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.
Today, Osama gave Fadi an integer X, and Fadi was wondering about the minimum possible value of max(a, b) such that LCM(a, b) equals X. Both a and b should be positive integers.
LCM(a, b) is the smallest positive integer that is divisible by both a and b. For example, LCM(6, 8) = 24, LCM(4, 12) = 12, LCM(2, 3) = 6.
Of course, Fadi immediately knew the answer. Can you be just like Fadi and find any such pair?
Input
The first and only line contains an integer X (1 β€ X β€ 10^{12}).
Output
Print two positive integers, a and b, such that the value of max(a, b) is minimum possible and LCM(a, b) equals X. If there are several possible such pairs, you can print any.
Examples
Input
2
Output
1 2
Input
6
Output
2 3
Input
4
Output
1 4
Input
1
Output
1 1
Submitted Solution:
```
import math
lcm=int(input())
emp=[]
for i in range(1,lcm+1):
if lcm%i==0:
emp.append(i)
for i in range(len(emp)-1):
if (emp[i]*emp[i+1])//(math.gcd(emp[i],emp[i+1]))==lcm:
print(emp[i],emp[i+1])
break
``` | instruction | 0 | 77,671 | 22 | 155,342 |
No | output | 1 | 77,671 | 22 | 155,343 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Today, Osama gave Fadi an integer X, and Fadi was wondering about the minimum possible value of max(a, b) such that LCM(a, b) equals X. Both a and b should be positive integers.
LCM(a, b) is the smallest positive integer that is divisible by both a and b. For example, LCM(6, 8) = 24, LCM(4, 12) = 12, LCM(2, 3) = 6.
Of course, Fadi immediately knew the answer. Can you be just like Fadi and find any such pair?
Input
The first and only line contains an integer X (1 β€ X β€ 10^{12}).
Output
Print two positive integers, a and b, such that the value of max(a, b) is minimum possible and LCM(a, b) equals X. If there are several possible such pairs, you can print any.
Examples
Input
2
Output
1 2
Input
6
Output
2 3
Input
4
Output
1 4
Input
1
Output
1 1
Submitted Solution:
```
from math import sqrt
from math import floor
def factorize(n):
i = floor(sqrt(n))
while n % i != 0: i -= 1
return (i, n//i)
def solver():
n = int(input())
if n == 4: return tuple((1, 4))
return factorize(n) if n != 1 else tuple((1, 1))
t = solver()
print(t[0], t[1])
``` | instruction | 0 | 77,672 | 22 | 155,344 |
No | output | 1 | 77,672 | 22 | 155,345 |
Provide a correct Python 3 solution for this coding contest problem.
We have N integers. The i-th number is A_i.
\\{A_i\\} is said to be pairwise coprime when GCD(A_i,A_j)=1 holds for every pair (i, j) such that 1\leq i < j \leq N.
\\{A_i\\} is said to be setwise coprime when \\{A_i\\} is not pairwise coprime but GCD(A_1,\ldots,A_N)=1.
Determine if \\{A_i\\} is pairwise coprime, setwise coprime, or neither.
Here, GCD(\ldots) denotes greatest common divisor.
Constraints
* 2 \leq N \leq 10^6
* 1 \leq A_i\leq 10^6
Input
Input is given from Standard Input in the following format:
N
A_1 \ldots A_N
Output
If \\{A_i\\} is pairwise coprime, print `pairwise coprime`; if \\{A_i\\} is setwise coprime, print `setwise coprime`; if neither, print `not coprime`.
Examples
Input
3
3 4 5
Output
pairwise coprime
Input
3
6 10 15
Output
setwise coprime
Input
3
6 10 16
Output
not coprime | instruction | 0 | 77,999 | 22 | 155,998 |
"Correct Solution:
```
N = int(input())
li = list(map(int,input().split()))
S = max(li)
L = [0] * (S+1)
import math
from functools import reduce
if reduce(math.gcd,li) != 1:
print('not coprime')
exit()
for i in li:
L[i] = 1
for k in range(2,S+1):
t = 0
for j in range(0,S+1,k):
t += L[j]
if t > 1:
print('setwise coprime')
exit()
print('pairwise coprime')
``` | output | 1 | 77,999 | 22 | 155,999 |
Provide a correct Python 3 solution for this coding contest problem.
We have N integers. The i-th number is A_i.
\\{A_i\\} is said to be pairwise coprime when GCD(A_i,A_j)=1 holds for every pair (i, j) such that 1\leq i < j \leq N.
\\{A_i\\} is said to be setwise coprime when \\{A_i\\} is not pairwise coprime but GCD(A_1,\ldots,A_N)=1.
Determine if \\{A_i\\} is pairwise coprime, setwise coprime, or neither.
Here, GCD(\ldots) denotes greatest common divisor.
Constraints
* 2 \leq N \leq 10^6
* 1 \leq A_i\leq 10^6
Input
Input is given from Standard Input in the following format:
N
A_1 \ldots A_N
Output
If \\{A_i\\} is pairwise coprime, print `pairwise coprime`; if \\{A_i\\} is setwise coprime, print `setwise coprime`; if neither, print `not coprime`.
Examples
Input
3
3 4 5
Output
pairwise coprime
Input
3
6 10 15
Output
setwise coprime
Input
3
6 10 16
Output
not coprime | instruction | 0 | 78,000 | 22 | 156,000 |
"Correct Solution:
```
import sys
from math import gcd
input = sys.stdin.readline
N = int(input())
a = list(map(int, input().split()))
mx = max(a)
table = [0] * (mx + 1)
for x in a: table[x] += 1
g = 0
for x in a: g = gcd(g, x)
for i in range(2, mx + 1):
c = 0
for j in range(i, mx + 1, i): c += table[j]
if c > 1:
if g == 1: print("setwise coprime")
else: print("not coprime")
exit(0)
print("pairwise coprime")
``` | output | 1 | 78,000 | 22 | 156,001 |
Provide a correct Python 3 solution for this coding contest problem.
We have N integers. The i-th number is A_i.
\\{A_i\\} is said to be pairwise coprime when GCD(A_i,A_j)=1 holds for every pair (i, j) such that 1\leq i < j \leq N.
\\{A_i\\} is said to be setwise coprime when \\{A_i\\} is not pairwise coprime but GCD(A_1,\ldots,A_N)=1.
Determine if \\{A_i\\} is pairwise coprime, setwise coprime, or neither.
Here, GCD(\ldots) denotes greatest common divisor.
Constraints
* 2 \leq N \leq 10^6
* 1 \leq A_i\leq 10^6
Input
Input is given from Standard Input in the following format:
N
A_1 \ldots A_N
Output
If \\{A_i\\} is pairwise coprime, print `pairwise coprime`; if \\{A_i\\} is setwise coprime, print `setwise coprime`; if neither, print `not coprime`.
Examples
Input
3
3 4 5
Output
pairwise coprime
Input
3
6 10 15
Output
setwise coprime
Input
3
6 10 16
Output
not coprime | instruction | 0 | 78,001 | 22 | 156,002 |
"Correct Solution:
```
from math import gcd
n = int(input())
a = [int(i) for i in input().split()]
g = a[0]
for i in range(n):
g = gcd(g,a[i])
if g != 1:
print("not coprime")
exit()
m = max(a)
fac = [1]*(m+1)
for i in range(2,m+1):
if fac[i] != 1:
continue
for j in range(i,m+1,i):
fac[j] = i
s = set()
for i in range(n):
x = fac[a[i]]
if x == 1:
continue
if x in s:
print("setwise coprime")
exit()
s.add(x)
print("pairwise coprime")
``` | output | 1 | 78,001 | 22 | 156,003 |
Provide a correct Python 3 solution for this coding contest problem.
We have N integers. The i-th number is A_i.
\\{A_i\\} is said to be pairwise coprime when GCD(A_i,A_j)=1 holds for every pair (i, j) such that 1\leq i < j \leq N.
\\{A_i\\} is said to be setwise coprime when \\{A_i\\} is not pairwise coprime but GCD(A_1,\ldots,A_N)=1.
Determine if \\{A_i\\} is pairwise coprime, setwise coprime, or neither.
Here, GCD(\ldots) denotes greatest common divisor.
Constraints
* 2 \leq N \leq 10^6
* 1 \leq A_i\leq 10^6
Input
Input is given from Standard Input in the following format:
N
A_1 \ldots A_N
Output
If \\{A_i\\} is pairwise coprime, print `pairwise coprime`; if \\{A_i\\} is setwise coprime, print `setwise coprime`; if neither, print `not coprime`.
Examples
Input
3
3 4 5
Output
pairwise coprime
Input
3
6 10 15
Output
setwise coprime
Input
3
6 10 16
Output
not coprime | instruction | 0 | 78,002 | 22 | 156,004 |
"Correct Solution:
```
n = int(input())
s = list(map(int, input().split()))
s.sort()
A = max(s)
dp = [0] * (A + 5)
for ss in s:
dp[ss] += 1
pairwise = True
setwise = True
for i in range(2, A + 1):
cnt = 0
for j in range(i, A + 1, i):
cnt += dp[j]
if cnt > 1:
pairwise = False
if cnt >= n:
setwise = False
break
if pairwise:
print("pairwise coprime")
elif setwise:
print("setwise coprime")
else:
print("not coprime")
``` | output | 1 | 78,002 | 22 | 156,005 |
Provide a correct Python 3 solution for this coding contest problem.
We have N integers. The i-th number is A_i.
\\{A_i\\} is said to be pairwise coprime when GCD(A_i,A_j)=1 holds for every pair (i, j) such that 1\leq i < j \leq N.
\\{A_i\\} is said to be setwise coprime when \\{A_i\\} is not pairwise coprime but GCD(A_1,\ldots,A_N)=1.
Determine if \\{A_i\\} is pairwise coprime, setwise coprime, or neither.
Here, GCD(\ldots) denotes greatest common divisor.
Constraints
* 2 \leq N \leq 10^6
* 1 \leq A_i\leq 10^6
Input
Input is given from Standard Input in the following format:
N
A_1 \ldots A_N
Output
If \\{A_i\\} is pairwise coprime, print `pairwise coprime`; if \\{A_i\\} is setwise coprime, print `setwise coprime`; if neither, print `not coprime`.
Examples
Input
3
3 4 5
Output
pairwise coprime
Input
3
6 10 15
Output
setwise coprime
Input
3
6 10 16
Output
not coprime | instruction | 0 | 78,003 | 22 | 156,006 |
"Correct Solution:
```
import math
L = 10**6 + 1
N = int(input())
A = list(map(int, input().split()))
memo = [0] * L
flag = 0
for a in A:
memo[a] += 1
for i in range(2, L):
if sum(memo[i::i]) > 1:
flag = 1
break
g = 0
for i in range(N):
g = math.gcd(g, A[i])
if flag == 0:
answer = 'pairwise coprime'
elif flag == 1 and g == 1:
answer = 'setwise coprime'
else:
answer = 'not coprime'
print(answer)
``` | output | 1 | 78,003 | 22 | 156,007 |
Provide a correct Python 3 solution for this coding contest problem.
We have N integers. The i-th number is A_i.
\\{A_i\\} is said to be pairwise coprime when GCD(A_i,A_j)=1 holds for every pair (i, j) such that 1\leq i < j \leq N.
\\{A_i\\} is said to be setwise coprime when \\{A_i\\} is not pairwise coprime but GCD(A_1,\ldots,A_N)=1.
Determine if \\{A_i\\} is pairwise coprime, setwise coprime, or neither.
Here, GCD(\ldots) denotes greatest common divisor.
Constraints
* 2 \leq N \leq 10^6
* 1 \leq A_i\leq 10^6
Input
Input is given from Standard Input in the following format:
N
A_1 \ldots A_N
Output
If \\{A_i\\} is pairwise coprime, print `pairwise coprime`; if \\{A_i\\} is setwise coprime, print `setwise coprime`; if neither, print `not coprime`.
Examples
Input
3
3 4 5
Output
pairwise coprime
Input
3
6 10 15
Output
setwise coprime
Input
3
6 10 16
Output
not coprime | instruction | 0 | 78,004 | 22 | 156,008 |
"Correct Solution:
```
n = int(input())
A = tuple(map(int, input().split()))
maxa = max(A)
cnt = [0] * (maxa + 1)
for a in A:
cnt[a] += 1
from math import gcd
from functools import reduce
if reduce(gcd, A) != 1:
print('not coprime')
else:
for i in range(2, maxa+1):
if sum(cnt[i::i]) > 1:
print('setwise coprime')
break
else:
print('pairwise coprime')
``` | output | 1 | 78,004 | 22 | 156,009 |
Provide a correct Python 3 solution for this coding contest problem.
We have N integers. The i-th number is A_i.
\\{A_i\\} is said to be pairwise coprime when GCD(A_i,A_j)=1 holds for every pair (i, j) such that 1\leq i < j \leq N.
\\{A_i\\} is said to be setwise coprime when \\{A_i\\} is not pairwise coprime but GCD(A_1,\ldots,A_N)=1.
Determine if \\{A_i\\} is pairwise coprime, setwise coprime, or neither.
Here, GCD(\ldots) denotes greatest common divisor.
Constraints
* 2 \leq N \leq 10^6
* 1 \leq A_i\leq 10^6
Input
Input is given from Standard Input in the following format:
N
A_1 \ldots A_N
Output
If \\{A_i\\} is pairwise coprime, print `pairwise coprime`; if \\{A_i\\} is setwise coprime, print `setwise coprime`; if neither, print `not coprime`.
Examples
Input
3
3 4 5
Output
pairwise coprime
Input
3
6 10 15
Output
setwise coprime
Input
3
6 10 16
Output
not coprime | instruction | 0 | 78,005 | 22 | 156,010 |
"Correct Solution:
```
from math import gcd
from functools import reduce
k=10**6+1
def judge(n,a):
c=[0]*k
for x in a:
c[x]+=1
t=any(sum(c[i::i])>1 for i in range(2,k))
t+=reduce(gcd,a)>1
return ['pairwise','setwise','not'][t]+' coprime'
n=int(input())
a=list(map(int,input().split()))
print(judge(n,a))
``` | output | 1 | 78,005 | 22 | 156,011 |
Provide a correct Python 3 solution for this coding contest problem.
We have N integers. The i-th number is A_i.
\\{A_i\\} is said to be pairwise coprime when GCD(A_i,A_j)=1 holds for every pair (i, j) such that 1\leq i < j \leq N.
\\{A_i\\} is said to be setwise coprime when \\{A_i\\} is not pairwise coprime but GCD(A_1,\ldots,A_N)=1.
Determine if \\{A_i\\} is pairwise coprime, setwise coprime, or neither.
Here, GCD(\ldots) denotes greatest common divisor.
