can do one of the following moves:
* decrease n by 1;
* divide n by k if n is divisible by k.
For example, if n = 27 and k = 3 you can do the following steps: 27 → 26 → 25 → 24 → 8 → 7 → 6 → 2 → 1 → 0.
You are asked to calculate the minimum number of steps to reach 0 from n.
Input
The first line contains one integer t (1 ≤ t ≤ 100) — the number of queries.
The only line of each query contains two integers n and k (1 ≤ n ≤ 10^{18}, 2 ≤ k ≤ 10^{18}).
Output
For each query print the minimum number of steps to reach 0 from n in single line.
Example
Input
2
59 3
1000000000000000000 10
Output
8
19
Note
Steps for the first test case are: 59 → 58 → 57 → 19 → 18 → 6 → 2 → 1 → 0.
In the second test case you have to divide n by k 18 times and then decrease n by 1. | instruction | 0 | 66,516 | 22 | 133,032 |
Tags: implementation, math
Correct Solution:
```
t = int(input())
cnt = 0
while t > 0:
cnt = 0
n, k = map(int, input().split())
while n > 0:
if n % k == 0:
n //= k
cnt += 1
else:
cnt += (n % k)
n -= (n % k)
t -= 1
print(cnt)
``` | output | 1 | 66,516 | 22 | 133,033 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an integer n and an integer k.
In one step you can do one of the following moves:
* decrease n by 1;
* divide n by k if n is divisible by k.
For example, if n = 27 and k = 3 you can do the following steps: 27 → 26 → 25 → 24 → 8 → 7 → 6 → 2 → 1 → 0.
You are asked to calculate the minimum number of steps to reach 0 from n.
Input
The first line contains one integer t (1 ≤ t ≤ 100) — the number of queries.
The only line of each query contains two integers n and k (1 ≤ n ≤ 10^{18}, 2 ≤ k ≤ 10^{18}).
Output
For each query print the minimum number of steps to reach 0 from n in single line.
Example
Input
2
59 3
1000000000000000000 10
Output
8
19
Note
Steps for the first test case are: 59 → 58 → 57 → 19 → 18 → 6 → 2 → 1 → 0.
In the second test case you have to divide n by k 18 times and then decrease n by 1. | instruction | 0 | 66,517 | 22 | 133,034 |
Tags: implementation, math
Correct Solution:
```
import math
def func(n, k):
s, e = 0, int(1e19)
while s <= e:
m = (s + e) // 2
#print(s, e, m, k * m)
#print(k*m ," and ",{n})
if k * m > n:
e = m - 1
else:
s = m + 1
#print(s, e, k*s, k*e)
return k * e
t = int(input())
for i in range(t):
n, k = map(int, input().split())
ans = 0
while n > 0:
f = func(n, k)
#print(n, f)
if f == 0:
ans += n
else:
ans += n - f + 1
n = f // k
print(ans)
``` | output | 1 | 66,517 | 22 | 133,035 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an integer n and an integer k.
In one step you can do one of the following moves:
* decrease n by 1;
* divide n by k if n is divisible by k.
For example, if n = 27 and k = 3 you can do the following steps: 27 → 26 → 25 → 24 → 8 → 7 → 6 → 2 → 1 → 0.
You are asked to calculate the minimum number of steps to reach 0 from n.
Input
The first line contains one integer t (1 ≤ t ≤ 100) — the number of queries.
The only line of each query contains two integers n and k (1 ≤ n ≤ 10^{18}, 2 ≤ k ≤ 10^{18}).
Output
For each query print the minimum number of steps to reach 0 from n in single line.
Example
Input
2
59 3
1000000000000000000 10
Output
8
19
Note
Steps for the first test case are: 59 → 58 → 57 → 19 → 18 → 6 → 2 → 1 → 0.
In the second test case you have to divide n by k 18 times and then decrease n by 1. | instruction | 0 | 66,518 | 22 | 133,036 |
Tags: implementation, math
Correct Solution:
```
def xdd(n: int, k: int):
i = 0
while n != 0:
i += n%k
i += 1
n = n//k
return i - 1
c = int(input())
A = []
i = 0
for xd in range(c):
a, b = input().split()
A.append([int(a), int(b)])
q = 0
for xd in range(c):
h = A[q][0]
j = A[q][1]
print(xdd(h, j))
q += 1
``` | output | 1 | 66,518 | 22 | 133,037 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an integer n and an integer k.
In one step you can do one of the following moves:
* decrease n by 1;
* divide n by k if n is divisible by k.
For example, if n = 27 and k = 3 you can do the following steps: 27 → 26 → 25 → 24 → 8 → 7 → 6 → 2 → 1 → 0.
You are asked to calculate the minimum number of steps to reach 0 from n.
Input
The first line contains one integer t (1 ≤ t ≤ 100) — the number of queries.
The only line of each query contains two integers n and k (1 ≤ n ≤ 10^{18}, 2 ≤ k ≤ 10^{18}).
Output
For each query print the minimum number of steps to reach 0 from n in single line.
Example
Input
2
59 3
1000000000000000000 10
Output
8
19
Note
Steps for the first test case are: 59 → 58 → 57 → 19 → 18 → 6 → 2 → 1 → 0.
In the second test case you have to divide n by k 18 times and then decrease n by 1. | instruction | 0 | 66,519 | 22 | 133,038 |
Tags: implementation, math
Correct Solution:
```
t = int(input())
for i in range(t):
s = 0
n, k = map(int, input().split())
while n > 0:
s += n % k + 1
n = n // k
print(s - 1)
``` | output | 1 | 66,519 | 22 | 133,039 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an integer n and an integer k.
In one step you can do one of the following moves:
* decrease n by 1;
* divide n by k if n is divisible by k.
For example, if n = 27 and k = 3 you can do the following steps: 27 → 26 → 25 → 24 → 8 → 7 → 6 → 2 → 1 → 0.
You are asked to calculate the minimum number of steps to reach 0 from n.
Input
The first line contains one integer t (1 ≤ t ≤ 100) — the number of queries.
The only line of each query contains two integers n and k (1 ≤ n ≤ 10^{18}, 2 ≤ k ≤ 10^{18}).
Output
For each query print the minimum number of steps to reach 0 from n in single line.
Example
Input
2
59 3
1000000000000000000 10
Output
8
19
Note
Steps for the first test case are: 59 → 58 → 57 → 19 → 18 → 6 → 2 → 1 → 0.
In the second test case you have to divide n by k 18 times and then decrease n by 1. | instruction | 0 | 66,520 | 22 | 133,040 |
Tags: implementation, math
Correct Solution:
```
t = int(input())
for _ in range(t):
n, k = map(int, (input().split()))
sol = 0
aux = n
while aux != 0:
if aux%k == 0:
aux //= k
sol += 1
else:
sol += aux%k
aux -= (aux%k)
print(sol)
``` | output | 1 | 66,520 | 22 | 133,041 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya and Petya are playing a simple game. Vasya thought of number x between 1 and n, and Petya tries to guess the number.
Petya can ask questions like: "Is the unknown number divisible by number y?".
The game is played by the following rules: first Petya asks all the questions that interest him (also, he can ask no questions), and then Vasya responds to each question with a 'yes' or a 'no'. After receiving all the answers Petya should determine the number that Vasya thought of.
Unfortunately, Petya is not familiar with the number theory. Help him find the minimum number of questions he should ask to make a guaranteed guess of Vasya's number, and the numbers yi, he should ask the questions about.
Input
A single line contains number n (1 ≤ n ≤ 103).
Output
Print the length of the sequence of questions k (0 ≤ k ≤ n), followed by k numbers — the questions yi (1 ≤ yi ≤ n).
If there are several correct sequences of questions of the minimum length, you are allowed to print any of them.
Examples
Input
4
Output
3
2 4 3
Input
6
Output
4
2 4 3 5
Note
The sequence from the answer to the first sample test is actually correct.
If the unknown number is not divisible by one of the sequence numbers, it is equal to 1.
If the unknown number is divisible by 4, it is 4.
If the unknown number is divisible by 3, then the unknown number is 3.
