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| description
stringlengths 29
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| source
int64 1
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| difficulty
int64 0
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| solution
stringlengths 7
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| language
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1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | t = int(input())
for i in range(t):
n = int(input())
divisorarr = []
for divisor in range(2, int(n ** 0.5) + 2):
if n % divisor == 0:
while n % divisor == 0:
n = n // divisor
divisorarr.append(divisor)
if n != 1:
divisorarr.append(n)
if len(divisorarr) < 3:
print('NO')
else:
a = divisorarr[0]
b = divisorarr[1]
index = 2
while b == a and index < len(divisorarr) - 1:
b *= divisorarr[index]
index += 1
c = 1
for divisor in range(index, len(divisorarr)):
c *= divisorarr[divisor]
if not a != b != c != a:
print('NO')
else:
print('YES')
print(a, b, c)
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | I = input
for _ in [0]*int(I()):
temp = int(I())
ans = []
for i in range(2,temp,1):
if(temp % i == 0):
ans.append(i)
break
elif i *i > temp:
break
if(len(ans) == 0 or ans[0]*ans[0] >= temp):
print("NO");continue
temp //= ans[0]
for i in range(ans[0],int(temp),1):
if(int(temp) % i == 0 and i != ans[0]):
ans.append(i)
temp //= i
break
elif i *i > temp:
break
if(len(ans) == 1):
print("NO");continue
if(int(temp) != ans[0] and int(temp) != ans[1]):
ans.append(int(temp))
print("YES")
for i in ans:
print(i,end = ' ')
print('')
else:
print("NO") | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int main() {
int t;
cin >> t;
while (t--) {
int ans[2], j = -1, i;
long long n;
cin >> n;
for (i = 2; i * i <= n && j != 1; i++) {
if (n % i == 0) {
ans[++j] = i;
n = n / i;
}
}
long long ch = ans[0] * ans[1];
if (j < 1 || (ans[0] == n || ans[1] == n))
cout << "NO" << endl;
else {
cout << "YES" << endl;
cout << ans[0] << " " << ans[1] << " " << n << endl;
}
}
}
| CPP |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | t = int(input())
for _ in range(t):
n = int(input())
nn = n
fac = []
i = 2
while i * i <= n:
cnt = 0
while n % i == 0:
cnt += 1
n //= i
if cnt:
fac.append((i, cnt))
i += 1
if n > 1:
fac.append((n, 1))
ok = False
ans = None
if len(fac) > 2 or len(fac) == 2 and fac[0][1] + fac[1][1] > 3:
ok = True
ans = (fac[0][0], fac[1][0])
elif len(fac) == 1 and fac[0][1] >= 6:
ok = True
ans = (fac[0][0], int(fac[0][0] ** 2))
if not ok:
print("NO")
else:
print("YES")
print(ans[0], ans[1], nn // ans[0] // ans[1]) | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | """
Author : Aman Thakur
mantra: chor kya hi kar sakte hai!!
"""
import math
class Solution:
def __init__(self):
self.arr = []
def solution(self):
n = int(input())
for i in range(2, n):
if n % i == 0:
self.arr.append(i)
n //= i
if len(self.arr) == 2 or n < i*i:
break
if len(self.arr) == 2 and n != self.arr[0] and n != self.arr[1] and n > 1:
print('YES')
print(self.arr[0], self.arr[1], n)
else:
print('NO')
if __name__ == '__main__':
for _ in range(int(input())):
ob = Solution()
ob.solution()
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | from math import *
for i in range(int(input())):
n=int(input())
j=2
count=0
num=[]
while j**2<=n:
if n%j==0:
if len(num)==1:
if n%(j*num[0])!=0:
j+=1
continue
num.append(j)
count+=1
if count==2:
break
j+=1
#print(num)
if count<2:
print("NO")
continue
if count==2:
val=n//(num[0]*num[1])
if val in num:
print("NO")
continue
if val*num[0]*num[1]!=n:
print("NO")
continue
print("YES")
print(num[0], num[1], val)
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | """
Author : Aman Thakur
mantra: chor kya hi kar sakte hai!!
"""
import math
class Solution:
def __init__(self):
self.c = 0
self.arr = []
def solution(self):
n = int(input())
for i in range(2, n):
if n % i == 0:
self.arr.append(i)
n //= i
if len(self.arr) == 2 or n < i*i:
break
if len(self.arr) == 2 and n != self.arr[0] and n != self.arr[1] and n > 1:
print('YES')
print(self.arr[0], self.arr[1], n)
else:
print('NO')
if __name__ == '__main__':
for _ in range(int(input())):
ob = Solution()
ob.solution()
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | for _ in range(int(input())):
n=int(input())
sq=int(n**0.5 +1)
a=[]
d={}
for i in range(2,sq):
if(n%i==0):
x=n//i
a.append(i)
d[i]=x
d[x]=i
ch=1
for i in range(len(a)):
for j in range(i+1,len(a)):
try:
x=d[ a[i]*a[j] ]
y=a[i]
z=a[j]
if(x!=y)and(x!=z):
ch=2
break
except:
continue
if(ch==2):
break
if(ch==2):
print("YES")
print(x,y,z)
else:
print("NO")
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import math
def ans(n):
a, b = -1, -1
for i in range(2, int(math.sqrt(n)) + 1):
if n % i == 0:
a = i
break
if a == -1:
print('NO')
return
for i in range(a + 1, int(math.sqrt(n)) + 1):
if (n % i == 0) & (n % (a * i) == 0):
b = i
break
if b == -1:
print('NO')
return
if (n / (a * b) != 1) & (n / (a * b) != a) & (n / (a * b) != b):
print('YES')
print(a, b, int(n / (a * b)))
return
print('NO')
return
t = int(input())
while(t):
t = t - 1
n = int(input())
ans(n) | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 |
import sys
import math
from math import gcd
from heapq import heappop
from heapq import heappush
from heapq import heapify
from bisect import insort
from bisect import bisect_right
from bisect import bisect_left
from sys import stdin,stdout
from collections import defaultdict, deque
from math import log2, ceil, floor
inp=lambda : int(input())
sip=lambda : input()
mulip =lambda : map(int,input().split())
lst=lambda : list(map(int,input().split()))
slst=lambda: list(sip())