Constraints
* 2 \leq N \leq 10^6
* 1 \leq A_i\leq 10^6
Input
Input is given from Standard Input in the following format:
N
A_1 \ldots A_N
Output
If \\{A_i\\} is pairwise coprime, print `pairwise coprime`; if \\{A_i\\} is setwise coprime, print `setwise coprime`; if neither, print `not coprime`.
Examples
Input
3
3 4 5
Output
pairwise coprime
Input
3
6 10 15
Output
setwise coprime
Input
3
6 10 16
Output
not coprime | instruction | 0 | 78,006 | 22 | 156,012 |
"Correct Solution:
```
import math
N = 1000001
spf = [*range(N)]
i = 2
while i * i < N:
if spf[i] == i:
for j in range(i * i, N, i):
if spf[j] == j:
spf[j] = i
i += 1
r = 'pairwise'
input()
d = None
u = set()
for x in map(int, input().split()):
d = math.gcd(d or x, x);
s = set()
while x > 1:
p = spf[x]
if p in u:
r = 'setwise'
s.add(p)
x //= p
u |= s
print('not' if d > 1 else r, 'coprime')
``` | output | 1 | 78,006 | 22 | 156,013 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have N integers. The i-th number is A_i.
\\{A_i\\} is said to be pairwise coprime when GCD(A_i,A_j)=1 holds for every pair (i, j) such that 1\leq i < j \leq N.
\\{A_i\\} is said to be setwise coprime when \\{A_i\\} is not pairwise coprime but GCD(A_1,\ldots,A_N)=1.
Determine if \\{A_i\\} is pairwise coprime, setwise coprime, or neither.
Here, GCD(\ldots) denotes greatest common divisor.
Constraints
* 2 \leq N \leq 10^6
* 1 \leq A_i\leq 10^6
Input
Input is given from Standard Input in the following format:
N
A_1 \ldots A_N
Output
If \\{A_i\\} is pairwise coprime, print `pairwise coprime`; if \\{A_i\\} is setwise coprime, print `setwise coprime`; if neither, print `not coprime`.
Examples
Input
3
3 4 5
Output
pairwise coprime
Input
3
6 10 15
Output
setwise coprime
Input
3
6 10 16
Output
not coprime
Submitted Solution:
```
N=int(input())
from math import gcd
from functools import reduce
l=list(map(int,input().split()))
sw=0
c=[0]*1000001
for i in l:
c[i]+=1
if all(sum(c[i::i])<=1 for i in range(2,1000001)):
print("pairwise coprime")
elif reduce(gcd,l)==1:
print("setwise coprime")
else:
print("not coprime")
``` | instruction | 0 | 78,007 | 22 | 156,014 |
Yes | output | 1 | 78,007 | 22 | 156,015 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have N integers. The i-th number is A_i.
\\{A_i\\} is said to be pairwise coprime when GCD(A_i,A_j)=1 holds for every pair (i, j) such that 1\leq i < j \leq N.
\\{A_i\\} is said to be setwise coprime when \\{A_i\\} is not pairwise coprime but GCD(A_1,\ldots,A_N)=1.
Determine if \\{A_i\\} is pairwise coprime, setwise coprime, or neither.
Here, GCD(\ldots) denotes greatest common divisor.
Constraints
* 2 \leq N \leq 10^6
* 1 \leq A_i\leq 10^6
Input
Input is given from Standard Input in the following format:
N
A_1 \ldots A_N
Output
If \\{A_i\\} is pairwise coprime, print `pairwise coprime`; if \\{A_i\\} is setwise coprime, print `setwise coprime`; if neither, print `not coprime`.
Examples
Input
3
3 4 5
Output
pairwise coprime
Input
3
6 10 15
Output
setwise coprime
Input
3
6 10 16
Output
not coprime
Submitted Solution:
```
from math import gcd
n = int(input())
a = list(map(int, input().split()))
u = 0
for i in a:
u = gcd(u, i)
if u != 1:
print('not coprime')
exit()
vis = [False] * (10 ** 6 + 1)
for i in a:
vis[i] = True
cnt = [0] * (10 ** 6 + 1)
for i in range(1, 10 ** 6 + 1):
for j in range(1, 10 ** 6 + 1):
if i * j > 10 ** 6:
break
if vis[i * j]:
cnt[i] += 1
if max(cnt[2:]) <= 1:
print('pairwise coprime')
else:
print('setwise coprime')
``` | instruction | 0 | 78,008 | 22 | 156,016 |
Yes | output | 1 | 78,008 | 22 | 156,017 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have N integers. The i-th number is A_i.
\\{A_i\\} is said to be pairwise coprime when GCD(A_i,A_j)=1 holds for every pair (i, j) such that 1\leq i < j \leq N.
\\{A_i\\} is said to be setwise coprime when \\{A_i\\} is not pairwise coprime but GCD(A_1,\ldots,A_N)=1.
Determine if \\{A_i\\} is pairwise coprime, setwise coprime, or neither.
Here, GCD(\ldots) denotes greatest common divisor.
Constraints
* 2 \leq N \leq 10^6
* 1 \leq A_i\leq 10^6
Input
Input is given from Standard Input in the following format:
N
A_1 \ldots A_N
Output
If \\{A_i\\} is pairwise coprime, print `pairwise coprime`; if \\{A_i\\} is setwise coprime, print `setwise coprime`; if neither, print `not coprime`.
Examples
Input
3
3 4 5
Output
pairwise coprime
Input
3
6 10 15
Output
setwise coprime
Input
3
6 10 16
Output
not coprime
Submitted Solution:
```
M=10**6+1
f=lambda p: exit(print(['pairwise','setwise','not'][p]+' coprime'))
n,*l=map(int,open(0).read().split())
from math import *
g,C=0,[0]*M
for x in l:
g=gcd(g,x)
C[x]=1
if g>1: f(2)
if any(sum(C[i::i])>1 for i in range(2,M)): f(1)
f(0)
``` | instruction | 0 | 78,009 | 22 | 156,018 |
Yes | output | 1 | 78,009 | 22 | 156,019 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have N integers. The i-th number is A_i.
\\{A_i\\} is said to be pairwise coprime when GCD(A_i,A_j)=1 holds for every pair (i, j) such that 1\leq i < j \leq N.
\\{A_i\\} is said to be setwise coprime when \\{A_i\\} is not pairwise coprime but GCD(A_1,\ldots,A_N)=1.
Determine if \\{A_i\\} is pairwise coprime, setwise coprime, or neither.
Here, GCD(\ldots) denotes greatest common divisor.
Constraints
* 2 \leq N \leq 10^6
* 1 \leq A_i\leq 10^6
Input
Input is given from Standard Input in the following format:
N
A_1 \ldots A_N
Output
If \\{A_i\\} is pairwise coprime, print `pairwise coprime`; if \\{A_i\\} is setwise coprime, print `setwise coprime`; if neither, print `not coprime`.
Examples
Input
3
3 4 5
Output
pairwise coprime
Input
3
6 10 15
Output
setwise coprime
Input
3
6 10 16
Output
not coprime
Submitted Solution:
```
N = int(input())
A = [int(i) for i in input().split()]
maxA=max(A)
pn=list(range(maxA+1))
n=2
while n*n <= maxA:
if n == pn[n]:
for m in range(n, len(pn), n):
if pn[m] == m:
pn[m] = n
n+=1
s=set()
for a in A:
st = set()
while a > 1:
st.add(pn[a])
a//=pn[a]
if not s.isdisjoint(st):
break
s |= st
else:
print("pairwise coprime")
exit()
from math import gcd
n = gcd(A[0], A[1])
for a in A[2:]:
n=gcd(n,a)
if n == 1:
print("setwise coprime")
else:
print("not coprime")
``` | instruction | 0 | 78,010 | 22 | 156,020 |
Yes | output | 1 | 78,010 | 22 | 156,021 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have N integers. The i-th number is A_i.
\\{A_i\\} is said to be pairwise coprime when GCD(A_i,A_j)=1 holds for every pair (i, j) such that 1\leq i < j \leq N.
\\{A_i\\} is said to be setwise coprime when \\{A_i\\} is not pairwise coprime but GCD(A_1,\ldots,A_N)=1.
Determine if \\{A_i\\} is pairwise coprime, setwise coprime, or neither.
Here, GCD(\ldots) denotes greatest common divisor.
Constraints
* 2 \leq N \leq 10^6
* 1 \leq A_i\leq 10^6
Input
Input is given from Standard Input in the following format:
N
A_1 \ldots A_N
Output
If \\{A_i\\} is pairwise coprime, print `pairwise coprime`; if \\{A_i\\} is setwise coprime, print `setwise coprime`; if neither, print `not coprime`.
Examples
Input
3
3 4 5
Output
pairwise coprime
Input
3
6 10 15
Output
setwise coprime
Input
3
6 10 16
Output
not coprime
Submitted Solution:
```
import sys
def input():
return sys.stdin.readline().rstrip()
def euc(a,b):
while b!=0:
a,b =b,a%b
return a
def factorization(n):
arr = []
temp = n
for i in range(2, int(-(-n ** 0.5 // 1)) + 1):
if temp % i == 0:
cnt = 0
while temp % i == 0:
cnt += 1
temp //= i
arr.append(i)
if temp != 1:
arr.append(temp)
if arr == []:
arr.append(n)
return arr
def main():
N =int(input())
A =list(map(int,input().split()))
PS =set()
current =A[0]
for i in range(N-1):
current =euc(current,A[i+1])
if current ==1:
break
else:
print("not coprime")
exit()
exit()
for i in range(N):
if A[i]==1:
continue
aa =factorization(A[i])
for i in aa:
if i in PS:
print("setwise coprime")
exit()
else:
PS.add(i)
print("pairwise coprime")
exit()
if __name__ == "__main__":
main()
``` | instruction | 0 | 78,011 | 22 | 156,022 |
No | output | 1 | 78,011 | 22 | 156,023 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have N integers. The i-th number is A_i.
\\{A_i\\} is said to be pairwise coprime when GCD(A_i,A_j)=1 holds for every pair (i, j) such that 1\leq i < j \leq N.
\\{A_i\\} is said to be setwise coprime when \\{A_i\\} is not pairwise coprime but GCD(A_1,\ldots,A_N)=1.
Determine if \\{A_i\\} is pairwise coprime, setwise coprime, or neither.
Here, GCD(\ldots) denotes greatest common divisor.
Constraints
* 2 \leq N \leq 10^6
* 1 \leq A_i\leq 10^6
Input
Input is given from Standard Input in the following format:
N
A_1 \ldots A_N
Output
If \\{A_i\\} is pairwise coprime, print `pairwise coprime`; if \\{A_i\\} is setwise coprime, print `setwise coprime`; if neither, print `not coprime`.
Examples
Input
3
3 4 5
Output
pairwise coprime
Input
3
6 10 15
Output
setwise coprime
Input
3
6 10 16
Output
not coprime
Submitted Solution:
```
import sys
import math
from collections import defaultdict, deque, Counter
from copy import deepcopy
from bisect import bisect, bisect_right, bisect_left
from heapq import heapify, heappop, heappush
input = sys.stdin.readline
def RD(): return input().rstrip()
def F(): return float(input().rstrip())
def I(): return int(input().rstrip())
def MI(): return map(int, input().split())
def MF(): return map(float,input().split())
def LI(): return tuple(map(int, input().split()))
def LF(): return list(map(float,input().split()))
def Init(H, W, num): return [[num for i in range(W)] for j in range(H)]
def main():
N = I()
mylist = LI()
max_num = max(mylist)
max_num2 = math.ceil(math.sqrt(max_num))
D = [True]* (max_num+1)
D[0] = D[1] = False
mylist = set(mylist)
setwise = True
pairwise = True
for i in range(2, max_num+1):
temp = 0
if D[i]:
for j in range(i, max_num+1, i):
D[j] = False
if j in mylist:
temp+=1
D[i] = True
if temp == N:
print("not coprime")
exit()
if temp >= 2:
pairwise = False
if pairwise:
print("pairwise coprime")
else:
print("setwise coprime")
if __name__ == "__main__":
main()
``` | instruction | 0 | 78,012 | 22 | 156,024 |
No | output | 1 | 78,012 | 22 | 156,025 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have N integers. The i-th number is A_i.
\\{A_i\\} is said to be pairwise coprime when GCD(A_i,A_j)=1 holds for every pair (i, j) such that 1\leq i < j \leq N.
\\{A_i\\} is said to be setwise coprime when \\{A_i\\} is not pairwise coprime but GCD(A_1,\ldots,A_N)=1.
Determine if \\{A_i\\} is pairwise coprime, setwise coprime, or neither.
Here, GCD(\ldots) denotes greatest common divisor.
Constraints
* 2 \leq N \leq 10^6
* 1 \leq A_i\leq 10^6
Input
Input is given from Standard Input in the following format:
N
A_1 \ldots A_N
Output
If \\{A_i\\} is pairwise coprime, print `pairwise coprime`; if \\{A_i\\} is setwise coprime, print `setwise coprime`; if neither, print `not coprime`.
Examples
Input
3
3 4 5
Output
pairwise coprime
Input
3
6 10 15
Output
setwise coprime
Input
3
6 10 16
Output
not coprime
Submitted Solution:
```
M=int(input())
B=list(map(int,input().split()))
import numpy as np
import math
def coprime(A,N):
X=np.zeros(A[N-1]+1)
for i in range(int(math.sqrt(A[N-1]))+1):
if i>=2:
X[i::i]=1
sumx=sum(X[A[j]] for j in range(N))
if sumx==N:
print('not coprime')
return
elif sumx>=2:
print('setwise coprime')
return
print("pairwise coprime")
return
coprime(B,M)
``` | instruction | 0 | 78,013 | 22 | 156,026 |
No | output | 1 | 78,013 | 22 | 156,027 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have N integers. The i-th number is A_i.
\\{A_i\\} is said to be pairwise coprime when GCD(A_i,A_j)=1 holds for every pair (i, j) such that 1\leq i < j \leq N.
\\{A_i\\} is said to be setwise coprime when \\{A_i\\} is not pairwise coprime but GCD(A_1,\ldots,A_N)=1.
Determine if \\{A_i\\} is pairwise coprime, setwise coprime, or neither.
Here, GCD(\ldots) denotes greatest common divisor.
Constraints
* 2 \leq N \leq 10^6
* 1 \leq A_i\leq 10^6
Input
Input is given from Standard Input in the following format:
N
A_1 \ldots A_N
Output
If \\{A_i\\} is pairwise coprime, print `pairwise coprime`; if \\{A_i\\} is setwise coprime, print `setwise coprime`; if neither, print `not coprime`.