Otherwise, it is equal to 2. Therefore, the sequence of questions allows you to guess the unknown number. It can be shown that there is no correct sequence of questions of length 2 or shorter. | instruction | 0 | 66,800 | 22 | 133,600 |
Tags: math, number theory
Correct Solution:
```
## necessary imports
import sys
input = sys.stdin.readline
from math import log2, log, ceil
# swap_array function
def swaparr(arr, a,b):
temp = arr[a];
arr[a] = arr[b];
arr[b] = temp
## gcd function
def gcd(a,b):
if a == 0:
return b
return gcd(b%a, a)
## nCr function efficient using Binomial Cofficient
def nCr(n, k):
if(k > n - k):
k = n - k
res = 1
for i in range(k):
res = res * (n - i)
res = res / (i + 1)
return res
## upper bound function code -- such that e in a[:i] e < x;
def upper_bound(a, x, lo=0):
hi = len(a)
while lo < hi:
mid = (lo+hi)//2
if a[mid] < x:
lo = mid+1
else:
hi = mid
return lo
## prime factorization
def primefs(n):
## if n == 1 ## calculating primes
primes = {}
while(n%2 == 0):
primes[2] = primes.get(2, 0) + 1
n = n//2
for i in range(3, int(n**0.5)+2, 2):
while(n%i == 0):
primes[i] = primes.get(i, 0) + 1
n = n//i
if n > 2:
primes[n] = primes.get(n, 0) + 1
## prime factoriazation of n is stored in dictionary
## primes and can be accesed. O(sqrt n)
return primes
## MODULAR EXPONENTIATION FUNCTION
def power(x, y, p):
res = 1
x = x % p
if (x == 0) :
return 0
while (y > 0) :
if ((y & 1) == 1) :
res = (res * x) % p
y = y >> 1
x = (x * x) % p
return res
## DISJOINT SET UNINON FUNCTIONS
def swap(a,b):
temp = a
a = b
b = temp
return a,b
# find function with path compression included (recursive)
# def find(x, link):
# if link[x] == x:
# return x
# link[x] = find(link[x], link);
# return link[x];
# find function with path compression (ITERATIVE)
def find(x, link):
p = x;
while( p != link[p]):
p = link[p];
while( x != p):
nex = link[x];
link[x] = p;
x = nex;
return p;
# the union function which makes union(x,y)
# of two nodes x and y
def union(x, y, link, size):
x = find(x, link)
y = find(y, link)
if size[x] < size[y]:
x,y = swap(x,y)
if x != y:
size[x] += size[y]
link[y] = x
## returns an array of boolean if primes or not USING SIEVE OF ERATOSTHANES
def sieve(n):
prime = [True for i in range(n+1)]
p = 2
while (p * p <= n):
if (prime[p] == True):
for i in range(p * p, n+1, p):
prime[i] = False
p += 1
return prime
#### PRIME FACTORIZATION IN O(log n) using Sieve ####
MAXN = int(1e6 + 5)
def spf_sieve():
spf[1] = 1;
for i in range(2, MAXN):
spf[i] = i;
for i in range(4, MAXN, 2):
spf[i] = 2;
for i in range(3, ceil(MAXN ** 0.5), 2):
if spf[i] == i:
for j in range(i*i, MAXN, i):
if spf[j] == j:
spf[j] = i;
## function for storing smallest prime factors (spf) in the array
################## un-comment below 2 lines when using factorization #################
# spf = [0 for i in range(MAXN)]
# spf_sieve()
def factoriazation(x):
ret = {};
while x != 1:
ret[spf[x]] = ret.get(spf[x], 0) + 1;
x = x//spf[x]
return ret
## this function is useful for multiple queries only, o/w use
## primefs function above. complexity O(log n)
## taking integer array input
def int_array():
return list(map(int, input().strip().split()))
## taking string array input
def str_array():
return input().strip().split();
#defining a couple constants
MOD = int(1e9)+7;
CMOD = 998244353;
INF = float('inf'); NINF = -float('inf');
################### ---------------- TEMPLATE ENDS HERE ---------------- ###################
def lcm(a,b):
return (a*b)//gcd(a,b);
n = int(input());
if n == 1:
print(0);
else:
mapx = [1]*(n+1);
ans = [2];
for i in range(3, n+1):
for j in ans:
x = lcm(i, j);
if x <= n and x != i:
mapx[x] = -1;
if mapx[i] != -1:
ans.append(i);
print(len(ans));
print(*ans);
``` | output | 1 | 66,800 | 22 | 133,601 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya and Petya are playing a simple game. Vasya thought of number x between 1 and n, and Petya tries to guess the number.
Petya can ask questions like: "Is the unknown number divisible by number y?".
The game is played by the following rules: first Petya asks all the questions that interest him (also, he can ask no questions), and then Vasya responds to each question with a 'yes' or a 'no'. After receiving all the answers Petya should determine the number that Vasya thought of.
Unfortunately, Petya is not familiar with the number theory. Help him find the minimum number of questions he should ask to make a guaranteed guess of Vasya's number, and the numbers yi, he should ask the questions about.
Input
A single line contains number n (1 ≤ n ≤ 103).
Output
Print the length of the sequence of questions k (0 ≤ k ≤ n), followed by k numbers — the questions yi (1 ≤ yi ≤ n).
If there are several correct sequences of questions of the minimum length, you are allowed to print any of them.
Examples
Input
4
Output
3
2 4 3
Input
6
Output
4
2 4 3 5
Note
The sequence from the answer to the first sample test is actually correct.
If the unknown number is not divisible by one of the sequence numbers, it is equal to 1.
If the unknown number is divisible by 4, it is 4.
If the unknown number is divisible by 3, then the unknown number is 3.
Otherwise, it is equal to 2. Therefore, the sequence of questions allows you to guess the unknown number. It can be shown that there is no correct sequence of questions of length 2 or shorter. | instruction | 0 | 66,801 | 22 | 133,602 |
Tags: math, number theory
Correct Solution:
```
from collections import Counter
def sieve(n):
prime = [True for i in range(n + 1)]
p = 2
while (p * p <= n):
if (prime[p] == True):
for i in range(p * 2, n + 1, p):
prime[i] = False
p += 1
prime[0]= False
prime[1]= False
for p in range(n + 1):
if prime[p]:
yield p
def main():
n = int(input())
asked = []
for p in sieve(n):
i = 1
pp = p ** i
while pp <= n:
asked.append(pp)
i += 1
pp = p ** i
print(len(asked))
print(*asked)
if __name__ == "__main__":
main()
``` | output | 1 | 66,801 | 22 | 133,603 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya and Petya are playing a simple game. Vasya thought of number x between 1 and n, and Petya tries to guess the number.
Petya can ask questions like: "Is the unknown number divisible by number y?".
The game is played by the following rules: first Petya asks all the questions that interest him (also, he can ask no questions), and then Vasya responds to each question with a 'yes' or a 'no'. After receiving all the answers Petya should determine the number that Vasya thought of.
Unfortunately, Petya is not familiar with the number theory. Help him find the minimum number of questions he should ask to make a guaranteed guess of Vasya's number, and the numbers yi, he should ask the questions about.
Input
A single line contains number n (1 ≤ n ≤ 103).
Output
Print the length of the sequence of questions k (0 ≤ k ≤ n), followed by k numbers — the questions yi (1 ≤ yi ≤ n).
If there are several correct sequences of questions of the minimum length, you are allowed to print any of them.
Examples
Input
4
Output
3
2 4 3
Input
6
Output
4
2 4 3 5
Note
The sequence from the answer to the first sample test is actually correct.
If the unknown number is not divisible by one of the sequence numbers, it is equal to 1.
If the unknown number is divisible by 4, it is 4.
If the unknown number is divisible by 3, then the unknown number is 3.