arr2d= lambda x: [[int(j) for j in input().split()] for i in range(x)]
odds = lambda l: len(list(filter(lambda x: x%2!=0, l)))
evens = lambda l: len(list(filter(lambda x: x%2==0, l)))
mod = pow(10,9)+7
#-------------------------------------------------------
Judge = 0
if Judge:
sys.stdin = open("input.in",'r')
#sys.stdout = open("output.in",'w')
for _ in range(inp()):
n = inp()
x = n
d = {}
i = 2
while i*i <= x:
cnt = 0
while(n%i==0 and n!=0):
n = n//i
#print(n,i)
cnt += 1
d[i] = cnt
i += 1
if n>2:
d[n] = 1
if len(d)==0:
print("NO")
elif len(d)==1:
l = list(d.keys())
if d[l[0]]>=6:
print("YES")
print(l[0],pow(l[0],2),pow(l[0],d[l[0]]-3))
else:
print("NO")
elif len(d)==2:
l = list(d.keys())
if d[l[0]]+d[l[1]]>=4:
print("YES")
print(l[0],l[1],x//(l[0]*l[1]))
else:
print("NO")
else:
l = list(d.keys())
print("YES")
print(pow(l[0],d[l[0]]),pow(l[1],d[l[1]]),end=" ")
ans = 1
for i in range(2,len(l)):
ans *= pow(l[i],d[l[i]])
print(ans)
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import math
for _ in range(int(input())):
n = int(input())
p = tuple((n, 1))
lt = []
while n % 2 == 0:
lt.append(2)
n = n / 2
for i in range(3,int(math.sqrt(n))+1,2):
while n % i== 0:
lt.append(i),
n = n / i
if n > 2:
lt.append(int(n))
#print(lt)
if len(lt)>5:
print('YES')
print(lt[0], end =' ')
print(lt[1]*lt[2], end =' ')
print(int(p[0]/(lt[0]*lt[1]*lt[2])))
if len(lt)<3:
print('NO')
continue
if len(lt)==3:
if (lt[0]==lt[1] or lt[1]==lt[2] or lt[0]==lt[2]):
print('NO')
continue
else:
print('YES')
print(lt[0], end=' ')
print(lt[1], end=' ')
print(lt[2])
continue
if len(lt)==4:
a=lt[0]
b=lt[1]*lt[2]
c=lt[3]
if a!=b and b!=c and c!=a:
print('YES')
print(a, end =' ')
print(b, end =' ')
print(c)
continue
else:
print('NO')
continue
if len(lt)==5:
if len(set(lt))==1:
print('NO')
continue
if len(set(lt))==2:
if lt[0]==lt[1]==lt[2]==lt[3]:
print('YES')
print(lt[0], end =' ')
print(lt[0]**3, end =' ')
print(lt[-1])
continue
elif lt[0]==lt[1]==lt[2] and lt[3]==lt[4]:
print('YES')
print(lt[0], lt[1]*lt[2], lt[3]*lt[4])
elif lt[0]==lt[1]and lt[2]==lt[3]==lt[4]:
print('YES')
print(lt[0]**2, lt[2], lt[2]**2)
else:
print('YES')
print(lt[0], end =' ')
print(lt[1]**3, end =' ')
print(lt[1])
continue
if len(set(lt))==3:
if lt[0]==lt[1]==lt[2]:
print('YES')
print(lt[0]**3, lt[-2], lt[-1])
continue
if lt[1]!=lt[0] and lt[2]==lt[1] and lt[3]==lt[4]:
print("YES")
print(lt[0], lt[1]*lt[2], lt[3]*lt[4])
continue
if lt[0]==lt[1] and lt[3]==lt[4]:
print('YES')
print(lt[0]*lt[1], lt[2], lt[3]*lt[4])
continue
if lt[0]==lt[1] and lt[3]==lt[2]:
print("YES")
print(lt[0]*lt[1], lt[2]*lt[3], lt[4])
if lt[1]==lt[2]==lt[3]:
print('YES')
print(lt[0], lt[1]**3, lt[-1])
continue
if lt[-1]==lt[-2]==lt[-3]:
print('YES')
print(lt[0], lt[1], lt[2]**3)
continue
if len(set(lt))==4:
print('YES')
print(lt[0], lt[2]*lt[1], lt[3]*lt[4])
continue
if len(set(lt))==5:
print('YES')
print(lt[0],lt[1], lt[2]*lt[3]*lt[4])
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | from collections import Counter
def min_del(n, mn):
for x in range(max(2, mn), int(n ** .5) + 1):
if n % x == 0:
return x
return n
for _ in range(int(input())):
n = int(input())
a = min_del(n, 1)
ab = n // a
b = min_del(ab, a + 1)
c = n // b // a
if c != a and c != b and c > 1:
print("YES")
print(a, b, c)
else:
print("NO")
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int main() {
long long t;
cin >> t;
while (t--) {
long long n, a = 2, b = 1, c = 1, t, cnt = 0;
cin >> n;
t = n;
while (cnt < 2 && n > 1) {
if (n % a == 0) {
if (cnt == 0) {
b = a;
} else {
c = a;
}
n = n / a;
cnt++;
a++;
} else if (a * a >= n) {
break;
} else
a++;
}
if (cnt == 2 && c != t / (b * c) && b != t / (b * c) && t / (b * c) != 1) {
cout << "YES\n";
cout << b << " " << c << " " << t / (b * c) << endl;
} else {
cout << "NO\n";
}
}
}
| CPP |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import java.io.*;
import java.util.*;
import java.math.*;
import java.lang.*;
import static java.lang.Math.*;
public class Numbers implements Runnable
{
static class InputReader
{
private InputStream stream;
private byte[] buf = new byte[1024];
private int curChar;
private int numChars;
private SpaceCharFilter filter;
private BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
public InputReader(InputStream stream)
{
this.stream = stream;
}
public int read()
{
if (numChars==-1)
throw new InputMismatchException();
if (curChar >= numChars)
{
curChar = 0;
try
{
numChars = stream.read(buf);
}
catch (IOException e)
{
throw new InputMismatchException();
}
if(numChars <= 0)
return -1;
}
return buf[curChar++];
}
public String nextLine()
{
String str = "";
try
{
str = br.readLine();
}
catch (IOException e)
{
e.printStackTrace();
}
return str;
}
public int nextInt()
{
int c = read();
while(isSpaceChar(c))
c = read();
int sgn = 1;
if (c == '-')
{
sgn = -1;
c = read();
}
int res = 0;
do
{
if(c<'0'||c>'9')
throw new InputMismatchException();
res *= 10;
res += c - '0';
c = read();
}
while (!isSpaceChar(c));
return res * sgn;
}
public long nextLong()
{
int c = read();
while (isSpaceChar(c))
c = read();
int sgn = 1;
if (c == '-')
{
sgn = -1;
c = read();
}
long res = 0;
do
{
if (c < '0' || c > '9')
throw new InputMismatchException();
res *= 10;
res += c - '0';
c = read();
}
while (!isSpaceChar(c));
return res * sgn;
}
public double nextDouble()
{
int c = read();
while (isSpaceChar(c))
c = read();
int sgn = 1;
if (c == '-')
{
sgn = -1;
c = read();
}
double res = 0;
while (!isSpaceChar(c) && c != '.') {
if (c == 'e' || c == 'E')
return res * Math.pow(10, nextInt());
if (c < '0' || c > '9')
throw new InputMismatchException();
res *= 10;
res += c - '0';
c = read();
}
if (c == '.')