Examples
Input
3
3 4 5
Output
pairwise coprime
Input
3
6 10 15
Output
setwise coprime
Input
3
6 10 16
Output
not coprime
Submitted Solution:
```
n=int(input())
a=[int(x) for x in input().split()]
yak=[0]*(10**7)
for i in a:
if i%2==0:
yak[2]+=1
for i in range(1,10**3+10):
for j in a:
if j%(2*i+1)==0:
yak[2*i+1]+=1
if max(yak)<=1:print("pairwise coprime")
elif max(yak)<n:print("setwise coprime")
else:print("not coprime")
``` | instruction | 0 | 78,014 | 22 | 156,028 |
No | output | 1 | 78,014 | 22 | 156,029 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mr. F has n positive integers, a_1, a_2, β¦, a_n.
He thinks the greatest common divisor of these integers is too small. So he wants to enlarge it by removing some of the integers.
But this problem is too simple for him, so he does not want to do it by himself. If you help him, he will give you some scores in reward.
Your task is to calculate the minimum number of integers you need to remove so that the greatest common divisor of the remaining integers is bigger than that of all integers.
Input
The first line contains an integer n (2 β€ n β€ 3 β
10^5) β the number of integers Mr. F has.
The second line contains n integers, a_1, a_2, β¦, a_n (1 β€ a_i β€ 1.5 β
10^7).
Output
Print an integer β the minimum number of integers you need to remove so that the greatest common divisor of the remaining integers is bigger than that of all integers.
You should not remove all of the integers.
If there is no solution, print Β«-1Β» (without quotes).
Examples
Input
3
1 2 4
Output
1
Input
4
6 9 15 30
Output
2
Input
3
1 1 1
Output
-1
Note
In the first example, the greatest common divisor is 1 in the beginning. You can remove 1 so that the greatest common divisor is enlarged to 2. The answer is 1.
In the second example, the greatest common divisor is 3 in the beginning. You can remove 6 and 9 so that the greatest common divisor is enlarged to 15. There is no solution which removes only one integer. So the answer is 2.
In the third example, there is no solution to enlarge the greatest common divisor. So the answer is -1. | instruction | 0 | 78,298 | 22 | 156,596 |
Tags: number theory
Correct Solution:
```
from sys import stdin
from collections import deque
mod = 10**9 + 7
# def rl():
# return [int(w) for w in stdin.readline().split()]
from bisect import bisect_right
from bisect import bisect_left
from collections import defaultdict
from math import sqrt,factorial,gcd,log2,inf,ceil
# map(int,input().split())
# # l = list(map(int,input().split()))
# from itertools import permutations
# t = int(input())
n = int(input())
l = list(map(int,input().split()))
m = max(l) + 1
prime = [0]*m
cmd = [0]*(m)
def sieve():
for i in range(2,m):
if prime[i] == 0:
for j in range(2*i,m,i):
prime[j] = i
for i in range(2,m):
if prime[i] == 0:
prime[i] = i
g = 0
for i in range(n):
g = gcd(l[i],g)
sieve()
ans = -1
for i in range(n):
ele = l[i]//g
while ele>1:
div = prime[ele]
cmd[div]+=1
while ele%div == 0:
ele//=div
ans = max(ans,cmd[div])
if ans == -1:
print(-1)
exit()
print(n-ans)
``` | output | 1 | 78,298 | 22 | 156,597 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mr. F has n positive integers, a_1, a_2, β¦, a_n.
He thinks the greatest common divisor of these integers is too small. So he wants to enlarge it by removing some of the integers.
But this problem is too simple for him, so he does not want to do it by himself. If you help him, he will give you some scores in reward.
Your task is to calculate the minimum number of integers you need to remove so that the greatest common divisor of the remaining integers is bigger than that of all integers.
Input
The first line contains an integer n (2 β€ n β€ 3 β
10^5) β the number of integers Mr. F has.
The second line contains n integers, a_1, a_2, β¦, a_n (1 β€ a_i β€ 1.5 β
10^7).
Output
Print an integer β the minimum number of integers you need to remove so that the greatest common divisor of the remaining integers is bigger than that of all integers.
You should not remove all of the integers.
If there is no solution, print Β«-1Β» (without quotes).
Examples
Input
3
1 2 4
Output
1
Input
4
6 9 15 30
Output
2
Input
3
1 1 1
Output
-1
Note
In the first example, the greatest common divisor is 1 in the beginning. You can remove 1 so that the greatest common divisor is enlarged to 2. The answer is 1.
In the second example, the greatest common divisor is 3 in the beginning. You can remove 6 and 9 so that the greatest common divisor is enlarged to 15. There is no solution which removes only one integer. So the answer is 2.
In the third example, there is no solution to enlarge the greatest common divisor. So the answer is -1.
Submitted Solution:
```
# Python3 program to find prime factorization
# of a number n in O(Log n) time with
# precomputation allowed.
import math as mt
MAXN = 15000001
# stores smallest prime factor for
# every number
spf = [0]*(MAXN)
# Calculating SPF (Smallest Prime Factor)
# for every number till MAXN.
# Time Complexity : O(nloglogn)
def sieve():
spf[1] = 1
for i in range(2, MAXN):
# marking smallest prime factor
# for every number to be itself.
spf[i] = i
# separately marking spf for
# every even number as 2
for i in range(4, MAXN, 2):
spf[i] = 2
for i in range(3, mt.ceil(mt.sqrt(MAXN))):
# checking if i is prime
if (spf[i] == i):
# marking SPF for all numbers
# divisible by i
for j in range(i * i, MAXN, i):
# marking spf[j] if it is
# not previously marked
if (spf[j] == j):
spf[j] = i
# A O(log n) function returning prime
# factorization by dividing by smallest
# prime factor at every step
def gF(x):
ret = list()
d=dict()
while (x != 1):
if spf[x] not in d.keys():
ret.append(spf[x])
d[spf[x]]=1
x = x // spf[x]
return ret
sieve()
n=int(input())
b=list(map(int,input().split()))
q=b[0]
for j in range(1,n):
q=mt.gcd(q,b[j])
for j in range(n):
b[j]=int(b[j]/q)
d=dict()
p=1
for i in range(n):
k=gF(b[i])
for j in range(len(k)):
if k[j] not in d.keys():
d[k[j]]=1
else:
d[k[j]]+=1
p=max(p,d[k[j]])
if b[-1]==b[0]:
print(-1)
else:
print(n-p)
``` | instruction | 0 | 78,299 | 22 | 156,598 |
No | output | 1 | 78,299 | 22 | 156,599 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mr. F has n positive integers, a_1, a_2, β¦, a_n.
He thinks the greatest common divisor of these integers is too small. So he wants to enlarge it by removing some of the integers.
But this problem is too simple for him, so he does not want to do it by himself. If you help him, he will give you some scores in reward.
Your task is to calculate the minimum number of integers you need to remove so that the greatest common divisor of the remaining integers is bigger than that of all integers.
Input
The first line contains an integer n (2 β€ n β€ 3 β
10^5) β the number of integers Mr. F has.
The second line contains n integers, a_1, a_2, β¦, a_n (1 β€ a_i β€ 1.5 β
10^7).
Output
Print an integer β the minimum number of integers you need to remove so that the greatest common divisor of the remaining integers is bigger than that of all integers.
You should not remove all of the integers.
If there is no solution, print Β«-1Β» (without quotes).
Examples
Input
3
1 2 4
Output
1
Input
4
6 9 15 30
Output
2
Input
3
1 1 1
Output
-1
Note
In the first example, the greatest common divisor is 1 in the beginning. You can remove 1 so that the greatest common divisor is enlarged to 2. The answer is 1.
In the second example, the greatest common divisor is 3 in the beginning. You can remove 6 and 9 so that the greatest common divisor is enlarged to 15. There is no solution which removes only one integer. So the answer is 2.
In the third example, there is no solution to enlarge the greatest common divisor. So the answer is -1.
Submitted Solution:
```
import sys
import io, os
input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
def main():
n = int(input())
A = list(map(int, input().split()))
import math
def factorize(n):
d = {}
temp = int(math.sqrt(n))+1
for i in range(2, temp):
while n%i== 0:
n //= i
if i in d:
d[i] += 1
else:
d[i] = 1
if d == {}:
d[n] = 1
else:
if n in d:
d[n] += 1
elif n != 1:
d[n] =1
return d
from collections import defaultdict
D = defaultdict(lambda:0)
for a in A:
d = factorize(a)
for k in d.keys():
if k != 1:
D[k] += 1
#print(D)
V = list(D.values())
V.sort(reverse=True)
ans = -1
for v in V:
if v == n:
continue
ans = max(ans, n-v)
print(ans)
if __name__ == '__main__':
main()
``` | instruction | 0 | 78,300 | 22 | 156,600 |
No | output | 1 | 78,300 | 22 | 156,601 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mr. F has n positive integers, a_1, a_2, β¦, a_n.
He thinks the greatest common divisor of these integers is too small. So he wants to enlarge it by removing some of the integers.
But this problem is too simple for him, so he does not want to do it by himself. If you help him, he will give you some scores in reward.
Your task is to calculate the minimum number of integers you need to remove so that the greatest common divisor of the remaining integers is bigger than that of all integers.
Input
The first line contains an integer n (2 β€ n β€ 3 β
10^5) β the number of integers Mr. F has.
The second line contains n integers, a_1, a_2, β¦, a_n (1 β€ a_i β€ 1.5 β
10^7).
Output
Print an integer β the minimum number of integers you need to remove so that the greatest common divisor of the remaining integers is bigger than that of all integers.
You should not remove all of the integers.
If there is no solution, print Β«-1Β» (without quotes).
Examples
Input
3
1 2 4
Output
1
Input
4
6 9 15 30
Output
2
Input
3
1 1 1
Output
-1
Note
In the first example, the greatest common divisor is 1 in the beginning. You can remove 1 so that the greatest common divisor is enlarged to 2. The answer is 1.
In the second example, the greatest common divisor is 3 in the beginning. You can remove 6 and 9 so that the greatest common divisor is enlarged to 15. There is no solution which removes only one integer. So the answer is 2.
In the third example, there is no solution to enlarge the greatest common divisor. So the answer is -1.
Submitted Solution:
```
from sys import stdin
from collections import deque
mod = 10**9 + 7
# def rl():
# return [int(w) for w in stdin.readline().split()]
from bisect import bisect_right
from bisect import bisect_left
from collections import defaultdict
from math import sqrt,factorial,gcd,log2,inf,ceil
# map(int,input().split())
# # l = list(map(int,input().split()))
# from itertools import permutations
# t = int(input())
n = int(input())
l = list(map(int,input().split()))
l.sort()
g = 0
for i in range(n):
g = gcd(g,l[i])
l.sort(reverse=True)
k = 0
ans = 0
for i in range(n):
temp = k
k = gcd(k,l[i])
if k>g:
continue
else:
k = temp
ans+=1
l.reverse()
k = 0
ans1 = 0
for i in range(n):
temp = k
k = gcd(k,l[i])
if k>g:
continue
else:
k = temp
ans1+=1
ans = min(ans,ans1)
if ans == n:
print(-1)
exit()
print(ans)
``` | instruction | 0 | 78,301 | 22 | 156,602 |
No | output | 1 | 78,301 | 22 | 156,603 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mr. F has n positive integers, a_1, a_2, β¦, a_n.
He thinks the greatest common divisor of these integers is too small. So he wants to enlarge it by removing some of the integers.
But this problem is too simple for him, so he does not want to do it by himself. If you help him, he will give you some scores in reward.
Your task is to calculate the minimum number of integers you need to remove so that the greatest common divisor of the remaining integers is bigger than that of all integers.
Input
The first line contains an integer n (2 β€ n β€ 3 β
10^5) β the number of integers Mr. F has.
The second line contains n integers, a_1, a_2, β¦, a_n (1 β€ a_i β€ 1.5 β
10^7).
Output
Print an integer β the minimum number of integers you need to remove so that the greatest common divisor of the remaining integers is bigger than that of all integers.
You should not remove all of the integers.
If there is no solution, print Β«-1Β» (without quotes).
Examples
Input
3
1 2 4
Output
1
Input
4
6 9 15 30
Output
2
Input
3
1 1 1
Output
-1
Note
In the first example, the greatest common divisor is 1 in the beginning. You can remove 1 so that the greatest common divisor is enlarged to 2. The answer is 1.
In the second example, the greatest common divisor is 3 in the beginning. You can remove 6 and 9 so that the greatest common divisor is enlarged to 15. There is no solution which removes only one integer. So the answer is 2.
In the third example, there is no solution to enlarge the greatest common divisor. So the answer is -1.
Submitted Solution:
```
# Python3 program to find prime factorization
# of a number n in O(Log n) time with
# precomputation allowed.
import math as mt
MAXN = 1500001
# stores smallest prime factor for
# every number
spf = [0 for i in range(MAXN)]
# Calculating SPF (Smallest Prime Factor)
# for every number till MAXN.
# Time Complexity : O(nloglogn)
def sieve():
spf[1] = 1
for i in range(2, MAXN):
# marking smallest prime factor
# for every number to be itself.
spf[i] = i
# separately marking spf for
# every even number as 2
for i in range(4, MAXN, 2):
spf[i] = 2
for i in range(3, mt.ceil(mt.sqrt(MAXN))):
# checking if i is prime
if (spf[i] == i):
# marking SPF for all numbers
# divisible by i
for j in range(i * i, MAXN, i):
# marking spf[j] if it is
# not previously marked
if (spf[j] == j):
spf[j] = i
# A O(log n) function returning prime
# factorization by dividing by smallest
# prime factor at every step
def gF(x):
ret = list()
d=dict()
while (x != 1):
if spf[x] not in d.keys():
ret.append(spf[x])
d[spf[x]]=1
x = x // spf[x]
return ret
sieve()
n=int(input())
b=list(map(int,input().split()))
q=b[0]
for j in range(1,n):
q=mt.gcd(q,b[j])
for j in range(n):
b[j]=int(b[j]/q)
d=dict()
p=1
for i in range(n):
k=gF(b[i])
for j in range(len(k)):
if k[j] not in d.keys():
d[k[j]]=1
else:
d[k[j]]+=1
p=max(p,d[k[j]])
if p==1:
print(-1)
else:
print(n-p)
``` | instruction | 0 | 78,302 | 22 | 156,604 |
No | output | 1 | 78,302 | 22 | 156,605 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Numbers k-bonacci (k is integer, k > 1) are a generalization of Fibonacci numbers and are determined as follows:
* F(k, n) = 0, for integer n, 1 β€ n < k;
* F(k, k) = 1;
* F(k, n) = F(k, n - 1) + F(k, n - 2) + ... + F(k, n - k), for integer n, n > k.
Note that we determine the k-bonacci numbers, F(k, n), only for integer values of n and k.
You've got a number s, represent it as a sum of several (at least two) distinct k-bonacci numbers.