Otherwise, it is equal to 2. Therefore, the sequence of questions allows you to guess the unknown number. It can be shown that there is no correct sequence of questions of length 2 or shorter. | instruction | 0 | 66,802 | 22 | 133,604 |
Tags: math, number theory
Correct Solution:
```
n = int(input())
def is_prime(n):
p = 2
while p*p <= n:
if n % p == 0:
return False
p += 1
return True
assert(is_prime(7))
assert(is_prime(47))
assert(is_prime(29))
assert(not is_prime(12))
assert(not is_prime(49))
asks = []
prime = 2
while prime <= n:
k = 1
while prime**k <= n:
asks.append(prime**k)
k += 1
prime += 1
while not is_prime(prime):
prime += 1
print(len(asks))
for e in asks:
print(e, end=" ")
print("")
``` | output | 1 | 66,802 | 22 | 133,605 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya and Petya are playing a simple game. Vasya thought of number x between 1 and n, and Petya tries to guess the number.
Petya can ask questions like: "Is the unknown number divisible by number y?".
The game is played by the following rules: first Petya asks all the questions that interest him (also, he can ask no questions), and then Vasya responds to each question with a 'yes' or a 'no'. After receiving all the answers Petya should determine the number that Vasya thought of.
Unfortunately, Petya is not familiar with the number theory. Help him find the minimum number of questions he should ask to make a guaranteed guess of Vasya's number, and the numbers yi, he should ask the questions about.
Input
A single line contains number n (1 ≤ n ≤ 103).
Output
Print the length of the sequence of questions k (0 ≤ k ≤ n), followed by k numbers — the questions yi (1 ≤ yi ≤ n).
If there are several correct sequences of questions of the minimum length, you are allowed to print any of them.
Examples
Input
4
Output
3
2 4 3
Input
6
Output
4
2 4 3 5
Note
The sequence from the answer to the first sample test is actually correct.
If the unknown number is not divisible by one of the sequence numbers, it is equal to 1.
If the unknown number is divisible by 4, it is 4.
If the unknown number is divisible by 3, then the unknown number is 3.
Otherwise, it is equal to 2. Therefore, the sequence of questions allows you to guess the unknown number. It can be shown that there is no correct sequence of questions of length 2 or shorter. | instruction | 0 | 66,803 | 22 | 133,606 |
Tags: math, number theory
Correct Solution:
```
def is_prime(n):
for i in range(2, n):
if ((n % i) == 0):
return 0
break
return 1
n = int(input())
if (n == 1):
print(0)
else:
q = []
for i in range(2, n + 1):
if (is_prime(i)):
q.append(i)
k = i
while(1):
k = k * i
if (k <= n):
q.append(k)
else:
break
print(len(q))
for i in q:
print(i, end=' ')
``` | output | 1 | 66,803 | 22 | 133,607 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya and Petya are playing a simple game. Vasya thought of number x between 1 and n, and Petya tries to guess the number.
Petya can ask questions like: "Is the unknown number divisible by number y?".
The game is played by the following rules: first Petya asks all the questions that interest him (also, he can ask no questions), and then Vasya responds to each question with a 'yes' or a 'no'. After receiving all the answers Petya should determine the number that Vasya thought of.
Unfortunately, Petya is not familiar with the number theory. Help him find the minimum number of questions he should ask to make a guaranteed guess of Vasya's number, and the numbers yi, he should ask the questions about.
Input
A single line contains number n (1 ≤ n ≤ 103).
Output
Print the length of the sequence of questions k (0 ≤ k ≤ n), followed by k numbers — the questions yi (1 ≤ yi ≤ n).
If there are several correct sequences of questions of the minimum length, you are allowed to print any of them.
Examples
Input
4
Output
3
2 4 3
Input
6
Output
4
2 4 3 5
Note
The sequence from the answer to the first sample test is actually correct.
If the unknown number is not divisible by one of the sequence numbers, it is equal to 1.
If the unknown number is divisible by 4, it is 4.
If the unknown number is divisible by 3, then the unknown number is 3.
Otherwise, it is equal to 2. Therefore, the sequence of questions allows you to guess the unknown number. It can be shown that there is no correct sequence of questions of length 2 or shorter. | instruction | 0 | 66,807 | 22 | 133,614 |
Tags: math, number theory
Correct Solution:
```
def SieveOfEratosthenes(n,ans):
prime = [True for i in range(n+1)]
p = 2
while (p * p <= n):
if (prime[p] == True):
for i in range(p * p, n+1, p):
prime[i] = False
p += 1
for p in range(2, n):
if prime[p]:
ans.append(p)
if __name__=='__main__':
n = int(input())
prime=[]
ans=[]
SieveOfEratosthenes(n+1,prime)
for i in prime:
k=1
pow=i
while(pow<=n):
ans.append(str(pow))
k+=1
pow=i**k
print(len(ans))
print(" ".join(ans))
``` | output | 1 | 66,807 | 22 | 133,615 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya and Petya are playing a simple game. Vasya thought of number x between 1 and n, and Petya tries to guess the number.
Petya can ask questions like: "Is the unknown number divisible by number y?".
The game is played by the following rules: first Petya asks all the questions that interest him (also, he can ask no questions), and then Vasya responds to each question with a 'yes' or a 'no'. After receiving all the answers Petya should determine the number that Vasya thought of.
Unfortunately, Petya is not familiar with the number theory. Help him find the minimum number of questions he should ask to make a guaranteed guess of Vasya's number, and the numbers yi, he should ask the questions about.
Input
A single line contains number n (1 ≤ n ≤ 103).
Output
Print the length of the sequence of questions k (0 ≤ k ≤ n), followed by k numbers — the questions yi (1 ≤ yi ≤ n).
If there are several correct sequences of questions of the minimum length, you are allowed to print any of them.
Examples
Input
4
Output
3
2 4 3
Input
6
Output
4
2 4 3 5
Note
The sequence from the answer to the first sample test is actually correct.
If the unknown number is not divisible by one of the sequence numbers, it is equal to 1.
If the unknown number is divisible by 4, it is 4.
If the unknown number is divisible by 3, then the unknown number is 3.
Otherwise, it is equal to 2. Therefore, the sequence of questions allows you to guess the unknown number. It can be shown that there is no correct sequence of questions of length 2 or shorter.
Submitted Solution:
```
n = int(input())
primes = []
is_prime = (n + 1) * [ True ]
is_prime[0] = is_prime[1] = False
pos = 0
while True:
while pos <= n and not is_prime[pos]:
pos += 1
if pos > n:
break
for i in range(2 * pos, n + 1, pos):
is_prime[i] = False
primes.append(pos)
pos += 1
def get_factors(x):
factors = []
for factor in primes:
if x % factor == 0:
factors.append(factor)
return factors
ask = []
for x in range(2, n + 1):
factors = get_factors(x)
if len(factors) == 1:
ask.append(x)
print(len(ask))
print(' '.join(map(str, ask)))
``` | instruction | 0 | 66,810 | 22 | 133,620 |
Yes | output | 1 | 66,810 | 22 | 133,621 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are well-known formulas: <image>, <image>, <image>. Also mathematicians found similar formulas for higher degrees.
Find the value of the sum <image> modulo 109 + 7 (so you should find the remainder after dividing the answer by the value 109 + 7).
Input
The only line contains two integers n, k (1 ≤ n ≤ 109, 0 ≤ k ≤ 106).
Output
Print the only integer a — the remainder after dividing the value of the sum by the value 109 + 7.
Examples
Input
4 1
Output
10
Input
4 2
Output
30
Input
4 3
Output
100
Input
4 0
Output
4
Submitted Solution:
```
n,k=map(int,input().split())
somme=0
for i in range(1,n+1):
somme+=i**k
print("{}".format(somme))
``` | instruction | 0 | 67,718 | 22 | 135,436 |
No | output | 1 | 67,718 | 22 | 135,437 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are well-known formulas: <image>, <image>, <image>. Also mathematicians found similar formulas for higher degrees.
Find the value of the sum <image> modulo 109 + 7 (so you should find the remainder after dividing the answer by the value 109 + 7).
Input
The only line contains two integers n, k (1 ≤ n ≤ 109, 0 ≤ k ≤ 106).
Output
Print the only integer a — the remainder after dividing the value of the sum by the value 109 + 7.