{
c = read();
double m = 1;
while (!isSpaceChar(c))
{
if (c == 'e' || c == 'E')
return res * Math.pow(10, nextInt());
if (c < '0' || c > '9')
throw new InputMismatchException();
m /= 10;
res += (c - '0') * m;
c = read();
}
}
return res * sgn;
}
public String readString()
{
int c = read();
while (isSpaceChar(c))
c = read();
StringBuilder res = new StringBuilder();
do
{
res.appendCodePoint(c);
c = read();
}
while (!isSpaceChar(c));
return res.toString();
}
public boolean isSpaceChar(int c)
{
if (filter != null)
return filter.isSpaceChar(c);
return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
}
public String next()
{
return readString();
}
public interface SpaceCharFilter
{
public boolean isSpaceChar(int ch);
}
}
// if modulo is required set value accordingly
public static long[][] matrixMultiply2dL(long t[][],long m[][])
{
long res[][]= new long[t.length][m[0].length];
for(int i=0;i<t.length;i++)
{
for(int j=0;j<m[0].length;j++)
{
res[i][j]=0;
for(int k=0;k<t[0].length;k++)
{
res[i][j]+=t[i][k]+m[k][j];
}
}
}
return res;
}
static long combination(long n,long r)
{
long ans=1;
for(long i=0;i<r;i++)
{
ans=(ans*(n-i))/(i+1);
}
return ans;
}
public static void main(String args[]) throws Exception
{
new Thread(null, new Numbers(),"Numbers",1<<27).start();
}
// **just change the name of class from Main to reuquired**
public void run()
{
InputReader sc = new InputReader(System.in);
PrintWriter w = new PrintWriter(System.out);
int tc=sc.nextInt();
while(tc-->0)
{
int n=sc.nextInt();
int a=0,b=0,c=0;
boolean f=false;
outer:
for(int i=2;i*i<=n;++i)
{
if(n%i==0)
{
int n2=n/i;
for(int j=i+1;j*j<=n2;++j)
{
if(n2%j==0)
{
c=n2/j;
if((i!=c) && (j!=c) && c>=2)
{
w.println("YES");
w.println(i+" "+j+" "+c);
f=true;
break outer;
}
}
}
}
}
if(!f) w.println("NO");
}
System.out.flush();
w.close();
}
}
class Factor
{
int num;
int pow;
Factor(int n, int p)
{
num=n;
pow=p;
}
} | JAVA |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | from itertools import combinations as com
from math import sqrt
for _ in range (int (input())):
k=[]
a=int(input())
for i in range (2,int(sqrt(a))+1):
if a%i ==0:
if a//i==i:
k.append(i)
else:
k.append(i)
k.append(a//i)
if(len(k)<3):
print("NO")
else:
p=com(k,3)
for i in p :
if i[0]*i[1]*i[2]==a:
print("YES")
print(i[0],i[1],i[2])
break
else:
print("NO")
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | from math import sqrt
n=int(input())
for i in range(n):
m=int(input())
m1=m
c=2
p=[]
while len(p)<3 and c<=sqrt(m1):
if m%c==0:
m=m//c
p.append(c)
else:
if c==2:
c=c+1
else:
c=c+2
if len(p)==3:
if p[0]==p[1]:
p[1]=p[1]*p[2]
p[2]=m1//p[0]//p[1]
if len(p)==2:
p.append(m1//p[0]//p[1])
if len(p)==3 and p[2]>1 and p[0]!=p[2] and p[1]!=p[2] and p[0]!=p[1]:
print ('YES')
print (*p)
else:
print ('NO') | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | from sys import stdin, stdout
from collections import defaultdict
import math
rl = lambda: stdin.readline()
rll = lambda: stdin.readline().split()
def main():
cases = int(input())
for line in stdin:
n = int(line)
ans = []
f1 = 2
while f1 <= math.sqrt(n):
if n % f1 == 0:
ans.append(f1)
break
f1 += 1
if len(ans) == 0:
stdout.write("NO\n")
continue
m = n//f1
f2 = f1 + 1
while f2 <= math.sqrt(m):
if m % f2 == 0:
ans.append(f2)
break
f2 += 1
if len(ans) == 1:
stdout.write("NO\n")
continue
f3 = n//(f1*f2)
if f3 not in {f1, f2}:
stdout.write("YES\n")
stdout.write(" ".join((str(x) for x in [f1, f2, f3])))
stdout.write("\n")
else:
stdout.write("NO\n")
if __name__ == "__main__":
main() | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | /*package whatever //do not write package name here */
import java.util.*;
public class Main {
public static void main (String[] args) {
Scanner s = new Scanner(System.in);
int t = s.nextInt();
for(int tt=0;tt<t;tt++)
{
int n = s.nextInt();
int no = 0;
ArrayList<Integer> x = new ArrayList<>();
for(int i=2;i<=Math.sqrt(n);i++)
{
if( n%i==0 )
{
x.add(i);
no++;
n = n/i;
if(no==2)
break;
}
}
if(n!=1&&no==2 && n!=x.get(0) && n!=x.get(1) )
{
System.out.println("YES");
System.out.println( x.get(0) + " " + x.get(1) + " " + n );
}
else
System.out.println("NO");
}
}
} | JAVA |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | from sys import stdin
from functools import reduce
def solve():
for _ in range(int(stdin.readline())):
n=int(stdin.readline())
#a=list(set(reduce(list.__add__,([i,n//i]for i in range(1,int(n**(.5))+1) if n%i==0))))
flag,a1,a2,a3,tmp=0,0,0,0,0
for i in range(2,int(n**.5)+1):
if flag==1:
break
if n%i==0:
tmp=n//i
a1=i
for j in range(2,int((tmp)**.5)+1):
if tmp%j==0:
a2=j
a3=tmp//j
if a1!=a2 and a2!=a3 and a3!=a1:
flag=1
break
if flag==1:
print('YES')
print(a1,a2,a3,sep=' ')
else:
print('NO')
solve()
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | #Author: ghoshashis545 -Ashis Ghosh
from __future__ import division, print_function
import os,sys
from io import BytesIO, IOBase
from math import ceil,sqrt
if sys.version_info[0] < 3:
from __builtin__ import xrange as range
from future_builtins import ascii, filter, hex, map, oct, zip
def ii(): return int(input())
def si(): return input()
def mi(): return map(int,input().strip().split(" "))
def li(): return list(mi())
mod=1000000007
#mod=998244353
inf = float("inf")
vow=['a','e','i','o','u']
dx,dy=[-1,1,0,0],[0,0,1,-1]
def getKey(item): return item[0]
def sort2(l):return sorted(l, key=getKey)
def d2(n,m,num):return [[num for x in range(m)] for y in range(n)]
def isPowerOfTwo (x): return (x and (not(x & (x - 1))) )
def decimalToBinary(n): return bin(n).replace("0b","")
def ntl(n):return [int(i) for i in str(n)]
abd={'a': 0, 'b': 1, 'c': 2, 'd': 3, 'e': 4, 'f': 5, 'g': 6, 'h': 7, 'i': 8, 'j': 9, 'k': 10, 'l': 11, 'm': 12, 'n': 13, 'o': 14, 'p': 15, 'q': 16, 'r': 17, 's': 18, 't': 19, 'u': 20, 'v': 21, 'w': 22, 'x': 23, 'y': 24, 'z': 25}
abc='abcdefghijklmnopqrstuvwxyz'
def powerMod(x,y,p):
res = 1
x %= p
while y > 0:
if y&1:
res = (res*x)%p
y = y>>1
x = (x*x)%p
return res
def gcd(x, y):
while y:
x, y = y, x % y
return x
def isPrime(n) : # Check Prime Number or not
if (n <= 1) : return False
if (n <= 3) : return True
if (n % 2 == 0 or n % 3 == 0) : return False
i = 5
while(i * i <= n) :
if (n % i == 0 or n % (i + 2) == 0) :
return False
i = i + 6
return True
def fact(n):
f=1
for i in range(1,n+1):
f*=i
return f
def f(n):
a=[]
a.append(1)
for i in range(2,int(sqrt(n))+1):
if(n%i==0):
if(i==n//i):
a.append(i)
else:
a.append(i)
a.append(n//i)
return a
''''''''''''''''''''''''''''''''''''''''''''''''''''''
t=ii()
while(t):
t-=1
n=ii()
f=0
for i in range(2,int(sqrt(n))+1):
if(n%i==0):
a=[]
a.append(i)
x=n//i
for j in range(2,int(sqrt(x))+1):
if(x%j==0 and j not in a):
a.append(j)
a.append(x//j)
if(len(set(a))==3):
print("YES")
print(*a)
f=1
break
if(f==1):
break
a=[]
a.append(n//i)
x=i
for j in range(2,int(sqrt(x))+1):
if(x%j==0 and j not in a):
a.append(j)
a.append(x//j)
if(len(set(a))==3):
print("YES")
print(*a)
f=1
break
if(f==1):
break
if(f==0):
print("NO")
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | t = int(input())
for i in range(t):
count = 0
flag = 0
a = []
n = int(input())
for j in range(2,int(n**0.5)+1):
if (count == 2 and n < j):
print("NO")
flag = 1
break
elif (count == 2 and n>j):
print("YES")
for x in a:
print(x,end = " ")
print(int(n))
count = count + 1
break
elif(count == 2 and n == j):
if j in a:
print("NO")
flag = 1
break
else:
print("YES")
for x in a:
print(x,end = " ")
print(int(n))
count = count + 1
break
if( n % j == 0):
if j not in a:
n = n/j
count += 1
a.append(j)
if (count <=2 and flag == 0):
print("NO") | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
while (n--) {
int flag = 0;
long long int q;
cin >> q;
for (long int i = 2; i <= sqrt(q); i++) {
if (q % i == 0) {
for (long int j = i + 1; j <= sqrt(q / i); j++) {
if ((q / i) % j == 0 && j != ((q / i) / j)) {
flag = 1;
cout << "YES" << endl;
cout << i << " " << j << " " << (q / i) / j << endl;
break;
}
}
if (flag == 1) break;
}
}
if (flag == 0) cout << "NO" << endl;
}
}
| CPP |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import sys
from math import sqrt
input=sys.stdin.readline
f=lambda :list(map(int, input().split()))
for _ in range(int(input())):
n=int(input())
a=0
for i in range(2, int(sqrt(n))+1):
if n%i==0:
a=i
break
b=0
if a:
for i in range(a+1, int(sqrt(n//a))+1):
if (n//a)%i==0:
b=i
break
c=0
if b and n%(a*b)==0:
c=n//(a*b)
if c and c!=1 and c!=a and c!=b:
print('YES\n', a, b, c)
else:
print('NO') | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | for i in range(int(input())):
n = int(input())
ans1 = 1
ans2 = 1
ans3 = 1
p = 2
while n % p == 0:
if ans1 == 1:
ans1 *= p
elif ans2 == 1 or ans2 == ans1:
ans2 *= p
else:
ans3 *= p
n //= p
p = 3
while p <= n ** 0.5:
while n % p == 0:
if ans1 == 1:
ans1 *= p
elif ans2 == 1 or ans2 == ans1:
ans2 *= p
else:
ans3 *= p
n //= p
p += 2
if n != 1:
if ans1 == 1:
ans1 *= n
elif ans2 == 1 or ans2 == ans1:
ans2 *= n
else:
ans3 *= n
if ans1 != 1 and ans2 != 1 and ans3 != 1:
if ans1 != ans2 and ans1 != ans3 and ans2 != ans3:
print('YES')
print(ans1, ans2, ans3)
else:
print('NO')
else:
print('NO') | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | from math import sqrt
t = int(input())
for _ in range(t):
n = int(input())
ans = []
for i in range(2, int(sqrt(n))+1):
if len(ans) == 2 and n>ans[-1]:
ans.append(n)
n = 1
break
elif n%i==0:
ans.append(i)
n//=i
if n==1:
break
if len(ans) and n<=ans[-1]:
break
if n!=1 or len(ans)<3:
print("NO")
else:
print("YES")
prod = 1
for i in range(2, len(ans)):
prod*=ans[i]
print(ans[0], ans[1], prod) | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
vector<int> printDivisors(int n) {
vector<int> v;
for (int i = 2; i <= sqrt(n); i++) {
if (n % i == 0) {
if (n / i == i)
v.push_back(i);
else {
v.push_back(n / i);
v.push_back(i);
}
}
}
return v;
}
int main() {
int t;
cin >> t;
for (int i = 0; i < t; i++) {
int n;
cin >> n;
bool p = false;
vector<int> v = printDivisors(n);
sort(v.begin(), v.end());
for (int r = 0; r < v.size(); r++) {
int j = v.at(r);
int n1 = n / j;
for (int k = 2; k <= sqrt(n1); k++) {
if (n1 % k == 0) {
if (n1 / k == k)
continue;
else {
if (j != k && j != n1 / k) {
cout << "YES" << endl;
cout << j << " " << k << " " << n1 / k << endl;
p = true;
break;
}
}
}
}
if (p) break;
}
if (!p) cout << "NO" << endl;
}
return 0;
}
| CPP |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | rr = raw_input
rri = lambda: int(raw_input())
rrm = lambda: map(int, raw_input().split())
MOD = 10 ** 9 + 7
INF = float('inf')
NINF = float('-inf')
YES, NO = "YES", "NO"
##
def factorize(n):
d = 2
facs = []
while d * d <= n:
e = 0
while n % d == 0:
n //= d
e += 1
if e:
facs.append((d,e))
d += 1 if d==2 else 2
if n > 1:
facs.append((n,1))
return facs
def solve(N):
# 2 <= a,b,c and abc = n or no
# a,b,c distinct
facs = factorize(N)
if len(facs) >= 3:
a = facs[0][0] ** facs[0][1]
b = facs[1][0] ** facs[1][1]
c = 1
for i in range(2, len(facs)):
c *= facs[i][0] ** facs[i][1]
return a,b,c
if len(facs) == 1:
p,e = facs[0]
if e < 6: return None
a,b,c= p, p*p, p**(e-3)
return a,b,c
(p1,e1),(p2,e2) = facs
if e1 == e2 == 1:
return None
a = p1
b = p2
c = p1**(e1-1)
c *= p2 **(e2-1)
if a != b != c != a:
return a,b,c
return None
##
T = rri()
for tc in xrange(T):
ans = solve(rri())
if ans is None:
print NO
else:
print YES
print "{} {} {}".format(*ans)
| PYTHON |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import math
from operator import mul
from functools import reduce
def isPrime(n) :
# Corner cases
if (n <= 1) :
return False
if (n <= 3) :
return True
# This is checked so that we can skip
# middle five numbers in below loop
if (n % 2 == 0 or n % 3 == 0) :
return False
i = 5
while(i * i <= n) :
if (n % i == 0 or n % (i + 2) == 0) :
return False
i = i + 6
return True
T = int(input())
for _ in range(T):
N = int(input())
n = N
zz = False
for i in range(2, math.floor(n**.5) + 1):
if n % i == 0:
n = n//i
for j in range(2, math.floor(n**.5) + 1):
if n % j == 0 and j != i:
n = n//j
if n != i and n != j and n > 1:
print ('YES')
print ('{} {} {}'.format(i, j, n))
zz=True
break
if zz:
break
if zz:
continue
print ('NO')
'''
# find the prime divisors of n
exp = []
while n > 1:
for m in range(2, n + 1):
if isPrime(m) and (n % m) == 0:
e = 1
while True:
if n % m**e == 0:
e += 1
else:
exp.append((m, e - 1))
break
break
n = n//(m**(e - 1))
if len(exp) >= 3:
break
abc = []
if len(exp) >= 3:
print ('YES')
#
abc.append(exp[0][0])
abc.append(exp[1][0])
abc.append(N//(abc[0]*abc[1]))
print (' '.join(map(str, abc)))
continue
if len(exp) == 2:
if sum([j for i, j in exp]) >= 4:
print ('YES')
#
abc.append(exp[0][0])
abc.append(exp[1][0])
abc.append(N//(abc[0]*abc[1]))
print (' '.join(map(str, abc)))
continue
if len(exp) == 1:
if sum([j for i, j in exp]) >= 6:
print('YES')
abc.append(exp[0][0])
abc.append(exp[0][0]**2)
abc.append(N//(abc[0]*abc[1]))
print (' '.join(map(str, abc)))
continue
print ('NO')
'''
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | // Working program using Reader Class
// Probably fastest
import java.io.DataInputStream;
import java.io.FileInputStream;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Scanner;
import java.util.StringTokenizer;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Random;
import java.util.StringTokenizer;
import java.util.List;
import java.util.Collections;
import java.util.Map;
import java.util.HashMap;
import java.util.Comparator;
import java.util.stream.IntStream;
import java.util.ArrayDeque;
import java.util.Set;
import java.util.HashSet;
import java.util.PriorityQueue;
public class Main1 {
private static PrintWriter out = new PrintWriter(System.out);
// private static Scanner sc = new Scanner(System.in);
public static void main(String[] args) {
int t = fs.nextInt();
outer :while(t-->0)
{
List<Integer> list = new ArrayList<>();
// Set<Integer> set = new HashSet<>();
int n = fs.nextInt();
int n1 =n;
int count =0;
// int arr[] = new int [1001];
int res[] = new int[2];