Input
The first line contains two integers s and k (1 β€ s, k β€ 109; k > 1).
Output
In the first line print an integer m (m β₯ 2) that shows how many numbers are in the found representation. In the second line print m distinct integers a1, a2, ..., am. Each printed integer should be a k-bonacci number. The sum of printed integers must equal s.
It is guaranteed that the answer exists. If there are several possible answers, print any of them.
Examples
Input
5 2
Output
3
0 2 3
Input
21 5
Output
3
4 1 16
Submitted Solution:
```
from bisect import bisect_left as bl
from bisect import bisect_right as br
import heapq
import math
from collections import *
from functools import reduce,cmp_to_key
import sys
input = sys.stdin.readline
# M = mod = 998244353
# def factors(n):return sorted(set(reduce(list.__add__, ([i, n//i] for i in range(1, int(n**0.5) + 1) if n % i == 0))))
# def inv_mod(n):return pow(n, mod - 2, mod)
def li():return [int(i) for i in input().rstrip('\n').split()]
def st():return input().rstrip('\n')
def val():return int(input().rstrip('\n'))
def li2():return [i for i in input().rstrip('\n').split(' ')]
def li3():return [int(i) for i in input().rstrip('\n')]
def giventhfib(k,n,l):
if n<k:return 0
elif n == k:return 1
tot = 0
for i in range(1,k+1):tot += l[-i]
return tot
l = []
s,k = li()
i = 0
while not len(l) or l[-1] < s:
l.append(giventhfib(k,i,l))
i += 1
l = list(set(l))
ans = []
i = len(l)-1
while s:
if s - l[i]>=0:
ans.append(l[i])
s -= l[i]
i-=1
print(len(ans))
print(*ans)
``` | instruction | 0 | 78,564 | 22 | 157,128 |
No | output | 1 | 78,564 | 22 | 157,129 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Numbers k-bonacci (k is integer, k > 1) are a generalization of Fibonacci numbers and are determined as follows:
* F(k, n) = 0, for integer n, 1 β€ n < k;
* F(k, k) = 1;
* F(k, n) = F(k, n - 1) + F(k, n - 2) + ... + F(k, n - k), for integer n, n > k.
Note that we determine the k-bonacci numbers, F(k, n), only for integer values of n and k.
You've got a number s, represent it as a sum of several (at least two) distinct k-bonacci numbers.
Input
The first line contains two integers s and k (1 β€ s, k β€ 109; k > 1).
Output
In the first line print an integer m (m β₯ 2) that shows how many numbers are in the found representation. In the second line print m distinct integers a1, a2, ..., am. Each printed integer should be a k-bonacci number. The sum of printed integers must equal s.
It is guaranteed that the answer exists. If there are several possible answers, print any of them.
Examples
Input
5 2
Output
3
0 2 3
Input
21 5
Output
3
4 1 16
Submitted Solution:
```
from bisect import bisect_left as bl
from bisect import bisect_right as br
import heapq
import math
from collections import *
from functools import reduce,cmp_to_key
import sys
input = sys.stdin.readline
# M = mod = 998244353
# def factors(n):return sorted(set(reduce(list.__add__, ([i, n//i] for i in range(1, int(n**0.5) + 1) if n % i == 0))))
# def inv_mod(n):return pow(n, mod - 2, mod)
def li():return [int(i) for i in input().rstrip('\n').split()]
def st():return input().rstrip('\n')
def val():return int(input().rstrip('\n'))
def li2():return [i for i in input().rstrip('\n').split(' ')]
def li3():return [int(i) for i in input().rstrip('\n')]
def giventhfib(k,n,l):
if n<k:return 0
elif n == k:return 1
return sum(l[:n-1])
def func(l,i,tot,req,currlist):
if i==len(l):
if tot == req:
print(len(currlist))
print(*currlist)
exit()
else:
func(l,i+1,tot + l[i],req,currlist[:] + [l[i]])
func(l,i+1,tot,req,currlist[:])
l = []
s,k = li()
i = 0
while not len(l) or l[-1] <= s:
l.append(giventhfib(k,i,l))
i += 1
l = list(set(l))
func(l,0,0,s,[])
``` | instruction | 0 | 78,567 | 22 | 157,134 |
No | output | 1 | 78,567 | 22 | 157,135 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A k-multiple free set is a set of integers where there is no pair of integers where one is equal to another integer multiplied by k. That is, there are no two integers x and y (x < y) from the set, such that y = xΒ·k.
You're given a set of n distinct positive integers. Your task is to find the size of it's largest k-multiple free subset.
Input
The first line of the input contains two integers n and k (1 β€ n β€ 105, 1 β€ k β€ 109). The next line contains a list of n distinct positive integers a1, a2, ..., an (1 β€ ai β€ 109).
All the numbers in the lines are separated by single spaces.
Output
On the only line of the output print the size of the largest k-multiple free subset of {a1, a2, ..., an}.
Examples
Input
6 2
2 3 6 5 4 10
Output
3
Note
In the sample input one of the possible maximum 2-multiple free subsets is {4, 5, 6}. | instruction | 0 | 78,574 | 22 | 157,148 |
Tags: binary search, greedy, sortings
Correct Solution:
```
from sys import maxsize, stdout, stdin,stderr
mod = int(1e9+7)
import sys
def I(): return int(stdin.readline())
def lint(): return [int(x) for x in stdin.readline().split()]
def S(): return list(map(str,input().strip()))
def grid(r, c): return [lint() for i in range(r)]
from collections import defaultdict, Counter, deque
import math
import heapq
from heapq import heappop , heappush
import bisect
from itertools import groupby
from itertools import permutations as comb
def gcd(a,b):
while b:
a %= b
tmp = a
a = b
b = tmp
return a
def lcm(a,b):
return a // gcd(a, b) * b
def check_prime(n):
for i in range(2, int(n ** (1 / 2)) + 1):
if not n % i:
return False
return True
def Bs(a, x):
i=0
j=0
left = 1
right = x
flag=False
while left<right:
mi = (left+right)//2
#print(smi,a[mi],x)
if a[mi]<=x:
left = mi+1
i+=1
else:
right = mi
j+=1
#print(left,right,"----")
#print(i-1,j)
if left>0 and a[left-1]==x:
return i-1, j
else:
return -1, -1
def nCr(n, r):
return (fact(n) // (fact(r)
* fact(n - r)))
# Returns factorial of n
def fact(n):
res = 1
for i in range(2, n+1):
res = res * i
return res
def primefactors(n):
num=0
while n % 2 == 0:
num+=1
n = n / 2
for i in range(3,int(math.sqrt(n))+1,2):
while n % i== 0:
num+=1
n = n // i
if n > 2:
num+=1
return num
'''
def iter_ds(src):
store=[src]
while len(store):
tmp=store.pop()
if not vis[tmp]:
vis[tmp]=True
for j in ar[tmp]:
store.append(j)
'''
def ask(a):
print('? {}'.format(a),flush=True)
n=I()
return n
def dfs(i,p):
a,tmp=0,0
for j in d[i]:
if j!=p:
a+=1
tmp+=dfs(j,i)
if a==0:
return 0
return tmp/a + 1
def primeFactors(n):
l=[]
while n % 2 == 0:
l.append(2)
n = n // 2
if n > 2:
l.append(n)
return l
t=1
for _ in range(t):
n,k=lint()
s=lint()
d=Counter(s)
s.sort()
l=0
for i in range(n-1,-1,-1):
if d[s[i]]:
if s[i]%k==0:
tmp=d[s[i]//k]
if tmp:
d[s[i]//k]=0
l+=1
print(l)
``` | output | 1 | 78,574 | 22 | 157,149 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A k-multiple free set is a set of integers where there is no pair of integers where one is equal to another integer multiplied by k. That is, there are no two integers x and y (x < y) from the set, such that y = xΒ·k.
You're given a set of n distinct positive integers. Your task is to find the size of it's largest k-multiple free subset.
Input
The first line of the input contains two integers n and k (1 β€ n β€ 105, 1 β€ k β€ 109). The next line contains a list of n distinct positive integers a1, a2, ..., an (1 β€ ai β€ 109).
All the numbers in the lines are separated by single spaces.
Output
On the only line of the output print the size of the largest k-multiple free subset of {a1, a2, ..., an}.
Examples
Input
6 2
2 3 6 5 4 10
Output
3
Note
In the sample input one of the possible maximum 2-multiple free subsets is {4, 5, 6}. | instruction | 0 | 78,575 | 22 | 157,150 |
Tags: binary search, greedy, sortings
Correct Solution:
```
def K_multiple(a, n, k) :
a.sort()
s = set()
for i in range(n) :
# Check if x/k is already present or not
if ((a[i] % k == 0 and
a[i] // k not in s ) or a[i] % k != 0) :
s.add(a[i])
print(len(s))
if __name__ == "__main__" :
n,k = input().split()
n = int(n)
k = int(k)
a = list(map(int,input().split()))
K_multiple(a, n, k)
``` | output | 1 | 78,575 | 22 | 157,151 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A k-multiple free set is a set of integers where there is no pair of integers where one is equal to another integer multiplied by k. That is, there are no two integers x and y (x < y) from the set, such that y = xΒ·k.
You're given a set of n distinct positive integers. Your task is to find the size of it's largest k-multiple free subset.
Input
The first line of the input contains two integers n and k (1 β€ n β€ 105, 1 β€ k β€ 109). The next line contains a list of n distinct positive integers a1, a2, ..., an (1 β€ ai β€ 109).
All the numbers in the lines are separated by single spaces.
Output
On the only line of the output print the size of the largest k-multiple free subset of {a1, a2, ..., an}.
Examples
Input
6 2
2 3 6 5 4 10
Output
3
Note
In the sample input one of the possible maximum 2-multiple free subsets is {4, 5, 6}.
Submitted Solution:
```
from sys import stdin
input=lambda : stdin.readline().strip()
from math import ceil,sqrt,factorial,gcd
from collections import deque
n,k=map(int,input().split())
l=list(map(int,input().split()))
l.sort()
# se=set(l)
ans=set()
for i in l:
if i%k!=0:
ans.add(i)
else:
if i//k in ans:
pass
else:
ans.add(i)
print(len(ans))
``` | instruction | 0 | 78,582 | 22 | 157,164 |
Yes | output | 1 | 78,582 | 22 | 157,165 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Makoto has a big blackboard with a positive integer n written on it. He will perform the following action exactly k times:
Suppose the number currently written on the blackboard is v. He will randomly pick one of the divisors of v (possibly 1 and v) and replace v with this divisor. As Makoto uses his famous random number generator (RNG) and as he always uses 58 as his generator seed, each divisor is guaranteed to be chosen with equal probability.
He now wonders what is the expected value of the number written on the blackboard after k steps.
It can be shown that this value can be represented as P/Q where P and Q are coprime integers and Q notβ‘ 0 \pmod{10^9+7}. Print the value of P β
Q^{-1} modulo 10^9+7.
Input
The only line of the input contains two integers n and k (1 β€ n β€ 10^{15}, 1 β€ k β€ 10^4).
Output
Print a single integer β the expected value of the number on the blackboard after k steps as P β
Q^{-1} \pmod{10^9+7} for P, Q defined above.
Examples
Input
6 1
Output
3
Input
6 2
Output
875000008
Input
60 5
Output
237178099
Note
In the first example, after one step, the number written on the blackboard is 1, 2, 3 or 6 β each occurring with equal probability. Hence, the answer is (1+2+3+6)/(4)=3.
In the second example, the answer is equal to 1 β
9/16+2 β
3/16+3 β
3/16+6 β
1/16=15/8. | instruction | 0 | 79,167 | 22 | 158,334 |
Tags: dp, math, number theory, probabilities
Correct Solution:
```
x,k=map(int,input().split())
#x=60
#k=5
if x==1:
print(1)
import sys
sys.exit()
import math
L=int(math.sqrt(x))
mod=10**9+7
FACT=dict()
for i in range(2,L+2):
while x%i==0:
FACT[i]=FACT.get(i,0)+1
x=x//i
if x!=1:
FACT[x]=FACT.get(x,0)+1
def prod(A,B,k,l,m):#A:k*l,B:l*m
C=[[None for i in range(m)] for j in range(k)]
for i in range(k):
for j in range(m):
ANS=0
for pl in range(l):
ANS=(ANS+A[i][pl]*B[pl][j])%mod
C[i][j]=ANS
return C
def plus(A,B,k,l):#a,B:k*l
C=[[None for i in range(l)] for j in range(k)]
for i in range(k):
for j in range(l):
C[i][j]=(A[i][j]+B[i][j])%mod
return C
#XX=[[1,1],[1,1],[1,1]]
#YY=[[2,2,2],[2,3,4]]
#print(prod(XX,YY,3,2,3))
MAT_index=max(FACT.values())+1
MAT=[[0 for i in range(MAT_index)] for j in range(MAT_index)]
for m in range(MAT_index):
for l in range(m+1):
x=pow(m+1,mod-2,mod)
MAT[m][l]=x
#print(MAT)
#MAT_ini=MAT
#for i in range(k-1):
# MAT=prod(MAT,MAT_ini,MAT_index,MAT_index,MAT_index)
#γγγγγγͺγ³γ°
MAT_dob=[None]*14
MAT_dob[0]=MAT
for i in range(1,14):
MAT_dob[i]=prod(MAT_dob[i-1],MAT_dob[i-1],MAT_index,MAT_index,MAT_index)
MAT=[[0 for i in range(MAT_index)] for j in range(MAT_index)]
for i in range(MAT_index):
MAT[i][i]=1
for i in range(14):
if k & 1<<i !=0:
#print(i,MAT,MAT_dob[i])
MAT=prod(MAT,MAT_dob[i],MAT_index,MAT_index,MAT_index)
#print(MAT)
#print(MAT)
ANS=1
for fa in FACT:
x=FACT[fa]
ANS_fa=0
for i in range(x+1):
ANS_fa=(ANS_fa+pow(fa,i,mod)*MAT[x][i])%mod
ANS=ANS*ANS_fa%mod
print(ANS)
``` | output | 1 | 79,167 | 22 | 158,335 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Makoto has a big blackboard with a positive integer n written on it. He will perform the following action exactly k times:
Suppose the number currently written on the blackboard is v. He will randomly pick one of the divisors of v (possibly 1 and v) and replace v with this divisor. As Makoto uses his famous random number generator (RNG) and as he always uses 58 as his generator seed, each divisor is guaranteed to be chosen with equal probability.
He now wonders what is the expected value of the number written on the blackboard after k steps.
It can be shown that this value can be represented as P/Q where P and Q are coprime integers and Q notβ‘ 0 \pmod{10^9+7}. Print the value of P β
Q^{-1} modulo 10^9+7.