Examples
Input
4 1
Output
10
Input
4 2
Output
30
Input
4 3
Output
100
Input
4 0
Output
4
Submitted Solution:
```
from fractions import Fraction as Fr
from functools import *
def bernoulli2():
A, m = [], 0
while True:
A.append(Fr(1, m+1))
for j in range(m, 0, -1):
A[j-1] = j*(A[j-1] - A[j])
yield A[0] # (which is Bm)
m += 1
bern_g = bernoulli2();
next(bern_g);next(bern_g)
bern_res = [1, -1/2]
bern_s = [0]
def bernoulli(n):
while bern_s[0] <= n:
bern_s[0]+=1
bern_res.append(next(bern_g))
return bern_res[n]
@lru_cache(maxsize=2**16)
#@profile
def binomial(n, r):
p = 1
for i in range(1, min(r, n - r) + 1):
p *= n
p //= i
n -= 1
return p
MOD = 10**9+7
#@profile
def poly(n, p):
s=0
for k in range(0,p+1):
#print(p,k)
#print((-1)**(p-k)*binomial(p,k)*Fr(bernoulli(p-k))*n**(k+1),(k+1))
s+=int(0.5+pow(-1,p-k,MOD)*binomial(p,k)*bernoulli(p-k)*pow(n,k+1,MOD)/(k+1)%MOD)
#print(s)
s%=MOD
return int(s)%MOD
print(poly(*map(int,input().split())))
#print(poly(4,100))
``` | instruction | 0 | 67,719 | 22 | 135,438 |
No | output | 1 | 67,719 | 22 | 135,439 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two integers n and k. Find k-th smallest divisor of n, or report that it doesn't exist.
Divisor of n is any such natural number, that n can be divided by it without remainder.
Input
The first line contains two integers n and k (1 ≤ n ≤ 1015, 1 ≤ k ≤ 109).
Output
If n has less than k divisors, output -1.
Otherwise, output the k-th smallest divisor of n.
Examples
Input
4 2
Output
2
Input
5 3
Output
-1
Input
12 5
Output
6
Note
In the first example, number 4 has three divisors: 1, 2 and 4. The second one is 2.
In the second example, number 5 has only two divisors: 1 and 5. The third divisor doesn't exist, so the answer is -1. | instruction | 0 | 67,765 | 22 | 135,530 |
Tags: math, number theory
Correct Solution:
```
def get_divisors(n):
limit = int(n ** 0.5) + 1
for i in range(1, limit):
if n % i == 0:
yield i
if i * i - n:
yield n // i
n, k = [int(i) for i in input().split()]
answer = list(sorted(get_divisors(n)))
print(answer[k - 1] if len(answer) >= k else -1)
``` | output | 1 | 67,765 | 22 | 135,531 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two integers n and k. Find k-th smallest divisor of n, or report that it doesn't exist.
Divisor of n is any such natural number, that n can be divided by it without remainder.
Input
The first line contains two integers n and k (1 ≤ n ≤ 1015, 1 ≤ k ≤ 109).
Output
If n has less than k divisors, output -1.
Otherwise, output the k-th smallest divisor of n.
Examples
Input
4 2
Output
2
Input
5 3
Output
-1
Input
12 5
Output
6
Note
In the first example, number 4 has three divisors: 1, 2 and 4. The second one is 2.
In the second example, number 5 has only two divisors: 1 and 5. The third divisor doesn't exist, so the answer is -1. | instruction | 0 | 67,766 | 22 | 135,532 |
Tags: math, number theory
Correct Solution:
```
n, k = map(int, input().split())
i=1
a = []
while i * i <= n:
if n % i == 0:
a.append(i)
if i * i != n:
a.append(n//i)
i+=1
a = sorted(a)
if k > len(a):
print(-1)
else:
print(a[k-1])
``` | output | 1 | 67,766 | 22 | 135,533 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two integers n and k. Find k-th smallest divisor of n, or report that it doesn't exist.
Divisor of n is any such natural number, that n can be divided by it without remainder.
Input
The first line contains two integers n and k (1 ≤ n ≤ 1015, 1 ≤ k ≤ 109).
Output
If n has less than k divisors, output -1.
Otherwise, output the k-th smallest divisor of n.
Examples
Input
4 2
Output
2
Input
5 3
Output
-1
Input
12 5
Output
6
Note
In the first example, number 4 has three divisors: 1, 2 and 4. The second one is 2.
In the second example, number 5 has only two divisors: 1 and 5. The third divisor doesn't exist, so the answer is -1. | instruction | 0 | 67,767 | 22 | 135,534 |
Tags: math, number theory
Correct Solution:
```
n,k = input().split()
n,k = int(n),int(k)
l1=[1]
if n != 1 :
l2 = [n]
else :
l2 = []
i = 2
while i*i <= n :
if n%i == 0 :
l1.append(i)
if i*i != n :
l2.append(int(n/i))
i += 1
l2.reverse()
l = l1+l2
if k> len(l) :
print(-1)
else :
print(l[k-1])
``` | output | 1 | 67,767 | 22 | 135,535 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two integers n and k. Find k-th smallest divisor of n, or report that it doesn't exist.
Divisor of n is any such natural number, that n can be divided by it without remainder.
Input
The first line contains two integers n and k (1 ≤ n ≤ 1015, 1 ≤ k ≤ 109).
Output
If n has less than k divisors, output -1.
Otherwise, output the k-th smallest divisor of n.
Examples
Input
4 2
Output
2
Input
5 3
Output
-1
Input
12 5
Output
6
Note
In the first example, number 4 has three divisors: 1, 2 and 4. The second one is 2.
In the second example, number 5 has only two divisors: 1 and 5. The third divisor doesn't exist, so the answer is -1. | instruction | 0 | 67,768 | 22 | 135,536 |
Tags: math, number theory
Correct Solution:
```
import math
n, k = input().split(" ")
n, k = int(n), int(k)
res = set()
for i in range(1, int(math.sqrt(n))+1):
if n%i == 0:
res.add(i)
res.add(int(n/i))
res = list(res)
if len(res) < k:
print(-1)
else:
res.sort()
print(res[k-1])
``` | output | 1 | 67,768 | 22 | 135,537 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two integers n and k. Find k-th smallest divisor of n, or report that it doesn't exist.
Divisor of n is any such natural number, that n can be divided by it without remainder.
Input
The first line contains two integers n and k (1 ≤ n ≤ 1015, 1 ≤ k ≤ 109).
Output
If n has less than k divisors, output -1.
Otherwise, output the k-th smallest divisor of n.
Examples
Input
4 2
Output
2
Input
5 3
Output
-1
Input
12 5
Output
6
Note
In the first example, number 4 has three divisors: 1, 2 and 4. The second one is 2.
In the second example, number 5 has only two divisors: 1 and 5. The third divisor doesn't exist, so the answer is -1. | instruction | 0 | 67,769 | 22 | 135,538 |
Tags: math, number theory
Correct Solution:
```
import collections
import itertools
n, k = map(int, input().split())
def prime_factors(n):
i = 2
while i * i <= n:
if n % i == 0:
n //= i
yield i
else:
i += 1
if n > 1:
yield n
def prod(iterable):
result = 1
for i in iterable:
result *= i
return result
def get_divisors(n):
pf = prime_factors(n)
pf_with_multiplicity = collections.Counter(pf)
powers = [
[factor ** i for i in range(count + 1)]
for factor, count in pf_with_multiplicity.items()
]
for prime_power_combo in itertools.product(*powers):
yield prod(prime_power_combo)
res = list(get_divisors(n))
res.sort()
if len(res) >= k:
print(res[k - 1])
else:
print(-1)
``` | output | 1 | 67,769 | 22 | 135,539 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two integers n and k. Find k-th smallest divisor of n, or report that it doesn't exist.
Divisor of n is any such natural number, that n can be divided by it without remainder.
Input
The first line contains two integers n and k (1 ≤ n ≤ 1015, 1 ≤ k ≤ 109).
Output
If n has less than k divisors, output -1.
Otherwise, output the k-th smallest divisor of n.
Examples
Input
4 2
Output
2
Input
5 3
Output
-1
Input
12 5
Output
6
Note
In the first example, number 4 has three divisors: 1, 2 and 4. The second one is 2.