int kl = 0;
// int res[] = new int[2];
for(int i=2;i*i<=n;i++)
{
if(n%i==0 &&kl<2)
{
count++;
res[kl] = i;
// System.out.println(i);
kl++;
}
while(n%i==0)
{
list.add(i);
n/=i;
}
}
if(n>1)
{
if(kl==1 && n != res[kl-1])
{
res[kl] = n;
count++;
}
list.add(n);
}
if(list.size()>=6)
{
out.println("YES");
int m = list.size();
int a1 = list.get(m-1);
int b1 = list.get(m-2)*list.get(m-3);
int c1 =1;
for(int i=0;i<(m-3);i++)
c1 *= list.get(i);
out.println(a1+" "+b1+" "+c1);
}
else if(count>=2)
{
int a = res[0];
int b = res[1];
// System.out.println(a+" "+b);
if(n1%(a*b)==0 && (n1/(a*b)) !=a && (n1/(a*b)) !=b && (n1/(a*b)) !=1)
{
out.println("YES");
out.println(a+" "+b+" "+(n1/(a*b)));
}
else out.println("NO");
}
else out.println("NO");
}
out.close();
}
interface Input {
public String next();
public String nextLine();
public int nextInt();
public long nextLong();
public double nextDouble();
}
static class StdIn implements Input {
final private int BUFFER_SIZE = 1 << 16;
final private int STRING_SIZE = 1 << 11;
private DataInputStream din;
private byte[] buffer;
private int bufferPointer, bytesRead;
public StdIn() {
din = new DataInputStream(System.in);
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public StdIn(String filename) {
try{
din = new DataInputStream(new FileInputStream(filename));
} catch(Exception e) {
throw new RuntimeException();
}
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public String next() {
byte[] buf = new byte[STRING_SIZE]; // string length
int cnt = 0, c;
while((c=read())!=-1&&(c==' '||c=='\n'||c=='\r'));
while (c != -1)
{
if (c == ' ' || c == '\n'||c=='\r')
break;
buf[cnt++] = (byte) c;
c=read();
}
return new String(buf, 0, cnt);
}
public String nextLine() {
byte[] buf = new byte[STRING_SIZE]; // line length
int cnt = 0, c;
while((c=read())!=-1&&(c==' '||c=='\n'||c=='\r'));
while (c != -1)
{
if (c == '\n'||c=='\r')
break;
buf[cnt++] = (byte) c;
c = read();
}
return new String(buf, 0, cnt);
}
public int nextInt() {
int ret = 0;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do
{
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public int[] readIntArray(int n) {
int[] ar = new int[n];
for(int i=0; i<n; ++i)
ar[i]=nextInt();
return ar;
}
public long nextLong() {
long ret = 0;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
}
while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public double nextDouble() {
double ret = 0, div = 1;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
}
while ((c = read()) >= '0' && c <= '9');
if (c == '.')
{
while ((c = read()) >= '0' && c <= '9')
{
ret += (c - '0') / (div *= 10);
}
}
if (neg)
return -ret;
return ret;
}
private void fillBuffer() throws IOException {
bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
if (bytesRead == -1)
buffer[0] = -1;
}
private byte read() {
try{
if (bufferPointer == bytesRead)
fillBuffer();
return buffer[bufferPointer++];
} catch(IOException e) {
throw new RuntimeException();
}
}
public void close() throws IOException {
if (din == null)
return;
din.close();
}
}
private static StdIn fs = new StdIn();
} | JAVA |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import math
def primeFactors(n):
temp = []
while n % 2 == 0:
temp.append(2)
n = n / 2
for i in range(3, int(math.sqrt(n))+1,2):
while n % i== 0:
temp.append(i)
n = n / i
if n > 2:
temp.append(int(n))
return temp
t = int(input())
while t:
t += -1
n = int(input())
prime = primeFactors(n)
if len(prime) < 3: print("NO")
else:
prime.sort()
ans = [prime[0]]
if prime[0] == prime[1]:
ans.append(prime[1] * prime[2])
p = 1
for i in range(3, len(prime)):
p *= prime[i]
if p != 1: ans.append(p)
if len(set(ans)) != 3: print("NO")
else:
print("YES")
print(*ans)
else:
ans.append(prime[1])
p = 1
for i in range(2, len(prime)):
p *= prime[i]
if p != 1: ans.append(p)
if len(set(ans)) != 3: print("NO")
else:
print("YES")
print(*ans) | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
const long long MX = 2e5 + 5;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int tc;
cin >> tc;
while (tc--) {
int n;
cin >> n;
vector<int> v;
for (int i = 2; i * i <= n; i++) {
if (n % i == 0) {
v.push_back(i);
n = n / i;
break;
}
}
for (int i = 2; i * i <= n; i++) {
if (n % i == 0 && i != v[0] && i != n / i) {
v.push_back(i);
v.push_back(n / i);
break;
}
}
if (v.size() == 3)
cout << "YES\n" << v[0] << ' ' << v[1] << ' ' << v[2] << endl;
else
cout << "NO\n";
}
cerr << "\ntime taken : " << (float)clock() / CLOCKS_PER_SEC << " secs\n";
return 0;
}
| CPP |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | // package Contest1294;
import java.util.*;
import java.lang.*;
import java.io.*;