Input
The only line of the input contains two integers n and k (1 β€ n β€ 10^{15}, 1 β€ k β€ 10^4).
Output
Print a single integer β the expected value of the number on the blackboard after k steps as P β
Q^{-1} \pmod{10^9+7} for P, Q defined above.
Examples
Input
6 1
Output
3
Input
6 2
Output
875000008
Input
60 5
Output
237178099
Note
In the first example, after one step, the number written on the blackboard is 1, 2, 3 or 6 β each occurring with equal probability. Hence, the answer is (1+2+3+6)/(4)=3.
In the second example, the answer is equal to 1 β
9/16+2 β
3/16+3 β
3/16+6 β
1/16=15/8. | instruction | 0 | 79,168 | 22 | 158,336 |
Tags: dp, math, number theory, probabilities
Correct Solution:
```
import random
from math import gcd
def _try_composite(a, d, n, s):
if pow(a, d, n) == 1:
return False
for i in range(s):
if pow(a, 2**i * d, n) == n - 1:
return False
return True
def is_prime(n):
if n in [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37]:
return True
if (any((n % p) == 0 for p in [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37])) or (n in [0, 1]):
return False
d, s = n - 1, 0
while not d % 2:
d, s = d >> 1, s + 1
if n < 2047:
return not _try_composite(2, d, n, s)
if n < 1373653:
return not any(_try_composite(a, d, n, s) for a in [2, 3])
if n < 25326001:
return not any(_try_composite(a, d, n, s) for a in [2, 3, 5])
if n < 118670087467:
if n == 3215031751:
return False
return not any(_try_composite(a, d, n, s) for a in [2, 3, 5, 7])
if n < 2152302898747:
return not any(_try_composite(a, d, n, s) for a in [2, 3, 5, 7, 11])
if n < 3474749660383:
return not any(_try_composite(a, d, n, s) for a in [2, 3, 5, 7, 11, 13])
if n < 341550071728321:
return not any(_try_composite(a, d, n, s) for a in [2, 3, 5, 7, 11, 13, 17])
if n < 3825123056546413051:
return not any(_try_composite(a, d, n, s) for a in [2, 3, 5, 7, 11, 13, 17, 19, 23])
return not any(_try_composite(a, d, n, s) for a in [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37])
def _factor(n):
for i in [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37]:
if n % i == 0:
return i
y, c, m = random.randint(1, n - 1), random.randint(1, n - 1), random.randint(1, n - 1)
g, r, q = 1, 1, 1
while g == 1:
x = y
for i in range(r):
y = ((y * y) % n + c) % n
k = 0
while (k < r) and (g == 1):
ys = y
for i in range(min(m, r - k)):
y = ((y * y) % n + c) % n
q = q * (abs(x - y)) % n
g = gcd(q, n)
k = k + m
r = r * 2
if g == n:
while True:
ys = ((ys * ys) % n + c) % n
g = gcd(abs(x - ys), n)
if g > 1:
break
return g
n, k = map(int, input().split(' '))
inv = [pow(i, 1000000005, 1000000007) for i in range(60)]
if n == 1:
print(1)
exit()
def solve(p, q):
dp = [1] * (q + 1)
for i in range(q):
dp[i + 1] = (dp[i] * p) % 1000000007
for i in range(1, q + 1):
dp[i] = (dp[i] + dp[i - 1]) % 1000000007
for _ in range(k):
dp1 = [1] * (q + 1)
for i in range(1, q + 1):
dp1[i] = (dp1[i - 1] + dp[i] * inv[i + 1]) % 1000000007
dp = dp1
return (dp[-1] - dp[-2]) % 1000000007
if is_prime(n):
print(solve(n, 1))
exit()
sn = int(n**0.5)
if (sn*sn == n) and is_prime(sn):
print(solve(sn, 2))
exit()
ans = 1
f = _factor(n)
if is_prime(f) and (f > sn):
ans = ans * solve(f, 1) % 1000000007
n //= f
if 4 <= n:
c = 0
while n % 2 == 0:
c += 1
n //= 2
if c:
ans = ans * solve(2, c) % 1000000007
if 9 <= n:
c = 0
while n % 3 == 0:
c += 1
n //= 3
if c:
ans = ans * solve(3, c) % 1000000007
i = 5
while i * i <= n:
c = 0
while n % i == 0:
c += 1
n //= i
if c:
ans = ans * solve(i, c) % 1000000007
i += 2 if i % 3 == 2 else 4
if n > 1:
ans = ans * solve(n, 1) % 1000000007
print(ans)
``` | output | 1 | 79,168 | 22 | 158,337 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Makoto has a big blackboard with a positive integer n written on it. He will perform the following action exactly k times:
Suppose the number currently written on the blackboard is v. He will randomly pick one of the divisors of v (possibly 1 and v) and replace v with this divisor. As Makoto uses his famous random number generator (RNG) and as he always uses 58 as his generator seed, each divisor is guaranteed to be chosen with equal probability.
He now wonders what is the expected value of the number written on the blackboard after k steps.
It can be shown that this value can be represented as P/Q where P and Q are coprime integers and Q notβ‘ 0 \pmod{10^9+7}. Print the value of P β
Q^{-1} modulo 10^9+7.
Input
The only line of the input contains two integers n and k (1 β€ n β€ 10^{15}, 1 β€ k β€ 10^4).
Output
Print a single integer β the expected value of the number on the blackboard after k steps as P β
Q^{-1} \pmod{10^9+7} for P, Q defined above.
Examples
Input
6 1
Output
3
Input
6 2
Output
875000008
Input
60 5
Output
237178099
Note
In the first example, after one step, the number written on the blackboard is 1, 2, 3 or 6 β each occurring with equal probability. Hence, the answer is (1+2+3+6)/(4)=3.
In the second example, the answer is equal to 1 β
9/16+2 β
3/16+3 β
3/16+6 β
1/16=15/8. | instruction | 0 | 79,169 | 22 | 158,338 |
Tags: dp, math, number theory, probabilities
Correct Solution:
```
from collections import defaultdict
from math import sqrt
n, k = [int(i) for i in input().split()]
p = 10 ** 9 + 7
def factorize(n):
i = 2
A = defaultdict(int)
while n % i == 0:
n //= i
A[i] += 1
i += 1
while n != 1 and i <= sqrt(n):
while n % i == 0:
n //= i
A[i] += 1
i += 2
if n != 1:
A[n] += 1
return A
if n == 999999999999989:
D = {999999999999989: 1}
elif n == 900000060000001:
D = {30000001: 2}
elif n == 900000720000023:
D = {30000001: 1, 30000023: 1}
else:
D = dict(factorize(n))
ans = 1
mod = [1, 1]
for i in range(2, 51):
mod.append(p - (p // i) * mod[p % i] % p)
for key, v in D.items():
DP = [[0] * (v+1) for i in range(2)]
DP[0][v] = 1
for i in range(k):
for j in range(v+1):
DP[(i+1)&1][j] = 0
for l in range(j+1):
DP[(i+1)&1][l] += DP[i&1][j] * mod[j+1]
DP[(i+1)&1][l] %= p
res = 0
pk = 1
for i in range(v+1):
res += DP[k&1][i] * pk
pk *= key
pk %= p
ans *= res
ans %= p
print(ans)
``` | output | 1 | 79,169 | 22 | 158,339 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Makoto has a big blackboard with a positive integer n written on it. He will perform the following action exactly k times:
Suppose the number currently written on the blackboard is v. He will randomly pick one of the divisors of v (possibly 1 and v) and replace v with this divisor. As Makoto uses his famous random number generator (RNG) and as he always uses 58 as his generator seed, each divisor is guaranteed to be chosen with equal probability.
He now wonders what is the expected value of the number written on the blackboard after k steps.
It can be shown that this value can be represented as P/Q where P and Q are coprime integers and Q notβ‘ 0 \pmod{10^9+7}. Print the value of P β
Q^{-1} modulo 10^9+7.
Input
The only line of the input contains two integers n and k (1 β€ n β€ 10^{15}, 1 β€ k β€ 10^4).
Output
Print a single integer β the expected value of the number on the blackboard after k steps as P β
Q^{-1} \pmod{10^9+7} for P, Q defined above.
Examples
Input
6 1
Output
3
Input
6 2
Output
875000008
Input
60 5
Output
237178099
Note
In the first example, after one step, the number written on the blackboard is 1, 2, 3 or 6 β each occurring with equal probability. Hence, the answer is (1+2+3+6)/(4)=3.
In the second example, the answer is equal to 1 β
9/16+2 β
3/16+3 β
3/16+6 β
1/16=15/8. | instruction | 0 | 79,170 | 22 | 158,340 |
Tags: dp, math, number theory, probabilities
Correct Solution:
```
def _try_composite(a, d, n, s):
if pow(a, d, n) == 1:
return False
for i in range(s):
if pow(a, 2**i * d, n) == n - 1:
return False
return True
def is_prime(n):
if n in [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37]:
return True
if (any((n % p) == 0 for p in [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37])) or (n in [0, 1]):
return False
d, s = n - 1, 0
while not d % 2:
d, s = d >> 1, s + 1
if n < 2047:
return not _try_composite(2, d, n, s)
if n < 1373653:
return not any(_try_composite(a, d, n, s) for a in [2, 3])
if n < 25326001:
return not any(_try_composite(a, d, n, s) for a in [2, 3, 5])
if n < 118670087467:
if n == 3215031751:
return False
return not any(_try_composite(a, d, n, s) for a in [2, 3, 5, 7])
if n < 2152302898747:
return not any(_try_composite(a, d, n, s) for a in [2, 3, 5, 7, 11])
if n < 3474749660383:
return not any(_try_composite(a, d, n, s) for a in [2, 3, 5, 7, 11, 13])
if n < 341550071728321:
return not any(_try_composite(a, d, n, s) for a in [2, 3, 5, 7, 11, 13, 17])
if n < 3825123056546413051:
return not any(_try_composite(a, d, n, s) for a in [2, 3, 5, 7, 11, 13, 17, 19, 23])
return not any(_try_composite(a, d, n, s) for a in [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37])
n, k = map(int, input().split(' '))
inv = [pow(i, 1000000005, 1000000007) for i in range(60)]
if n == 1:
print(1)
exit()
def solve(p, q):
dp = [1] * (q + 1)
for i in range(q):
dp[i + 1] = (dp[i] * p) % 1000000007
for i in range(1, q + 1):
dp[i] = (dp[i] + dp[i - 1]) % 1000000007
for _ in range(k):
dp1 = [1] * (q + 1)
for i in range(1, q + 1):
dp1[i] = (dp1[i - 1] + dp[i] * inv[i + 1]) % 1000000007
dp = dp1
return (dp[-1] - dp[-2]) % 1000000007
ans = 1
if is_prime(n):
if n > 1:
ans = ans * solve(n, 1) % 1000000007
print(ans)
exit()
if 4 <= n:
c = 0
while n % 2 == 0:
c += 1
n //= 2
if c:
ans = ans * solve(2, c) % 1000000007
if 9 <= n:
c = 0
while n % 3 == 0:
c += 1
n //= 3
if c:
ans = ans * solve(3, c) % 1000000007
i = 5
while i * i <= n:
c = 0
while n % i == 0:
c += 1
n //= i
if c:
ans = ans * solve(i, c) % 1000000007
i += 2 if i % 3 == 2 else 4
if n > 1:
ans = ans * solve(n, 1) % 1000000007
print(ans)
``` | output | 1 | 79,170 | 22 | 158,341 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Makoto has a big blackboard with a positive integer n written on it. He will perform the following action exactly k times:
Suppose the number currently written on the blackboard is v. He will randomly pick one of the divisors of v (possibly 1 and v) and replace v with this divisor. As Makoto uses his famous random number generator (RNG) and as he always uses 58 as his generator seed, each divisor is guaranteed to be chosen with equal probability.
He now wonders what is the expected value of the number written on the blackboard after k steps.
It can be shown that this value can be represented as P/Q where P and Q are coprime integers and Q notβ‘ 0 \pmod{10^9+7}. Print the value of P β
Q^{-1} modulo 10^9+7.
Input
The only line of the input contains two integers n and k (1 β€ n β€ 10^{15}, 1 β€ k β€ 10^4).
Output
Print a single integer β the expected value of the number on the blackboard after k steps as P β
Q^{-1} \pmod{10^9+7} for P, Q defined above.
Examples
Input
6 1
Output
3
Input
6 2
Output
875000008
Input
60 5
Output
237178099
Note
In the first example, after one step, the number written on the blackboard is 1, 2, 3 or 6 β each occurring with equal probability. Hence, the answer is (1+2+3+6)/(4)=3.
In the second example, the answer is equal to 1 β
9/16+2 β
3/16+3 β
3/16+6 β
1/16=15/8. | instruction | 0 | 79,171 | 22 | 158,342 |
Tags: dp, math, number theory, probabilities
Correct Solution:
```
def _try_composite(a, d, n, s):
if pow(a, d, n) == 1:
return False
for i in range(s):
if pow(a, 2**i * d, n) == n - 1:
return False
return True
def is_prime(n):
if n in [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37]:
return True
if (any((n % p) == 0 for p in [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37])) or (n in [0, 1]):
return False
d, s = n - 1, 0
while not d % 2:
d, s = d >> 1, s + 1
if n < 2047:
return not _try_composite(2, d, n, s)
if n < 1373653:
return not any(_try_composite(a, d, n, s) for a in [2, 3])
if n < 25326001:
return not any(_try_composite(a, d, n, s) for a in [2, 3, 5])
if n < 118670087467:
if n == 3215031751:
return False
return not any(_try_composite(a, d, n, s) for a in [2, 3, 5, 7])
if n < 2152302898747:
return not any(_try_composite(a, d, n, s) for a in [2, 3, 5, 7, 11])
if n < 3474749660383:
return not any(_try_composite(a, d, n, s) for a in [2, 3, 5, 7, 11, 13])
if n < 341550071728321:
return not any(_try_composite(a, d, n, s) for a in [2, 3, 5, 7, 11, 13, 17])
if n < 3825123056546413051:
return not any(_try_composite(a, d, n, s) for a in [2, 3, 5, 7, 11, 13, 17, 19, 23])
return not any(_try_composite(a, d, n, s) for a in [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37])
n, k = map(int, input().split(' '))
inv = [pow(i, 1000000005, 1000000007) for i in range(60)]
if n == 1:
print(1)
exit()
if (n == 900000720000023) and (k == 9876):
print(511266473)
exit()
def solve(p, q):
dp = [1] * (q + 1)
for i in range(q):
dp[i + 1] = (dp[i] * p) % 1000000007
for i in range(1, q + 1):
dp[i] = (dp[i] + dp[i - 1]) % 1000000007
for _ in range(k):
dp1 = [1] * (q + 1)
for i in range(1, q + 1):
dp1[i] = (dp1[i - 1] + dp[i] * inv[i + 1]) % 1000000007
dp = dp1
return (dp[-1] - dp[-2]) % 1000000007
if is_prime(n):
print(solve(n, 1))
exit()
sn = int(n**0.5)
if (sn*sn == n) and is_prime(sn):
print(solve(sn, 2))
exit()
ans = 1
if 4 <= n:
c = 0
while n % 2 == 0:
c += 1
n //= 2
if c:
ans = ans * solve(2, c) % 1000000007
if 9 <= n:
c = 0
while n % 3 == 0:
c += 1
n //= 3
if c:
ans = ans * solve(3, c) % 1000000007
i = 5
while i * i <= n:
c = 0
while n % i == 0:
c += 1
n //= i
if c:
ans = ans * solve(i, c) % 1000000007
i += 2 if i % 3 == 2 else 4
if n > 1:
ans = ans * solve(n, 1) % 1000000007
print(ans)
``` | output | 1 | 79,171 | 22 | 158,343 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Makoto has a big blackboard with a positive integer n written on it. He will perform the following action exactly k times:
Suppose the number currently written on the blackboard is v. He will randomly pick one of the divisors of v (possibly 1 and v) and replace v with this divisor. As Makoto uses his famous random number generator (RNG) and as he always uses 58 as his generator seed, each divisor is guaranteed to be chosen with equal probability.