In the second example, number 5 has only two divisors: 1 and 5. The third divisor doesn't exist, so the answer is -1. | instruction | 0 | 67,770 | 22 | 135,540 |
Tags: math, number theory
Correct Solution:
```
def primes(n):
sieve = [True] * n
for i in range(3, int(n ** 0.5) + 1, 2):
if sieve[i]:
sieve[i * i::2 * i] = [False] * ((n - i * i - 1) // (2 * i) + 1)
return [2] + [i for i in range(3, n, 2) if sieve[i]]
def binomial_coefficient(n, k):
if 0 <= k <= n:
ntok = 1
ktok = 1
for t in range(1, min(k, n - k) + 1):
ntok *= n
ktok *= t
n -= 1
return ntok // ktok
else:
return 0
def powerOfK(k, max):
if k == 1:
return [1]
if k == -1:
return [-1, 1]
result = []
n = 1
while n <= max:
result.append(n)
n *= k
return result
def prefixSum(arr):
for i in range(1, len(arr)):
arr[i] = arr[i] + arr[i - 1]
return arr
def divisors(n):
i = 1
result = []
while i * i <= n:
if n % i == 0:
if n / i == i:
result.append(i)
else:
result.append(i)
result.append(n / i)
i += 1
return result
def kadane(a, size):
max_so_far = 0
max_ending_here = 0
for i in range(0, size):
max_ending_here = max_ending_here + a[i]
if (max_so_far < max_ending_here):
max_so_far = max_ending_here
if max_ending_here < 0:
max_ending_here = 0
return max_so_far
# @lru_cache(maxsize=None)
def digitsSum(n):
if n == 0:
return 0
r = 0
while n > 0:
r += n % 10
n //= 10
return r
def print_grid(grid):
for line in grid:
print("".join(line))
# n=int(input())
# n,k=map(int,input().split())
# arr=list(map(int,input().split()))
#ls=list(map(int,input().split()))
#for i in range(m):
# for _ in range(int(input())):
from collections import Counter
#from fractions import Fraction
#n = int(input())
# for _ in range(int(input())):
# import math
import sys
# from collections import deque
# from collections import Counter
# ls=list(map(int,input().split()))
# for i in range(m):
# for i in range(int(input())):
# n,k= map(int, input().split())
# arr=list(map(int,input().split()))
# n=sys.stdin.readline()
# n=int(n)
# n,k= map(int, input().split())
# arr=list(map(int,input().split()))
# n=int(input())
import math
#for _ in range(int(input())):
n,k= map(int, input().split())
ls=[]
for i in range(1,int(n**0.5)+1):
if n%i==0:
if i*i!=n:
ls+=[i]
ls+=[n//i]
else:
ls+=[i]
if len(ls)<k:
print(-1)
else:
print(sorted(ls)[k-1])
``` | output | 1 | 67,770 | 22 | 135,541 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two integers n and k. Find k-th smallest divisor of n, or report that it doesn't exist.
Divisor of n is any such natural number, that n can be divided by it without remainder.
Input
The first line contains two integers n and k (1 ≤ n ≤ 1015, 1 ≤ k ≤ 109).
Output
If n has less than k divisors, output -1.
Otherwise, output the k-th smallest divisor of n.
Examples
Input
4 2
Output
2
Input
5 3
Output
-1
Input
12 5
Output
6
Note
In the first example, number 4 has three divisors: 1, 2 and 4. The second one is 2.
In the second example, number 5 has only two divisors: 1 and 5. The third divisor doesn't exist, so the answer is -1. | instruction | 0 | 67,771 | 22 | 135,542 |
Tags: math, number theory
Correct Solution:
```
n,k=map(int,input().split())
ct=0
num=int(n**(0.5))
i=1
ct=0
arr=[]
ch=0
l=0
while(i<=num):
if(n%i==0):
if(i*i!=n):
arr.append(n//i)
l+=1
arr.append(i)
ct+=1
l+=1
if(ct==k):
print(i)
ch=1
break
i+=1
if(ch==0):
if(l<k):
print(-1)
else:
arr.sort()
print(arr[k-1])
``` | output | 1 | 67,771 | 22 | 135,543 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two integers n and k. Find k-th smallest divisor of n, or report that it doesn't exist.
Divisor of n is any such natural number, that n can be divided by it without remainder.
Input
The first line contains two integers n and k (1 ≤ n ≤ 1015, 1 ≤ k ≤ 109).
Output
If n has less than k divisors, output -1.
Otherwise, output the k-th smallest divisor of n.
Examples
Input
4 2
Output
2
Input
5 3
Output
-1
Input
12 5
Output
6
Note
In the first example, number 4 has three divisors: 1, 2 and 4. The second one is 2.
In the second example, number 5 has only two divisors: 1 and 5. The third divisor doesn't exist, so the answer is -1. | instruction | 0 | 67,772 | 22 | 135,544 |
Tags: math, number theory
Correct Solution:
```
n,k=map(int,input().split())
f=[]
for i in range(1,int(n**.5)+1):
if n%i==0:
f.append(i)
if i*i!=n: f+=[n//i]
f.sort()
print(-1 if len(f)<k else f[k-1])
``` | output | 1 | 67,772 | 22 | 135,545 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two integers n and k. Find k-th smallest divisor of n, or report that it doesn't exist.
Divisor of n is any such natural number, that n can be divided by it without remainder.
Input
The first line contains two integers n and k (1 ≤ n ≤ 1015, 1 ≤ k ≤ 109).
Output
If n has less than k divisors, output -1.
Otherwise, output the k-th smallest divisor of n.
Examples
Input
4 2
Output
2
Input
5 3
Output
-1
Input
12 5
Output
6
Note
In the first example, number 4 has three divisors: 1, 2 and 4. The second one is 2.
In the second example, number 5 has only two divisors: 1 and 5. The third divisor doesn't exist, so the answer is -1.
Submitted Solution:
```
from math import ceil,sqrt
def divisors(n,k):
begin= []
end = []
for i in range(1, int(sqrt(n))+1):
if n % i is 0:
if n//i is i:
begin.append(i)
else:
begin.append(i)
end.append(n//i)
begin += end[::-1]
size = len(begin)
if size < k:
print(-1)
else:
print(begin[k-1])
n,k = input().split()
n,k = int(n), int(k)
divisors(n,k)
``` | instruction | 0 | 67,773 | 22 | 135,546 |
Yes | output | 1 | 67,773 | 22 | 135,547 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two integers n and k. Find k-th smallest divisor of n, or report that it doesn't exist.
Divisor of n is any such natural number, that n can be divided by it without remainder.
Input
The first line contains two integers n and k (1 ≤ n ≤ 1015, 1 ≤ k ≤ 109).
Output
If n has less than k divisors, output -1.
Otherwise, output the k-th smallest divisor of n.
Examples
Input
4 2
Output
2
Input
5 3
Output
-1
Input
12 5
Output
6
Note
In the first example, number 4 has three divisors: 1, 2 and 4. The second one is 2.
In the second example, number 5 has only two divisors: 1 and 5. The third divisor doesn't exist, so the answer is -1.
Submitted Solution:
```
n,k=map(int,input().split())
if n==1:
if k==1:
print(1)
else:
print(-1)
exit()
from math import sqrt as S
res=[]
res.append(1)
res.append(n)
for i in range(2,int(S(n))+1):
if n%i==0:
if i==n//i:
res.append(i)
else:
res.append(i)
res.append(n//i)
res.sort()
if k>len(res):
print(-1)
else:
print(res[k-1])
``` | instruction | 0 | 67,774 | 22 | 135,548 |
Yes | output | 1 | 67,774 | 22 | 135,549 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two integers n and k. Find k-th smallest divisor of n, or report that it doesn't exist.
Divisor of n is any such natural number, that n can be divided by it without remainder.
Input
The first line contains two integers n and k (1 ≤ n ≤ 1015, 1 ≤ k ≤ 109).
Output
If n has less than k divisors, output -1.
Otherwise, output the k-th smallest divisor of n.
Examples
Input
4 2
Output
2
Input
5 3
Output
-1
Input
12 5
Output
6
Note
In the first example, number 4 has three divisors: 1, 2 and 4. The second one is 2.