public class C {
private static FastReader fr;
private static OutputStream out;
private static int mod = (int)(1e9+7);
private void solve() {
int t = fr.nextInt();
while(t-- > 0) {
long n = fr.nextLong();
ArrayList<Long> factors = factors(n);
if(factors.size() < 3) {
println("NO");
continue;
}
Collections.sort(factors);
boolean bool = false;
long a = factors.get(0);
long b = 1;
long c = 1;
// for(long val: factors) {
// print(val + " ");
// }
for(int i=1;i<factors.size();i++) {
b = factors.get(i);
if(n % (b * a) == 0) {
c = n / (a * b);
}else {
c = 1;
}
if(a > 1 && b > 1 && c > 1 && a != b && a != c && b != c && a * b * c == n) {
bool = true;
break;
}
}
if(bool) {
println("YES");
println(a + " " + b + " " + c);
}else {
println("NO");
}
}
}
public ArrayList<Long> factors(long num){
ArrayList<Long> arr = new ArrayList<>();
for(long i=2;i*i<=num;i++) {
if(num % i == 0) {
arr.add(i);
if(num / i != i) {
arr.add(num / i);
}
}
}
return arr;
}
public static void main(String args[]) throws IOException{
new C().run();
}
private ArrayList<Integer> factors(int n, boolean include){
ArrayList<Integer> factors = new ArrayList<>();
if(n < 0)
return factors;
if(include) {
factors.add(1);
if(n > 1)
factors.add(n);
}
int i = 2;
for(;i*i<n;i++) {
if(n % i == 0) {
factors.add(i);
factors.add(n / i);
}
}
if(i * i == n) {
factors.add(i);
}
return factors;
}
private ArrayList<Integer> PrimeFactors(int n) {
ArrayList<Integer> primes = new ArrayList<>();
int i = 2;
while (i * i <= n) {
if (n % i == 0) {
primes.add(i);
while (n % i == 0) {
n /= i;
}
}
i++;
}
if (n > 1) {
primes.add(n);
}
return primes;
}
private boolean isPrime(int n) {
if(n == 0 || n == 1) {
return false;
}
if(n % 2 == 0) {
return false;
}
for(int i=3;i*i<=n;i+=2) {
if(n % i == 0) {
return false;
}
}
return true;
}
private ArrayList<Integer> Sieve(int n){
boolean bool[] = new boolean[n+1];
Arrays.fill(bool, true);
bool[0] = bool[1] = false;
for(int i=2;i*i<=n;i++) {
if(bool[i]) {
int j = 2;
while(i*j <= n) {
bool[i*j] = false;
j++;
}
}
}
ArrayList<Integer> primes = new ArrayList<>();
for(int i=2;i<=n;i++) {
if(bool[i])
primes.add(i);
}
return primes;
}
private HashMap<Integer, Integer> addToHashMap(HashMap<Integer, Integer> map, int arr[]){
for(int val: arr) {
if(!map.containsKey(val)) {
map.put(val, 1);
}else {
map.put(val, map.get(val) + 1);
}
}
return map;
}
private int factorial(int n) {
long fac = 1;
for(int i=2;i<=n;i++) {
fac *= i;
fac %= mod;
}
return (int)(fac % mod);
}
private static int pow(int base,int exp){
if(exp == 0){
return 1;
}else if(exp == 1){
return base;
}
int a = pow(base,exp/2);
a = ((a % mod) * (a % mod)) % mod;
if(exp % 2 == 1) {
a = ((a % mod) * (base % mod));
}
return a;
}
private static int gcd(int a,int b){
if(a == 0){
return b;
}
return gcd(b%a,a);
}
private static int lcm(int a,int b){
return (a * b)/gcd(a,b);
}
private void run() throws IOException{
fr = new FastReader();
out = new BufferedOutputStream(System.out);
solve();
out.flush();
out.close();
}
private static class FastReader{
private static BufferedReader br;
private static StringTokenizer st;
public FastReader() {
br = new BufferedReader(new InputStreamReader(System.in));
out = new BufferedOutputStream(System.out);
}
public String next() {
while (st == null || !st.hasMoreElements())
{
try
{
st = new StringTokenizer(br.readLine());
}
catch (IOException e)
{
e.printStackTrace();
}
}
return st.nextToken();
}
public String nextLine() {
String str = "";
try
{
str = br.readLine();
}
catch (IOException e)
{
e.printStackTrace();
}
return str;
}
public int nextInt() {
return Integer.parseInt(next());
}
public long nextLong() {
return Long.parseLong(next());
}
public double nextDouble() {
return Double.parseDouble(next());
}
public char[] nextCharArray() {
return next().toCharArray();
}
public int[] nextIntArray(int n) {
int arr[] = new int[n];
for(int i=0;i<n;i++) {
arr[i] = nextInt();
}
return arr;
}
public long[] nextLongArray(int n) {
long arr[] = new long[n];
for(int i=0;i<n;i++) {
arr[i] = nextLong();
}
return arr;
}
public String[] nextStringArray(int n) {
String arr[] = new String[n];
for(int i=0;i<n;i++) {
arr[i] = next();
}
return arr;
}
}
public static void print(Object str) {
try {
out.write(str.toString().getBytes());
} catch (IOException e) {
e.printStackTrace();
}
}
public static void println(Object str) {
try {
out.write((str.toString() + "\n").getBytes());
} catch (IOException e) {
e.printStackTrace();
}
}
public static void println() {
println("");
}
public static void printArray(Object str[]){
for(Object s : str) {
try {
out.write(str.toString().getBytes());
} catch (IOException e) {
e.printStackTrace();