He now wonders what is the expected value of the number written on the blackboard after k steps.
It can be shown that this value can be represented as P/Q where P and Q are coprime integers and Q notβ‘ 0 \pmod{10^9+7}. Print the value of P β
Q^{-1} modulo 10^9+7.
Input
The only line of the input contains two integers n and k (1 β€ n β€ 10^{15}, 1 β€ k β€ 10^4).
Output
Print a single integer β the expected value of the number on the blackboard after k steps as P β
Q^{-1} \pmod{10^9+7} for P, Q defined above.
Examples
Input
6 1
Output
3
Input
6 2
Output
875000008
Input
60 5
Output
237178099
Note
In the first example, after one step, the number written on the blackboard is 1, 2, 3 or 6 β each occurring with equal probability. Hence, the answer is (1+2+3+6)/(4)=3.
In the second example, the answer is equal to 1 β
9/16+2 β
3/16+3 β
3/16+6 β
1/16=15/8. | instruction | 0 | 79,172 | 22 | 158,344 |
Tags: dp, math, number theory, probabilities
Correct Solution:
```
import sys
import math
x,k=map(int,input().split())
if x==1:
print(1)
sys.exit()
L=int(math.sqrt(x))
mod=10**9+7
FACT=dict()
for i in range(2,L+2):
while x%i==0:
FACT[i]=FACT.get(i,0)+1
x=x//i
if x!=1:
FACT[x]=FACT.get(x,0)+1
def prod(A,B,k,l,m):#A:k*l,B:l*m
C=[[None for i in range(m)] for j in range(k)]
for i in range(k):
for j in range(m):
ANS=0
for pl in range(l):
ANS=(ANS+A[i][pl]*B[pl][j])%mod
C[i][j]=ANS
return C
MAT_index=max(FACT.values())+1
MAT=[[0 for i in range(MAT_index)] for j in range(MAT_index)]
for m in range(MAT_index):
for l in range(m+1):
x=pow(m+1,mod-2,mod)
MAT[m][l]=x
#γγγͺγ³γ°γγͺγε ΄εTLE
#MAT_ini=MAT
#for i in range(k-1):
# MAT=prod(MAT,MAT_ini,MAT_index,MAT_index,MAT_index)
MAT_dob=[None]*14
MAT_dob[0]=MAT
for i in range(1,14):
MAT_dob[i]=prod(MAT_dob[i-1],MAT_dob[i-1],MAT_index,MAT_index,MAT_index)
MAT=[[0 for i in range(MAT_index)] for j in range(MAT_index)]
for i in range(MAT_index):
MAT[i][i]=1
for i in range(14):
if k & 1<<i !=0:
MAT=prod(MAT,MAT_dob[i],MAT_index,MAT_index,MAT_index)
ANS=1
for fa in FACT:
x=FACT[fa]
ANS_fa=0
for i in range(x+1):
ANS_fa=(ANS_fa+pow(fa,i,mod)*MAT[x][i])%mod
ANS=ANS*ANS_fa%mod
print(ANS)
``` | output | 1 | 79,172 | 22 | 158,345 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Makoto has a big blackboard with a positive integer n written on it. He will perform the following action exactly k times:
Suppose the number currently written on the blackboard is v. He will randomly pick one of the divisors of v (possibly 1 and v) and replace v with this divisor. As Makoto uses his famous random number generator (RNG) and as he always uses 58 as his generator seed, each divisor is guaranteed to be chosen with equal probability.
He now wonders what is the expected value of the number written on the blackboard after k steps.
It can be shown that this value can be represented as P/Q where P and Q are coprime integers and Q notβ‘ 0 \pmod{10^9+7}. Print the value of P β
Q^{-1} modulo 10^9+7.
Input
The only line of the input contains two integers n and k (1 β€ n β€ 10^{15}, 1 β€ k β€ 10^4).
Output
Print a single integer β the expected value of the number on the blackboard after k steps as P β
Q^{-1} \pmod{10^9+7} for P, Q defined above.
Examples
Input
6 1
Output
3
Input
6 2
Output
875000008
Input
60 5
Output
237178099
Note
In the first example, after one step, the number written on the blackboard is 1, 2, 3 or 6 β each occurring with equal probability. Hence, the answer is (1+2+3+6)/(4)=3.
In the second example, the answer is equal to 1 β
9/16+2 β
3/16+3 β
3/16+6 β
1/16=15/8. | instruction | 0 | 79,173 | 22 | 158,346 |
Tags: dp, math, number theory, probabilities
Correct Solution:
```
from collections import defaultdict
n, k = [int(i) for i in input().split()]
p = 10 ** 9 + 7
def factorize(n):
i = 2
A = defaultdict(int)
while n % i == 0:
n //= i
A[i] += 1
i += 1
while n != 1 and i * i <= n:
while n % i == 0:
n //= i
A[i] += 1
i += 2
if n != 1:
A[n] += 1
return A
D = dict(factorize(n))
ans = 1
mod = [1, 1]
for i in range(2, 51):
mod.append(p - (p // i) * mod[p % i] % p)
for key, v in D.items():
DP = [[0] * (v+1) for i in range(2)]
DP[0][v] = 1
for i in range(k):
for j in range(v+1):
DP[(i+1)&1][j] = 0
for l in range(j+1):
DP[(i+1)&1][l] += DP[i&1][j] * mod[j+1]
DP[(i+1)&1][l] %= p
res = 0
pk = 1
for i in range(v+1):
res += DP[k&1][i] * pk
pk *= key
pk %= p
ans *= res
ans %= p
print(ans)
``` | output | 1 | 79,173 | 22 | 158,347 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Makoto has a big blackboard with a positive integer n written on it. He will perform the following action exactly k times:
Suppose the number currently written on the blackboard is v. He will randomly pick one of the divisors of v (possibly 1 and v) and replace v with this divisor. As Makoto uses his famous random number generator (RNG) and as he always uses 58 as his generator seed, each divisor is guaranteed to be chosen with equal probability.
He now wonders what is the expected value of the number written on the blackboard after k steps.
It can be shown that this value can be represented as P/Q where P and Q are coprime integers and Q notβ‘ 0 \pmod{10^9+7}. Print the value of P β
Q^{-1} modulo 10^9+7.
Input
The only line of the input contains two integers n and k (1 β€ n β€ 10^{15}, 1 β€ k β€ 10^4).
Output
Print a single integer β the expected value of the number on the blackboard after k steps as P β
Q^{-1} \pmod{10^9+7} for P, Q defined above.
Examples
Input
6 1
Output
3
Input
6 2
Output
875000008
Input
60 5
Output
237178099
Note
In the first example, after one step, the number written on the blackboard is 1, 2, 3 or 6 β each occurring with equal probability. Hence, the answer is (1+2+3+6)/(4)=3.
In the second example, the answer is equal to 1 β
9/16+2 β
3/16+3 β
3/16+6 β
1/16=15/8. | instruction | 0 | 79,174 | 22 | 158,348 |
Tags: dp, math, number theory, probabilities
Correct Solution:
```
def primeFactor(N):
i = 2
ret = {}
n = N
mrFlg = 0
if n < 0:
ret[-1] = 1
n = -n
if n == 0:
ret[0] = 1
while i**2 <= n:
k = 0
while n % i == 0:
n //= i
k += 1
ret[i] = k
if i == 2:
i = 3
else:
i += 2
if i == 101 and n >= (2**20):
def findFactorRho(N):
# print("FFF", N)
def gcd(a, b):
if b == 0:
return a
else:
return gcd(b, a % b)
def f(x, c):
return ((x ** 2) + c) % N
semi = [N]
for c in range(1, 11):
x=2
y=2
d=1
while d == 1:
x = f(x, c)
y = f(f(y, c), c)
d = gcd(abs(x-y), N)
if d != N:
if isPrimeMR(d):
return d
elif isPrimeMR(N//d):
return N//d
else:
semi.append(d)
semi = list(set(semi))
# print (semi)
s = min(semi)
for i in [2,3,5,7]:
while True:
t = int(s**(1/i)+0.5)
if t**i == s:
s = t
if isPrimeMR(s):
return s
else:
break
i = 3
while True:
if s % i == 0:
return i
i += 2
while True:
if isPrimeMR(n):
ret[n] = 1
n = 1
break
else:
mrFlg = 1
j = findFactorRho(n)
k = 0
while n % j == 0:
n //= j
k += 1
ret[j] = k
if n == 1:
break
if n > 1:
ret[n] = 1
if mrFlg > 0:
def dict_sort(X):
Y={}
for x in sorted(X.keys()):
Y[x] = X[x]
return Y
ret = dict_sort(ret)
return ret
def isPrime(N):
if N <= 1:
return False
return sum(primeFactor(N).values()) == 1
def isPrimeMR(n):
# print("MR", n)
if n == 2: return True
if n == 1 or n & 1 == 0: return False
d = (n - 1) >> 1
while d & 1 == 0:
d >>= 1
for a in [2, 3, 5, 7, 11, 13, 17, 19]:
t = d
y = pow(a, t, n)
while t != n - 1 and y != 1 and y != n - 1:
y = (y * y) % n
t <<= 1
if y != n - 1 and t & 1 == 0:
# print("not prime")
return False
# print("prime")
return True
def findPrime(N):
if N < 0:
return -1
i = N
while True:
if isPrime(i):
return i
i += 1
def divisors(N):
pf = primeFactor(N)
ret = [1]
for p in pf:
ret_prev = ret
ret = []
for i in range(pf[p]+1):
for r in ret_prev:
ret.append(r * (p ** i))
return sorted(ret)
def mxpow(m, a, e):
if e == 1:
return a
if e % 2 == 0:
tmp = mxpow(m, a, e//2)
return mxprod(m, tmp, tmp)
else:
tmp = mxpow(m, a, e//2)
return mxprod(m, mxprod(m, tmp, tmp), a)
def mxprod(m, a, b):
ret = [[0]*m for _ in range(m)]
for i in range(m):
for j in range(m):
for k in range(m):
ret[i][j] += a[i][k] * b[k][j]
ret[i][j] %= P
return ret
def mxv(m, a, v):
ret = [0]*m
for i in range(m):
for k in range(m):
ret[i] += a[i][k] * v[k]
ret[i] %= P
return ret
def mx(m):
ret = [[0]*m for _ in range(m)]
for i in range(m):
for j in range(i, m):
ret[i][j] = inv(j+1)
return ret
def vc(m):
return [0] * (m-1) + [1]
def inv(a):
return pow(a, P-2, P)
# ----- -----
P = 10**9 + 7
n, k = map(int, input().split())
# n = 6
# k = 2
pf = primeFactor(n)
# print(pf)
ans = 1
for p in pf:
m = pf[p] + 1
vvv = mxv(m, mxpow(m, mx(m), k), vc(m))
t = 0
for i in range(m):
t += (vvv[i] * p ** i) % P
t %= P
ans *= t
ans %= P
print(ans)
``` | output | 1 | 79,174 | 22 | 158,349 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Makoto has a big blackboard with a positive integer n written on it. He will perform the following action exactly k times:
Suppose the number currently written on the blackboard is v. He will randomly pick one of the divisors of v (possibly 1 and v) and replace v with this divisor. As Makoto uses his famous random number generator (RNG) and as he always uses 58 as his generator seed, each divisor is guaranteed to be chosen with equal probability.
He now wonders what is the expected value of the number written on the blackboard after k steps.
It can be shown that this value can be represented as P/Q where P and Q are coprime integers and Q notβ‘ 0 \pmod{10^9+7}. Print the value of P β
Q^{-1} modulo 10^9+7.
Input
The only line of the input contains two integers n and k (1 β€ n β€ 10^{15}, 1 β€ k β€ 10^4).
Output
Print a single integer β the expected value of the number on the blackboard after k steps as P β
Q^{-1} \pmod{10^9+7} for P, Q defined above.
Examples
Input
6 1
Output
3
Input
6 2
Output
875000008
Input
60 5
Output
237178099
Note
In the first example, after one step, the number written on the blackboard is 1, 2, 3 or 6 β each occurring with equal probability. Hence, the answer is (1+2+3+6)/(4)=3.
In the second example, the answer is equal to 1 β
9/16+2 β
3/16+3 β
3/16+6 β
1/16=15/8.
Submitted Solution:
```
import time
mod = int(1e9 + 7)
def fast_expo(x, k):
ans = 1
while k > 0:
if (k & 1) > 0:
ans = (ans * x) % mod
x = (x * x) % mod
k >>= 1
return ans
def inverse(x):
return fast_expo(x, mod - 2)
N, K = [int(_) for _ in input().split()]
start_time = time.time()
prime_factors = {}
i = 2
curr = N
while i * i <= curr:
if curr % i == 0:
prime_factors[i] = 0
while curr % i == 0:
prime_factors[i] += 1
curr //= i
i += 1
if curr > 1:
prime_factors[curr] = 1
# print(time.time() - start_time)
inv = {i: inverse(i) for i in range(60)}
ans = 1
for p in prime_factors:
size = prime_factors[p]
dp = [0] * (size + 1)
dp[0] = 1
pre = [1] * (size + 1)
for e in range(1, size + 1):
dp[e] = (p * dp[e - 1]) % mod
pre[e] = (pre[e - 1] + dp[e]) % mod
for it in range(K):
# print(dp, pre)
dpNext = [0] * (size + 1)
preNext = [1] * (size + 1)
for e in range(1, size + 1):
dpNext[e] = (inv[e + 1] * pre[e]) % mod
preNext[e] = (preNext[e - 1] + dpNext[e]) % mod
dp = dpNext
pre = preNext
ans = (ans * dp[size]) % mod
# print(time.time() - start_time)
# because the dp is multiplicative, we can merge the answer
print(ans)
``` | instruction | 0 | 79,175 | 22 | 158,350 |
Yes | output | 1 | 79,175 | 22 | 158,351 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Makoto has a big blackboard with a positive integer n written on it. He will perform the following action exactly k times:
Suppose the number currently written on the blackboard is v. He will randomly pick one of the divisors of v (possibly 1 and v) and replace v with this divisor. As Makoto uses his famous random number generator (RNG) and as he always uses 58 as his generator seed, each divisor is guaranteed to be chosen with equal probability.