In the second example, number 5 has only two divisors: 1 and 5. The third divisor doesn't exist, so the answer is -1.
Submitted Solution:
```
import math
from sys import stdin
n,m=[int(x)for x in stdin.readline().split()]
f=0
a=[]
count=0
if m>n:
print(-1)
elif m==n:
if m==1:
print(1)
else:
print(-1)
else:
x=math.sqrt(n)
x=int(x)
for i in range(1,x+1):
if n%i==0:
a.append(i)
if i!= n//i:
a.append(n//i)
count+=2
else:
count+=1
# print(a,count)
# b=set(a)
# a=list(b)
# print(a,count)
if count<m:
print(-1)
else:
a.sort()
print(a[m-1])
``` | instruction | 0 | 67,775 | 22 | 135,550 |
Yes | output | 1 | 67,775 | 22 | 135,551 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two integers n and k. Find k-th smallest divisor of n, or report that it doesn't exist.
Divisor of n is any such natural number, that n can be divided by it without remainder.
Input
The first line contains two integers n and k (1 ≤ n ≤ 1015, 1 ≤ k ≤ 109).
Output
If n has less than k divisors, output -1.
Otherwise, output the k-th smallest divisor of n.
Examples
Input
4 2
Output
2
Input
5 3
Output
-1
Input
12 5
Output
6
Note
In the first example, number 4 has three divisors: 1, 2 and 4. The second one is 2.
In the second example, number 5 has only two divisors: 1 and 5. The third divisor doesn't exist, so the answer is -1.
Submitted Solution:
```
ll=lambda:map(int,input().split())
t=lambda:int(input())
ss=lambda:input()
#from math import log10 ,log2,ceil,factorial as f,gcd
#from itertools import combinations_with_replacement as cs
#from functools import reduce
#from math import gcd
n,k=ll()
i=2
l=set()
l.add(1)
l.add(n)
while i*i<=n:
if n%i==0:
l.add(i)
l.add(n//i)
i+=1
if len(list(l))<k:
print(-1)
else:
print(sorted(list(l))[k-1])
``` | instruction | 0 | 67,776 | 22 | 135,552 |
Yes | output | 1 | 67,776 | 22 | 135,553 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two integers n and k. Find k-th smallest divisor of n, or report that it doesn't exist.
Divisor of n is any such natural number, that n can be divided by it without remainder.
Input
The first line contains two integers n and k (1 ≤ n ≤ 1015, 1 ≤ k ≤ 109).
Output
If n has less than k divisors, output -1.
Otherwise, output the k-th smallest divisor of n.
Examples
Input
4 2
Output
2
Input
5 3
Output
-1
Input
12 5
Output
6
Note
In the first example, number 4 has three divisors: 1, 2 and 4. The second one is 2.
In the second example, number 5 has only two divisors: 1 and 5. The third divisor doesn't exist, so the answer is -1.
Submitted Solution:
```
def divisores(num):
div = []
old_Big = n
i = 1
counter = 0
while counter < k and i < n and i < old_Big:
#print(i, old_Big)
if(num%i == 0):
div.append(i)
counter += 1
if(num/i != i):
div.append(int(num/i))
old_Big = int(num/i)
div.sort()
i += 1
#print(div)
return div
values = input()
n, k = values.split()
n = int(n)
k = int(k)
div = divisores(n)
if(k > len(div)):
print(-1)
else:
print(div[k-1])
``` | instruction | 0 | 67,777 | 22 | 135,554 |
No | output | 1 | 67,777 | 22 | 135,555 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two integers n and k. Find k-th smallest divisor of n, or report that it doesn't exist.
Divisor of n is any such natural number, that n can be divided by it without remainder.
Input
The first line contains two integers n and k (1 ≤ n ≤ 1015, 1 ≤ k ≤ 109).
Output
If n has less than k divisors, output -1.
Otherwise, output the k-th smallest divisor of n.
Examples
Input
4 2
Output
2
Input
5 3
Output
-1
Input
12 5
Output
6
Note
In the first example, number 4 has three divisors: 1, 2 and 4. The second one is 2.
In the second example, number 5 has only two divisors: 1 and 5. The third divisor doesn't exist, so the answer is -1.
Submitted Solution:
```
import math
n,k=(int(i) for i in input().split())
if(k==1):
print(1)
else:
l=[1]
c=1
k1=0
k2=2
for i in range(2,int(math.sqrt(n))+1):
if(n%i==0):
c+=1
l+=[i]
k2+=2
if(c==k):
print(i)
k1=1
break
if(math.sqrt(n)==int(math.sqrt(n))):
k2-=1
if(k2<k):
print(-1)
else:
if(k1==0):
print(n//l[k-c-1])
``` | instruction | 0 | 67,778 | 22 | 135,556 |
No | output | 1 | 67,778 | 22 | 135,557 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two integers n and k. Find k-th smallest divisor of n, or report that it doesn't exist.
Divisor of n is any such natural number, that n can be divided by it without remainder.
Input
The first line contains two integers n and k (1 ≤ n ≤ 1015, 1 ≤ k ≤ 109).
Output
If n has less than k divisors, output -1.
Otherwise, output the k-th smallest divisor of n.
Examples
Input
4 2
Output
2
Input
5 3
Output
-1
Input
12 5
Output
6
Note
In the first example, number 4 has three divisors: 1, 2 and 4. The second one is 2.
In the second example, number 5 has only two divisors: 1 and 5. The third divisor doesn't exist, so the answer is -1.
Submitted Solution:
```
values = input()
n, k = values.split()
n = int(n)
k = int(k)
counter = 0
i = 1
big = n
div = []
while counter < k and i < big:
if(n%i == 0):
counter += 1
div.append(int(i))
if(n/i != i):
div.append(int(n/i))
big = n/i
div.sort()
i+= 1
if(k >= len(div)):
print(-1)
elif(k < len(div)):
print(div[k-1])
``` | instruction | 0 | 67,779 | 22 | 135,558 |
No | output | 1 | 67,779 | 22 | 135,559 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two integers n and k. Find k-th smallest divisor of n, or report that it doesn't exist.
Divisor of n is any such natural number, that n can be divided by it without remainder.
Input
The first line contains two integers n and k (1 ≤ n ≤ 1015, 1 ≤ k ≤ 109).
Output
If n has less than k divisors, output -1.
Otherwise, output the k-th smallest divisor of n.
Examples
Input
4 2
Output
2
Input
5 3
Output
-1
Input
12 5
Output
6
Note
In the first example, number 4 has three divisors: 1, 2 and 4. The second one is 2.
In the second example, number 5 has only two divisors: 1 and 5. The third divisor doesn't exist, so the answer is -1.
Submitted Solution:
```
n, k = map(int, input().split())
count = 0
for i in range(1, int(n**0.5)+1):
if n%i == 0:
count += 1
if k == count:
print(i)
break
else:
for i in range(int(n**0.5)-1, 0, -1):
if n % i == 0:
count += 1
if k == count:
print(n//i)
break
else:
print(-1)
``` | instruction | 0 | 67,780 | 22 | 135,560 |
No | output | 1 | 67,780 | 22 | 135,561 |
Provide a correct Python 3 solution for this coding contest problem.
We arrange the numbers between 1 and N (1 <= N <= 10000) in increasing order and decreasing order like this:
1 2 3 4 5 6 7 8 9 . . . N
N . . . 9 8 7 6 5 4 3 2 1
Two numbers faced each other form a pair. Your task is to compute the number of pairs P such that both numbers in the pairs are prime.
Input
Input contains several test cases. Each test case consists of an integer N in one line.
Output
For each line of input, output P .
Example
Input
1
4
7
51
Output
0
2
2
6 | instruction | 0 | 68,057 | 22 | 136,114 |
"Correct Solution:
```
while True:
try:
n = eval(input())
is_prime = [True for _ in range(n+1)]
is_prime[1] = False
for i in range(2,int((n+1)/2)):
if is_prime[i]:
for j in range(i*2,n+1,i):
is_prime[j] = False
is_prime_rev = list(is_prime)
is_prime_rev.reverse()
count = 0
for i in range(1,n+1):
count = count + (1 if is_prime[i] and is_prime_rev[i-1] else 0)
print(count)
except EOFError:
break
``` | output | 1 | 68,057 | 22 | 136,115 |
Provide a correct Python 3 solution for this coding contest problem.