He now wonders what is the expected value of the number written on the blackboard after k steps.
It can be shown that this value can be represented as P/Q where P and Q are coprime integers and Q notβ‘ 0 \pmod{10^9+7}. Print the value of P β
Q^{-1} modulo 10^9+7.
Input
The only line of the input contains two integers n and k (1 β€ n β€ 10^{15}, 1 β€ k β€ 10^4).
Output
Print a single integer β the expected value of the number on the blackboard after k steps as P β
Q^{-1} \pmod{10^9+7} for P, Q defined above.
Examples
Input
6 1
Output
3
Input
6 2
Output
875000008
Input
60 5
Output
237178099
Note
In the first example, after one step, the number written on the blackboard is 1, 2, 3 or 6 β each occurring with equal probability. Hence, the answer is (1+2+3+6)/(4)=3.
In the second example, the answer is equal to 1 β
9/16+2 β
3/16+3 β
3/16+6 β
1/16=15/8.
Submitted Solution:
```
MOD = 10 ** 9 + 7
inv = [pow(i, MOD - 2, MOD) for i in range(60)]
n, k = map(int, input().split())
def solve(p, q):
dp = [1]
for i in range(q):
dp.append(dp[-1] * p % MOD)
for i in range(1, q + 1):
dp[i] = (dp[i] + dp[i - 1]) % MOD
for _ in range(k):
dp1 = [1] * (q + 1)
for i in range(1, q + 1):
dp1[i] = (dp1[i - 1] + dp[i] * inv[i + 1]) % MOD
dp = dp1
return (dp[-1] - dp[-2]) % MOD
ans = 1
i = 2
while i * i <= n:
c = 0
while n % i == 0:
c += 1
n //= i
if c:
ans = ans * solve(i, c) % MOD
i += 1
if n > 1:
ans = ans * solve(n, 1) % MOD
print(ans)
``` | instruction | 0 | 79,176 | 22 | 158,352 |
Yes | output | 1 | 79,176 | 22 | 158,353 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Makoto has a big blackboard with a positive integer n written on it. He will perform the following action exactly k times:
Suppose the number currently written on the blackboard is v. He will randomly pick one of the divisors of v (possibly 1 and v) and replace v with this divisor. As Makoto uses his famous random number generator (RNG) and as he always uses 58 as his generator seed, each divisor is guaranteed to be chosen with equal probability.
He now wonders what is the expected value of the number written on the blackboard after k steps.
It can be shown that this value can be represented as P/Q where P and Q are coprime integers and Q notβ‘ 0 \pmod{10^9+7}. Print the value of P β
Q^{-1} modulo 10^9+7.
Input
The only line of the input contains two integers n and k (1 β€ n β€ 10^{15}, 1 β€ k β€ 10^4).
Output
Print a single integer β the expected value of the number on the blackboard after k steps as P β
Q^{-1} \pmod{10^9+7} for P, Q defined above.
Examples
Input
6 1
Output
3
Input
6 2
Output
875000008
Input
60 5
Output
237178099
Note
In the first example, after one step, the number written on the blackboard is 1, 2, 3 or 6 β each occurring with equal probability. Hence, the answer is (1+2+3+6)/(4)=3.
In the second example, the answer is equal to 1 β
9/16+2 β
3/16+3 β
3/16+6 β
1/16=15/8.
Submitted Solution:
```
from collections import defaultdict
from math import sqrt
n, k = [int(i) for i in input().split()]
p = 10 ** 9 + 7
last = 0
lastrt = 0
def rt(n):
global last, lastrt
if last == n:
return lastrt
last = n
lastrt = sqrt(n)
return lastrt
def factorize(n):
i = 2
A = defaultdict(int)
while n % i == 0:
n //= i
A[i] += 1
i += 1
while n != 1 and i <= rt(n):
while n % i == 0:
n //= i
A[i] += 1
i += 2
if n != 1:
A[n] += 1
return A
D = dict(factorize(n))
ans = 1
mod = [1, 1]
for i in range(2, 51):
mod.append(p - (p // i) * mod[p % i] % p)
for key, v in D.items():
DP = [[0] * (v+1) for i in range(2)]
DP[0][v] = 1
for i in range(k):
for j in range(v+1):
DP[(i+1)&1][j] = 0
for l in range(j+1):
DP[(i+1)&1][l] += DP[i&1][j] * mod[j+1]
DP[(i+1)&1][l] %= p
res = 0
pk = 1
for i in range(v+1):
res += DP[k&1][i] * pk
pk *= key
pk %= p
ans *= res
ans %= p
print(ans)
``` | instruction | 0 | 79,177 | 22 | 158,354 |
Yes | output | 1 | 79,177 | 22 | 158,355 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Makoto has a big blackboard with a positive integer n written on it. He will perform the following action exactly k times:
Suppose the number currently written on the blackboard is v. He will randomly pick one of the divisors of v (possibly 1 and v) and replace v with this divisor. As Makoto uses his famous random number generator (RNG) and as he always uses 58 as his generator seed, each divisor is guaranteed to be chosen with equal probability.
He now wonders what is the expected value of the number written on the blackboard after k steps.
It can be shown that this value can be represented as P/Q where P and Q are coprime integers and Q notβ‘ 0 \pmod{10^9+7}. Print the value of P β
Q^{-1} modulo 10^9+7.
Input
The only line of the input contains two integers n and k (1 β€ n β€ 10^{15}, 1 β€ k β€ 10^4).
Output
Print a single integer β the expected value of the number on the blackboard after k steps as P β
Q^{-1} \pmod{10^9+7} for P, Q defined above.
Examples
Input
6 1
Output
3
Input
6 2
Output
875000008
Input
60 5
Output
237178099
Note
In the first example, after one step, the number written on the blackboard is 1, 2, 3 or 6 β each occurring with equal probability. Hence, the answer is (1+2+3+6)/(4)=3.
In the second example, the answer is equal to 1 β
9/16+2 β
3/16+3 β
3/16+6 β
1/16=15/8.
Submitted Solution:
```
n, k = map(int, input().split(' '))
inv = [pow(i, 1000000005, 1000000007) for i in range(60)]
def solve(p, q):
dp = [1] * (q + 1)
for i in range(q):
dp[i + 1] = (dp[i] * p) % 1000000007
for i in range(1, q + 1):
dp[i] = (dp[i] + dp[i - 1]) % 1000000007
for _ in range(k):
dp1 = [1] * (q + 1)
for i in range(1, q + 1):
dp1[i] = (dp1[i - 1] + dp[i] * inv[i + 1]) % 1000000007
dp = dp1
return (dp[-1] - dp[-2]) % 1000000007
ans = 1
if 4 <= n:
c = 0
while n % 2 == 0:
c += 1
n //= 2
if c:
ans = ans * solve(2, c) % 1000000007
if 9 <= n:
c = 0
while n % 3 == 0:
c += 1
n //= 3
if c:
ans = ans * solve(3, c) % 1000000007
i = 5
while i * i <= n:
c = 0
while n % i == 0:
c += 1
n //= i
if c:
ans = ans * solve(i, c) % 1000000007
i += 2 if i % 3 == 2 else 4
if n > 1:
ans = ans * solve(n, 1) % 1000000007
print(ans)
``` | instruction | 0 | 79,178 | 22 | 158,356 |
Yes | output | 1 | 79,178 | 22 | 158,357 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Makoto has a big blackboard with a positive integer n written on it. He will perform the following action exactly k times:
Suppose the number currently written on the blackboard is v. He will randomly pick one of the divisors of v (possibly 1 and v) and replace v with this divisor. As Makoto uses his famous random number generator (RNG) and as he always uses 58 as his generator seed, each divisor is guaranteed to be chosen with equal probability.
He now wonders what is the expected value of the number written on the blackboard after k steps.
It can be shown that this value can be represented as P/Q where P and Q are coprime integers and Q notβ‘ 0 \pmod{10^9+7}. Print the value of P β
Q^{-1} modulo 10^9+7.
Input
The only line of the input contains two integers n and k (1 β€ n β€ 10^{15}, 1 β€ k β€ 10^4).
Output
Print a single integer β the expected value of the number on the blackboard after k steps as P β
Q^{-1} \pmod{10^9+7} for P, Q defined above.
Examples
Input
6 1
Output
3
Input
6 2
Output
875000008
Input
60 5
Output
237178099
Note
In the first example, after one step, the number written on the blackboard is 1, 2, 3 or 6 β each occurring with equal probability. Hence, the answer is (1+2+3+6)/(4)=3.
In the second example, the answer is equal to 1 β
9/16+2 β
3/16+3 β
3/16+6 β
1/16=15/8.
Submitted Solution:
```
[n,k] = list(map(int, input().split()))
MMI = lambda A, n,s=1,t=0,N=0: (n < 2 and t%N or MMI(n, A%n, t, s-A//n*t, N or n),-1)[n<1]
mod = 1000000007
if [n,k] == [1000000000000000,10000]:
print(215514159)
quit()
if [n, k] == [260858031033600,9696]:
print(692221824)
quit()
if [n, k] == [866421317361600,10000]:
print(692221824)
quit()
def primeFactors(n):
prime_factors = []
prime_set = set()
while(n % 2 == 0):
if 2 not in prime_set:
prime_set.add(2)
prime_factors.append(2)
n //= 2
i = 3
while i*i <= n:
while(n % i == 0):
if i not in prime_set:
prime_factors.append(i)
prime_set.add(i)
n //= i
i += 2
if n > 2:
prime_factors.append(n)
return prime_factors
prime_factors = primeFactors(n)
def getPowers(n):
powers = [0] * len(prime_factors)
for i, prime in enumerate(prime_factors):
while(n % prime == 0):
powers[i] += 1
n //= prime
return powers
n_powers = getPowers(n)
def getFactors(power_tuple):
factors = []
if len(power_tuple) == 0:
return [()]
else:
other_factors = getFactors(power_tuple[1:])
for i in range(power_tuple[0]+1):
for other_factor in other_factors:
factors.append(tuple([i]) + other_factor)
return factors
factors = getFactors(n_powers)
index = {}
for i,f in enumerate(factors):
index[f] = i
w = len(factors)
h = w
T= [[0 for x in range(w)] for y in range(h)]
smallerTuples = {}
def getSmallerTuple(t):
if t in smallerTuples:
return smallerTuples[t]
smaller = []
if len(t) == 0:
return [()]
if len(t) == 1:
for i in range(t[0]+1):
smaller.append(tuple([i]))
else:
for i in range(t[0]+1):
for s in getSmallerTuple(t[1:]):
smaller.append(tuple([i]) + s)
smallerTuples[t] = smaller
return smaller
MMImemo = {}
for i,f in enumerate(factors):
num_factors = 1
for x in f:
num_factors *= (1+x)
for f_2 in getSmallerTuple(f):
i_2 = index[f_2]
if not num_factors in MMImemo:
MMImemo[num_factors] = MMI(num_factors,mod)
T[i][i_2] = MMImemo[num_factors]
def multiply(T1, T2):
#print("MULTIPLYING ", w)
T = [[0 for x in range(w)] for y in range(w)]
for i in range(w):
for k in range(w):
for j in range(w):
T[i][j] += (T1[i][k]*T2[k][j]) % mod
T[i][j] %= mod
return T
powerMemo = {}
def matrixPower(n):
if n in powerMemo:
return powerMemo[n]
if n == 0:
I = [[0 for x in range(w)] for y in range(h)]
for i in range(w):
I[i][i] = 1
if n == 1:
return T
elif n == 2:
powerMemo[n] = multiply(T,T)
elif n % 2 == 0:
sroot = matrixPower(n//2)
powerMemo[n] = multiply(sroot, sroot)
else:
powerMemo[n] = multiply(T, matrixPower(n-1))
return powerMemo[n]
expectedState = matrixPower(k)[-1]
e = 0
for i,f in enumerate(factors):
value = 1
for ii, ff in enumerate(f):
value *= prime_factors[ii]**ff
e += value*expectedState[i]
e %= mod
print(e)
``` | instruction | 0 | 79,179 | 22 | 158,358 |
No | output | 1 | 79,179 | 22 | 158,359 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Makoto has a big blackboard with a positive integer n written on it. He will perform the following action exactly k times:
Suppose the number currently written on the blackboard is v. He will randomly pick one of the divisors of v (possibly 1 and v) and replace v with this divisor. As Makoto uses his famous random number generator (RNG) and as he always uses 58 as his generator seed, each divisor is guaranteed to be chosen with equal probability.
He now wonders what is the expected value of the number written on the blackboard after k steps.
It can be shown that this value can be represented as P/Q where P and Q are coprime integers and Q notβ‘ 0 \pmod{10^9+7}. Print the value of P β
Q^{-1} modulo 10^9+7.
Input
The only line of the input contains two integers n and k (1 β€ n β€ 10^{15}, 1 β€ k β€ 10^4).
Output
Print a single integer β the expected value of the number on the blackboard after k steps as P β
Q^{-1} \pmod{10^9+7} for P, Q defined above.
Examples
Input
6 1
Output
3
Input
6 2
Output
875000008
Input
60 5
Output
237178099
Note
In the first example, after one step, the number written on the blackboard is 1, 2, 3 or 6 β each occurring with equal probability. Hence, the answer is (1+2+3+6)/(4)=3.
In the second example, the answer is equal to 1 β
9/16+2 β
3/16+3 β
3/16+6 β
1/16=15/8.
Submitted Solution:
```
import math
from fractions import Fraction
n, m = list(map(int, input().split()))
def divisorGenerator(n):
large_divisors = []
for i in range(1, int(math.sqrt(n) + 1)):
if n % i == 0:
yield i
if i*i != n:
large_divisors.append(n / i)
for divisor in reversed(large_divisors):
yield int(divisor)
divsCache = {}
total = [0]
def calc(divs,prob,m):
length = len(divs)
if m == 0:
for el in divs:
total[0] += el*prob/length
return
for num in divs:
if num in divsCache:
calc(divsCache[num], prob/length, m-1,)
else:
divsCache[num] = list(divisorGenerator(num))
calc(divsCache[num], prob/length, m-1)
calc([n], 1, m)
p = 10**9+7
k = len(divsCache[n])**m
z = k*total[0]
while z > int(z):
z *= 10
k *= 10
print(int((z*(pow(k, p-2, p)))%p))
``` | instruction | 0 | 79,180 | 22 | 158,360 |
No | output | 1 | 79,180 | 22 | 158,361 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Makoto has a big blackboard with a positive integer n written on it. He will perform the following action exactly k times:
Suppose the number currently written on the blackboard is v. He will randomly pick one of the divisors of v (possibly 1 and v) and replace v with this divisor. As Makoto uses his famous random number generator (RNG) and as he always uses 58 as his generator seed, each divisor is guaranteed to be chosen with equal probability.
He now wonders what is the expected value of the number written on the blackboard after k steps.
It can be shown that this value can be represented as P/Q where P and Q are coprime integers and Q notβ‘ 0 \pmod{10^9+7}. Print the value of P β
Q^{-1} modulo 10^9+7.
Input
The only line of the input contains two integers n and k (1 β€ n β€ 10^{15}, 1 β€ k β€ 10^4).
Output
Print a single integer β the expected value of the number on the blackboard after k steps as P β
Q^{-1} \pmod{10^9+7} for P, Q defined above.
Examples
Input
6 1
Output
3
Input
6 2
Output
875000008
Input
60 5
Output
237178099
Note
In the first example, after one step, the number written on the blackboard is 1, 2, 3 or 6 β each occurring with equal probability. Hence, the answer is (1+2+3+6)/(4)=3.
In the second example, the answer is equal to 1 β
9/16+2 β
3/16+3 β
3/16+6 β
1/16=15/8.
Submitted Solution:
```
x,k=map(int,input().split())
#x=60
#k=5
if x==1:
print(1)
import sys
sys.exit()
import math
L=int(math.sqrt(x))
mod=10**9+7
FACT=dict()
for i in range(2,L+2):
while x%i==0:
FACT[i]=FACT.get(i,0)+1
x=x//i
if x!=1:
FACT[x]=FACT.get(x,0)+1
def prod(A,B,k,l,m):#A:k*l,B:l*m
C=[[None for i in range(m)] for j in range(k)]
for i in range(k):
for j in range(m):
ANS=0
for pl in range(l):
ANS=(ANS+A[i][pl]*B[pl][j])%mod
C[i][j]=ANS
return C
def plus(A,B,k,l):#a,B:k*l
C=[[None for i in range(l)] for j in range(k)]
for i in range(k):
for j in range(l):
C[i][j]=(A[i][j]+B[i][j])%mod
return C
#XX=[[1,1],[1,1],[1,1]]
#YY=[[2,2,2],[2,3,4]]
#print(prod(XX,YY,3,2,3))
MAT_index=max(FACT.values())+1
MAT=[[0 for i in range(MAT_index)] for j in range(MAT_index)]
for m in range(MAT_index):
for l in range(m+1):
x=pow(m+1,mod-2,mod)
MAT[m][l]=x
#print(MAT)
#MAT_ini=MAT
#for i in range(k-1):
# MAT=prod(MAT,MAT_ini,MAT_index,MAT_index,MAT_index)
#γγγγγγͺγ³γ°
MAT_dob=[None]*13
MAT_dob[0]=MAT
for i in range(1,13):
MAT_dob[i]=prod(MAT_dob[i-1],MAT_dob[i-1],MAT_index,MAT_index,MAT_index)
MAT=[[0 for i in range(MAT_index)] for j in range(MAT_index)]
for i in range(MAT_index):
MAT[i][i]=1
for i in range(13):
if k & 1<<i !=0:
#print(i,MAT,MAT_dob[i])
MAT=prod(MAT,MAT_dob[i],MAT_index,MAT_index,MAT_index)
#print(MAT)
#print(MAT)
ANS=1
for fa in FACT:
x=FACT[fa]
ANS_fa=0
for i in range(x+1):
ANS_fa=(ANS_fa+pow(fa,i,mod)*MAT[x][i])%mod
ANS=ANS*ANS_fa%mod
print(ANS)
``` | instruction | 0 | 79,181 | 22 | 158,362 |
No | output | 1 | 79,181 | 22 | 158,363 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Makoto has a big blackboard with a positive integer n written on it. He will perform the following action exactly k times:
Suppose the number currently written on the blackboard is v. He will randomly pick one of the divisors of v (possibly 1 and v) and replace v with this divisor. As Makoto uses his famous random number generator (RNG) and as he always uses 58 as his generator seed, each divisor is guaranteed to be chosen with equal probability.
He now wonders what is the expected value of the number written on the blackboard after k steps.
It can be shown that this value can be represented as P/Q where P and Q are coprime integers and Q notβ‘ 0 \pmod{10^9+7}. Print the value of P β
Q^{-1} modulo 10^9+7.
Input
The only line of the input contains two integers n and k (1 β€ n β€ 10^{15}, 1 β€ k β€ 10^4).
Output
Print a single integer β the expected value of the number on the blackboard after k steps as P β
Q^{-1} \pmod{10^9+7} for P, Q defined above.
Examples
Input
6 1
Output
3
Input
6 2
Output
875000008
Input
60 5
Output
237178099
Note
In the first example, after one step, the number written on the blackboard is 1, 2, 3 or 6 β each occurring with equal probability. Hence, the answer is (1+2+3+6)/(4)=3.
In the second example, the answer is equal to 1 β
9/16+2 β
3/16+3 β
3/16+6 β
1/16=15/8.
Submitted Solution:
```
n, k = [int(i) for i in input().split(" ")]
def fctr(n) :
f = list()
p = 1
while p*p<=n :
if n%p==0 :
f.append(p)
f.append(n//p)
p += 1
return f
def fctrdict(n) :
return {i:fctr(i)*(len(fctr(n))//len(fctr(i))) for i in fctr(n)}
d = fctrdict(n)
p = []
def recur(n,k,s=0,a=n) :
if s==k :
p.append(a)
else :
for i in d.get(a) :
recur(n,k,s+1,i)
recur(n,k)
dd = {i:p.count(i) for i in set(p)}
smp = sum(dd.get(i) for i in dd)
exp = sum(i*dd.get(i) for i in dd)
c = 10**9+5
print((exp*pow(smp,c,c+2))%(c+2))
``` | instruction | 0 | 79,182 | 22 | 158,364 |
No | output | 1 | 79,182 | 22 | 158,365 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a consisting of n integers.
Your task is to say the number of such positive integers x such that x divides each number from the array. In other words, you have to find the number of common divisors of all elements in the array.
For example, if the array a will be [2, 4, 6, 2, 10], then 1 and 2 divide each number from the array (so the answer for this test is 2).
Input
The first line of the input contains one integer n (1 β€ n β€ 4 β
10^5) β the number of elements in a.
The second line of the input contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ 10^{12}), where a_i is the i-th element of a.
Output
Print one integer β the number of such positive integers x such that x divides each number from the given array (in other words, the answer is the number of common divisors of all elements in the array).
Examples
Input
5
1 2 3 4 5
Output
1
Input
6
6 90 12 18 30 18
Output
4 | instruction | 0 | 79,235 | 22 | 158,470 |
Tags: implementation, math
Correct Solution:
```
import math
n = int(input())
nub = list(map(int, input().split()))
temp = 0
for elem in nub:
temp = math.gcd(temp, elem)
count = 0
i = 1
while (i*i) <= temp:
if temp%i == 0:
count += 1
if (i*i) != temp:
count += 1
i += 1
print(count)
``` | output | 1 | 79,235 | 22 | 158,471 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a consisting of n integers.
Your task is to say the number of such positive integers x such that x divides each number from the array. In other words, you have to find the number of common divisors of all elements in the array.
For example, if the array a will be [2, 4, 6, 2, 10], then 1 and 2 divide each number from the array (so the answer for this test is 2).
Input
The first line of the input contains one integer n (1 β€ n β€ 4 β
10^5) β the number of elements in a.
The second line of the input contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ 10^{12}), where a_i is the i-th element of a.
Output
Print one integer β the number of such positive integers x such that x divides each number from the given array (in other words, the answer is the number of common divisors of all elements in the array).
Examples
Input
5
1 2 3 4 5
Output
1
Input
6
6 90 12 18 30 18
Output
4 | instruction | 0 | 79,236 | 22 | 158,472 |
Tags: implementation, math
Correct Solution:
```
import math
n = int(input())
a = list(map(int,input().split()))
g = 0
for i in range(n):
g = math.gcd(g,a[i])
i = 1
count = 0
while (i*i<=g):
if (i*i==g):
count += 1
elif (g%i==0):
count += 2
i += 1
print (count)
``` | output | 1 | 79,236 | 22 | 158,473 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a consisting of n integers.
Your task is to say the number of such positive integers x such that x divides each number from the array. In other words, you have to find the number of common divisors of all elements in the array.
For example, if the array a will be [2, 4, 6, 2, 10], then 1 and 2 divide each number from the array (so the answer for this test is 2).
Input
The first line of the input contains one integer n (1 β€ n β€ 4 β
10^5) β the number of elements in a.
The second line of the input contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ 10^{12}), where a_i is the i-th element of a.
Output
Print one integer β the number of such positive integers x such that x divides each number from the given array (in other words, the answer is the number of common divisors of all elements in the array).
Examples
Input
5
1 2 3 4 5
Output
1
Input
6
6 90 12 18 30 18
Output
4 | instruction | 0 | 79,237 | 22 | 158,474 |
Tags: implementation, math
Correct Solution:
```
n=int(input())
A=list(map(int,input().split()))
GCD=0
import math
for a in A:
GCD=math.gcd(GCD,a)
x=GCD
L=int(math.sqrt(x))
FACT=dict()
for i in range(2,L+2):
while x%i==0:
FACT[i]=FACT.get(i,0)+1
x=x//i
if x!=1:
FACT[x]=FACT.get(x,0)+1
ANS=1
for f in FACT:
ANS*=FACT[f]+1
print(ANS)
``` | output | 1 | 79,237 | 22 | 158,475 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a consisting of n integers.
Your task is to say the number of such positive integers x such that x divides each number from the array. In other words, you have to find the number of common divisors of all elements in the array.
For example, if the array a will be [2, 4, 6, 2, 10], then 1 and 2 divide each number from the array (so the answer for this test is 2).
Input
The first line of the input contains one integer n (1 β€ n β€ 4 β
10^5) β the number of elements in a.
The second line of the input contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ 10^{12}), where a_i is the i-th element of a.
Output
Print one integer β the number of such positive integers x such that x divides each number from the given array (in other words, the answer is the number of common divisors of all elements in the array).
Examples
Input
5
1 2 3 4 5
Output
1
Input
6
6 90 12 18 30 18
Output
4 | instruction | 0 | 79,238 | 22 | 158,476 |
Tags: implementation, math
Correct Solution:
```
import math
n = int(input())
x = 0
for q in map(int,input().split()):
x = math.gcd(x,q)
ans = 0
for q in range(1,int(math.sqrt(x))+1):
ans+= 2*(x%q==0)
print(ans-(int(math.sqrt(x))**2==x))
``` | output | 1 | 79,238 | 22 | 158,477 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a consisting of n integers.
Your task is to say the number of such positive integers x such that x divides each number from the array. In other words, you have to find the number of common divisors of all elements in the array.
For example, if the array a will be [2, 4, 6, 2, 10], then 1 and 2 divide each number from the array (so the answer for this test is 2).
Input
The first line of the input contains one integer n (1 β€ n β€ 4 β
10^5) β the number of elements in a.
The second line of the input contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ 10^{12}), where a_i is the i-th element of a.
Output
Print one integer β the number of such positive integers x such that x divides each number from the given array (in other words, the answer is the number of common divisors of all elements in the array).
Examples
Input
5
1 2 3 4 5
Output
1
Input
6
6 90 12 18 30 18
Output
4 | instruction | 0 | 79,239 | 22 | 158,478 |
Tags: implementation, math
Correct Solution:
```
from functools import reduce
def sieve(n):
numbers = list(range(2, n+1))
i = 0
while i < len(numbers) and (numbers[i] * 2) - 2 <= len(numbers):
if (numbers[i] > 0):
prime = numbers[i]
for j in range(prime * prime - 2, len(numbers), prime):
numbers[j] = -1
i += 1
primes = list(filter(lambda x: x != -1, numbers))
return primes
def factor(number, primes):
factors = {}
for p in primes:
while number % p == 0:
if p in factors: factors[p] += 1
else: factors[p] = 1
number //= p
if number == 1:
return factors
else:
factors[number] = 1
return factors
def gcd(a, b):
if a == 0: return b
if b == 0: return a
return gcd(b, a % b)
n = int(input())
a = list(set(map(int, input().split())))
s = sieve(10**6)
if (len(a) > 1):
d = gcd(a[0], a[1])
for i in range(2, len(a)):
d = gcd(d, a[i])
f = factor(d, s)
if f == {}:
print(1)
else:
print(reduce(lambda x, y: x*y, map(lambda x: x+1, f.values())))
else:
f = factor(a[0], s)
if f == {}:
print(1)
else:
print(reduce(lambda x, y: x*y, map(lambda x: x+1, f.values())))
``` | output | 1 | 79,239 | 22 | 158,479 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a consisting of n integers.
Your task is to say the number of such positive integers x such that x divides each number from the array. In other words, you have to find the number of common divisors of all elements in the array.
For example, if the array a will be [2, 4, 6, 2, 10], then 1 and 2 divide each number from the array (so the answer for this test is 2).
Input
The first line of the input contains one integer n (1 β€ n β€ 4 β
10^5) β the number of elements in a.
The second line of the input contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ 10^{12}), where a_i is the i-th element of a.
Output
Print one integer β the number of such positive integers x such that x divides each number from the given array (in other words, the answer is the number of common divisors of all elements in the array).
Examples
Input
5
1 2 3 4 5
Output
1
Input
6
6 90 12 18 30 18
Output
4 | instruction | 0 | 79,240 | 22 | 158,480 |
Tags: implementation, math
Correct Solution:
```
# code by RAJ BHAVSAR
def helper(n):
ans = 0
i = 1
while(i*i <= n):
if(n%i == 0):
ans += 1
if(n//i != i):
ans += 1
i += 1
return ans
from math import gcd
n = int(input())
arr = list(map(int,input().split()))
if(n == 1):
print(helper(arr[0]))
else:
ans = gcd(arr[0],arr[1])
for i in range(2,n):
ans = gcd(ans,arr[i])
print(helper(ans))
``` | output | 1 | 79,240 | 22 | 158,481 |
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