We arrange the numbers between 1 and N (1 <= N <= 10000) in increasing order and decreasing order like this:
1 2 3 4 5 6 7 8 9 . . . N
N . . . 9 8 7 6 5 4 3 2 1
Two numbers faced each other form a pair. Your task is to compute the number of pairs P such that both numbers in the pairs are prime.
Input
Input contains several test cases. Each test case consists of an integer N in one line.
Output
For each line of input, output P .
Example
Input
1
4
7
51
Output
0
2
2
6 | instruction | 0 | 68,058 | 22 | 136,116 |
"Correct Solution:
```
# 演習1-15
#時間がオーバーしたのでmathパッケージに頼り、素数の性質を利用
#素数判定機
import math as math
def prime_checker(limit):
n = 0
prime_index = []
for k in range(2, limit+1):
factor = 0
# 2以外の偶数は素数ではないので無視する
if k % 2 == 0 and k != 2:
continue
# 繰り返しの上限を対象の平方根にする
for divisor in range(2, math.floor(math.sqrt(k))+1):
if k % divisor == 0:
factor += 1
if factor == 0:
prime_index.append(k)
return prime_index
def prime_checker2(limit):
n = 0
prime_index = []
for k in range(9754, limit+1):
factor = 0
# 2以外の偶数は素数ではないので無視する(test caseに合わせた)
if k % 2 == 0 and k != 2:
continue
# 繰り返しの上限を対象の平方根にする
for divisor in range(2, math.floor(math.sqrt(k))+1):
if k % divisor == 0:
factor += 1
if factor == 0:
prime_index.append(k)
return prime_index
all_primes = prime_checker(10000)
while True:
try:
limit = int(input())
if limit == 1:
print(0)
else:
li = list(range(1,limit+1,1))
li_r = list(reversed(li))
#マッチするインデックスを返す
index_r = []
match_primes = [i for i in all_primes if i <= limit]
index = match_primes
index_r = [limit+1-j for j in match_primes]
#インデックス番号が一致するものを抽出する
print(len(list(set(index) & set(index_r))))
except:
break
``` | output | 1 | 68,058 | 22 | 136,117 |
Provide a correct Python 3 solution for this coding contest problem.
We arrange the numbers between 1 and N (1 <= N <= 10000) in increasing order and decreasing order like this:
1 2 3 4 5 6 7 8 9 . . . N
N . . . 9 8 7 6 5 4 3 2 1
Two numbers faced each other form a pair. Your task is to compute the number of pairs P such that both numbers in the pairs are prime.
Input
Input contains several test cases. Each test case consists of an integer N in one line.
Output
For each line of input, output P .
Example
Input
1
4
7
51
Output
0
2
2
6 | instruction | 0 | 68,059 | 22 | 136,118 |
"Correct Solution:
```
primes = [0, 0] + [1] * 9999
for i in range(2, 101):
if primes[i]:
for j in range(i*i, 10001, i):
primes[j] = 0
while True:
try:
N = int(input())
except:
break
print(sum(primes[i] & primes[N-i+1] for i in range(1, N+1)))
``` | output | 1 | 68,059 | 22 | 136,119 |
Provide a correct Python 3 solution for this coding contest problem.
We arrange the numbers between 1 and N (1 <= N <= 10000) in increasing order and decreasing order like this:
1 2 3 4 5 6 7 8 9 . . . N
N . . . 9 8 7 6 5 4 3 2 1
Two numbers faced each other form a pair. Your task is to compute the number of pairs P such that both numbers in the pairs are prime.
Input
Input contains several test cases. Each test case consists of an integer N in one line.
Output
For each line of input, output P .
Example
Input
1
4
7
51
Output
0
2
2
6 | instruction | 0 | 68,060 | 22 | 136,120 |
"Correct Solution:
```
primes = [0, 0] + [1] * 9999
for i in range(2, 101):
if primes[i]:
for j in range(i*i, 10001, i):
primes[j] = 0
while True:
try:
N = int(input())
except:
break
print(sum(primes[i] and primes[N-i+1] for i in range(1, N+1)))
``` | output | 1 | 68,060 | 22 | 136,121 |
Provide a correct Python 3 solution for this coding contest problem.
We arrange the numbers between 1 and N (1 <= N <= 10000) in increasing order and decreasing order like this:
1 2 3 4 5 6 7 8 9 . . . N
N . . . 9 8 7 6 5 4 3 2 1
Two numbers faced each other form a pair. Your task is to compute the number of pairs P such that both numbers in the pairs are prime.
Input
Input contains several test cases. Each test case consists of an integer N in one line.
Output
For each line of input, output P .
Example
Input
1
4
7
51
Output
0
2
2
6 | instruction | 0 | 68,061 | 22 | 136,122 |
"Correct Solution:
```
import sys
import math
M = 10001
size = (M-1)//2
l = [1] * size
b = int(math.sqrt(M)+1)
for i in range(3,b,2):
index = (i-3)//2
if l[index]:
l[index+i::i] = [0] * len(l[index+i::i])
m = []
for i in range(size):
if l[i]:
m.append(i)
s = ""
p = [0,0,0,1]
for i in sys.stdin:
n = int(i)
if n <= 3:
print(p[n])
elif n % 2 == 0:
if l[(n-3)//2]:
print(2)
else:
print(0)
else:
ct = 0
n2 = (n-5)//2
for j in m:
if j > n2:
break
if l[n2-j]:
ct += 1
print(ct)
``` | output | 1 | 68,061 | 22 | 136,123 |
Provide a correct Python 3 solution for this coding contest problem.
We arrange the numbers between 1 and N (1 <= N <= 10000) in increasing order and decreasing order like this:
1 2 3 4 5 6 7 8 9 . . . N
N . . . 9 8 7 6 5 4 3 2 1
Two numbers faced each other form a pair. Your task is to compute the number of pairs P such that both numbers in the pairs are prime.
Input
Input contains several test cases. Each test case consists of an integer N in one line.
Output
For each line of input, output P .
Example
Input
1
4
7
51
Output
0
2
2
6 | instruction | 0 | 68,062 | 22 | 136,124 |
"Correct Solution:
```
import math
def ifprime(num):
if num == 1:
return False
for i in range(2,math.floor(num ** 0.5) + 1):
if num % i:
continue
else:
return False
return True
while(True):
try:
num = int(input())
except:
break
count = 0
for i in range(1,num+1):
if ifprime(i) and ifprime(num+1-i):
count += 1
print(count)
``` | output | 1 | 68,062 | 22 | 136,125 |
Provide a correct Python 3 solution for this coding contest problem.
We arrange the numbers between 1 and N (1 <= N <= 10000) in increasing order and decreasing order like this:
1 2 3 4 5 6 7 8 9 . . . N
N . . . 9 8 7 6 5 4 3 2 1
Two numbers faced each other form a pair. Your task is to compute the number of pairs P such that both numbers in the pairs are prime.
Input
Input contains several test cases. Each test case consists of an integer N in one line.
Output
For each line of input, output P .
Example
Input
1
4
7
51
Output
0
2
2
6 | instruction | 0 | 68,063 | 22 | 136,126 |
"Correct Solution:
```
n=10000
p=[1]*(n+1)
p[0],p[1]=0,0
for i in range(2,int(n**0.5)+1):
if p[i]:
for j in range(i*i,n+1,i):
p[j]=0
#p=[i for i in range(n+1) if p[i]==1]
while 1:
try:n=int(input())
except:break
c=0
for i in range(2,n):
if (p[i],p[n-i+1])==(1,1):c+=1
print(c)
``` | output | 1 | 68,063 | 22 | 136,127 |
Provide a correct Python 3 solution for this coding contest problem.
We arrange the numbers between 1 and N (1 <= N <= 10000) in increasing order and decreasing order like this:
1 2 3 4 5 6 7 8 9 . . . N
N . . . 9 8 7 6 5 4 3 2 1
Two numbers faced each other form a pair. Your task is to compute the number of pairs P such that both numbers in the pairs are prime.
Input
Input contains several test cases. Each test case consists of an integer N in one line.
Output
For each line of input, output P .
Example
Input
1
4
7
51
Output
0
2
2
6 | instruction | 0 | 68,064 | 22 | 136,128 |
"Correct Solution:
```
import sys
def is_prime(x):
if((x % 2 == 0 and x > 2) or x < 2):
return 0
elif(x <= 5):
return 1
a = 3
while(a * a <= x):
if(x % a == 0):
return 0
a += 2
return 1
l = []
for i in sys.stdin:
l.append(int(i))
for data in l:
count = 0
data1 = [i for i in range(1,data+1)]
data2 = [i for i in range(data,0,-1)]
for i in range(0,data):
count += 1 if(is_prime(data1[i]) == 1 and is_prime(data2[i]) == 1) else 0
print(count)
``` | output | 1 | 68,064 | 22 | 136,129 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We arrange the numbers between 1 and N (1 <= N <= 10000) in increasing order and decreasing order like this:
1 2 3 4 5 6 7 8 9 . . . N
N . . . 9 8 7 6 5 4 3 2 1
Two numbers faced each other form a pair. Your task is to compute the number of pairs P such that both numbers in the pairs are prime.
Input
Input contains several test cases. Each test case consists of an integer N in one line.
Output
For each line of input, output P .
Example
Input
1
4
7
51
Output
0
2
2
6
Submitted Solution:
```
def isPrime(n):
if n < 2:
return False
if n == 2:
return True
for p in range(2, int(n**0.5)+1):
if n % p == 0:
return False
return True
while True:
try:
n=int(input())
except:
break
oup=0
if n <=1:
print(0)
else:
if isPrime(n-1)==True:
if n-1==2:
oup+=1
else:
oup+=2
for i in range(3,n//2+2,2):
if isPrime(i)!=True:
continue
j=n-i+1
if i>j:
continue
elif isPrime(j)==True:
if i==j:
oup+=1
else:
oup+=2
print(oup)
``` | instruction | 0 | 68,065 | 22 | 136,130 |
Yes | output | 1 | 68,065 | 22 | 136,131 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We arrange the numbers between 1 and N (1 <= N <= 10000) in increasing order and decreasing order like this:
1 2 3 4 5 6 7 8 9 . . . N
N . . . 9 8 7 6 5 4 3 2 1
Two numbers faced each other form a pair. Your task is to compute the number of pairs P such that both numbers in the pairs are prime.
Input
Input contains several test cases. Each test case consists of an integer N in one line.
Output
For each line of input, output P .
Example
Input
1
4
7
51
Output
0
2
2
6
Submitted Solution:
```
max = 10000
j = 3
l = [1] * (max + 1)
l[1] = 0
for i in range(4,max + 1, 2):
l[i] = 0
for i in range(3,max + 1, 2):
if ( l[i] == 0 ):
continue
j = i + i
while ( j <= max ):
l[j] = 0
j = j + i
while True:
try:
n = int(input())
except EOFError:
break
cnt = 0
for i in range(1,n+1):
if (l[i] + l[n+1-i] == 2 ):
cnt += 1
print(cnt)
``` | instruction | 0 | 68,066 | 22 | 136,132 |
Yes | output | 1 | 68,066 | 22 | 136,133 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We arrange the numbers between 1 and N (1 <= N <= 10000) in increasing order and decreasing order like this:
1 2 3 4 5 6 7 8 9 . . . N
N . . . 9 8 7 6 5 4 3 2 1
Two numbers faced each other form a pair. Your task is to compute the number of pairs P such that both numbers in the pairs are prime.
Input
Input contains several test cases. Each test case consists of an integer N in one line.
Output
For each line of input, output P .
Example
Input
1
4
7
51
Output
0
2
2
6
Submitted Solution:
```
primes = [0, 0] + [1]*9999
for i in range(2, 101):
if primes[i]:
for j in range(i*i, 10001, i):
primes[j] = 0
while True:
try:
n = int(input())
print(sum([1 for a, b in zip(list(range(1,n+1)), list(range(1,n+1)[::-1])) if primes[a] and primes[b]]))
except:
break
``` | instruction | 0 | 68,067 | 22 | 136,134 |
Yes | output | 1 | 68,067 | 22 | 136,135 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We arrange the numbers between 1 and N (1 <= N <= 10000) in increasing order and decreasing order like this:
1 2 3 4 5 6 7 8 9 . . . N
N . . . 9 8 7 6 5 4 3 2 1
Two numbers faced each other form a pair. Your task is to compute the number of pairs P such that both numbers in the pairs are prime.
Input
Input contains several test cases. Each test case consists of an integer N in one line.
Output
For each line of input, output P .
Example
Input
1
4
7
51
Output
0
2
2
6
Submitted Solution:
```
# AOJ 1004: Pair of Primes
# Python3 2018.7.4 bal4u
import sys
from sys import stdin
input = stdin.readline
MAX = 10000
SQRT = 100 # sqrt(MAX)
notprime = [False]*MAX
def sieve():
notprime[1] = True
notprime[2] = False
for i in range(3, SQRT, 2):
if notprime[i] == False:
for j in range(i*i, MAX, i): notprime[j] = True
sieve()
while True:
try: k = int(input())
except: break
if (k&1) == 0: print(0 if notprime[k-1] else 2)
elif k == 3: print(1)
else:
ans = 0
s, p, q = 3, 3, k-2
while q >= s:
if not notprime[p] and not notprime[q]: ans += 1
p, q = p+2, q-2
print(ans)
``` | instruction | 0 | 68,068 | 22 | 136,136 |
Yes | output | 1 | 68,068 | 22 | 136,137 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We arrange the numbers between 1 and N (1 <= N <= 10000) in increasing order and decreasing order like this:
1 2 3 4 5 6 7 8 9 . . . N
N . . . 9 8 7 6 5 4 3 2 1
Two numbers faced each other form a pair. Your task is to compute the number of pairs P such that both numbers in the pairs are prime.
Input
Input contains several test cases. Each test case consists of an integer N in one line.
Output
For each line of input, output P .
Example
Input
1
4
7
51
Output
0
2
2
6
Submitted Solution:
```
import sys
import math
# primes = []
# for n in range(2, 10001):
# is_prime = True
# for prime in primes:
# if prime > math.sqrt(n):
# break
# if n % prime == 0:
# is_prime = False
# break
# if is_prime:
# primes.append(n)
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]
def __main__():
ns = [[int(n) for n in line.strip().split()][0] for line in sys.stdin]
for n in ns:
print(len([m for m in range(n // 2, 0, -1) if n -
m + 1 in primes and m in primes]) * 2)
if __name__ == "__main__":
__main__()
``` | instruction | 0 | 68,069 | 22 | 136,138 |
No | output | 1 | 68,069 | 22 | 136,139 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We arrange the numbers between 1 and N (1 <= N <= 10000) in increasing order and decreasing order like this:
1 2 3 4 5 6 7 8 9 . . . N
N . . . 9 8 7 6 5 4 3 2 1
Two numbers faced each other form a pair. Your task is to compute the number of pairs P such that both numbers in the pairs are prime.
Input
Input contains several test cases. Each test case consists of an integer N in one line.
Output
For each line of input, output P .
Example
Input
1
4
7
51
Output
0
2
2
6
Submitted Solution:
```
import sys
import math
primes = []
for n in range(2, 10001):
is_prime = True
for prime in primes:
if prime > math.sqrt(n):
break
if n % prime == 0:
is_prime = False
break
if is_prime:
primes.append(n)
def __main__():
ns = [[int(n) for n in line.strip().split()][0] for line in sys.stdin]
p = 0
for n in ns:
pairs = [m for m in range(n, 0, -1) if n-m+1 in primes and m in primes]
print(len(pairs))
if __name__ == "__main__":
__main__()
``` | instruction | 0 | 68,070 | 22 | 136,140 |
No | output | 1 | 68,070 | 22 | 136,141 |
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