}
}
}
} | JAVA |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 |
import java.util.*;
public class vc2 {
// public static int[]arr=new int[100001];
public static void main(String[] args) {
Scanner scn=new Scanner(System.in);
int t=scn.nextInt();
// prime();
while(t-- !=0) {
int n=scn.nextInt();
int i=2;
boolean found=false;
int count=0;int[]a=new int[2];
for(;i<Math.sqrt(n) && !found;i++) {
if(n%i==0) {
int m=n/i;
for(int j=i+1;j<Math.sqrt(m);j++) {
if(m%j==0) {
System.out.println("YES");
System.out.println(i+" "+j+" "+(m/j));
found=true;
break;
}
}
}
}
if(!found)
System.out.println("NO");
}
}
// public static void prime() {
// for(int i=2;i<100000;i++) {
// if(arr[i]==0) {
// for(int j=i+1;j<100000;j++) {
// if(j%i==0 && arr[j]==0)
// arr[j]=i;
// }
// }
// }
// }
}
| JAVA |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import math
t=int(input())
for i in range(0,t):
n=int(input())
n1=0
n2=0
n3=0
for j in range(2,int(math.ceil((math.sqrt(n))))):
if n%j==0:
n1=j
n3=n/j
break
else:
continue
z=0
for j in range(n1+1,(int(math.ceil((math.sqrt(n3)))))):
if n3%j==0 and j!=n3/j and (n3/j)!=n1:
z=1
n2=j
n3=int(n3/j)
break
else:
continue
if z==0:
print("NO")
else:
print("YES")
print(n1,n2,n3)
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
long long myCeil(long long num, long long divider) {
long long result = num / divider;
if ((result * divider) == num) return result;
return result + 1;
}
long long gcd(long long a, long long b) {
if (!b) return a;
return gcd(b, a % b);
}
int main() {
ios_base::sync_with_stdio(false);
long long tt;
cin >> tt;
while (tt--) {
long long n;
cin >> n;
long long limit = sqrt(n) + 1;
vector<long long> V;
long long tempN = n;
for (long long i = 2; i <= limit; i++) {
if (tempN % i == 0) {
V.push_back(i);
tempN /= i;
if (V.size() == 2) {
V.push_back(tempN);
break;
}
}
}
if (V.size() == 3) {
if (V[1] < V[2]) {
cout << "YES"
<< "\n";
for (long long i = 0; i < 3; i++) {
cout << V[i] << " ";
}
cout << "\n";
continue;
}
}
cout << "NO"
<< "\n";
}
return 0;
}
| CPP |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | #include <bits/stdc++.h>
int main() {
long long int n, t, m, i, a, b, c, k;
scanf("%lld", &t);
for (m = 1; m <= t; m++) {
k = 0;
scanf("%lld", &n);
for (i = 2; i <= sqrt(n); i++) {
if (n % i == 0) {
a = i;
n = n / i;
k++;
break;
}
}
if (k > 0) {
for (i = a + 1; i <= sqrt(n); i++) {
if (n % i == 0) {
break;
}
}
if (i != n / i && n / i != a && i * (n / i) == n && n / i != 1) {
b = i;
c = n / i;
printf("YES\n");
printf("%lld %lld %lld\n", a, b, c);
} else {
printf("NO\n");
}
}
if (k == 0) {
printf("NO\n");
}
}
return 0;
}
| CPP |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | t = int(input())
for y in range(t):
n = int(input())
i = 2
a = n
ans = []
while(i*i <= n):
if(a%i == 0):
ans.append(i)
a //= i
i += 1
if len(ans) == 2: break
if(len(ans) != 2):
print("NO")
elif(a == ans[0] or a == ans[1]):
print("NO")
else:
print("YES")
print(ans[0],ans[1],a) | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import sys
from collections import defaultdict
import math
dic=defaultdict(int)
t=int(sys.stdin.readline())
for _ in range(t):
dic=defaultdict(int)
m=int(sys.stdin.readline())
d=m
l=[]
while m%2==0:
m=m//2
dic[2]+=1
#print(m,'mm')
for i in range(3,int(math.sqrt(d))+1,2):
#print(i,m)
while m%i==0:
#print('yes',i)
dic[i]+=1
m=m//i
if m>2:
dic[m]+=1
#print(dic,'dic')
if len(dic)>=3:
print("YES")
l=[]
for i in dic:
l.append(i)
if len(l)==2:
break
l.append(d//(l[0]*l[1]))
print(*l)
elif len(dic)==2:
a=[]
for i in dic:
a.append(i)
z=True
if dic[a[0]]>2 and z:
l=[a[0],a[0]**(dic[a[0]]-1)]
l.append(d//(l[0]*l[1]))
print("YES")
print(*l)
z=False
elif dic[a[1]]>2 and z:
l=[a[1],a[1]**(dic[a[1]]-1)]
l.append(d//(l[0]*l[1]))
print("YES")
print(*l)
z=False
elif dic[a[1]]==dic[a[0]]==2 and z:
l=[a[0],a[1],a[0]*a[1]]
print("YES")
print(*l)
z=False
else:
print("NO")
elif len(dic)==1:
for i in dic:
if dic[i]>5:
print("YES")
l=[i,i**2,d//(i**3)]
print(*l)
else:
print("NO")
else:
print("NO")
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 |
import java.util.*;
import java.io.*;
import java.math.*;
import javafx.util.Pair;
public class Fff {
/**
* @param args the command line arguments
*/
public static void main(String[] args) throws IOException {
BufferedReader bf=new BufferedReader(new InputStreamReader(System.in));
StringBuffer out = new StringBuffer();
int t=Integer.parseInt(bf.readLine());
while (t-- > 0) {
int n = Integer.parseInt(bf.readLine());
ArrayList<Integer> al = new ArrayList();
StringBuffer out1 = new StringBuffer();
for (int i = 2; i * i <= n; i++) {
if (n % i == 0) {
al.add(i);
al.add(n / i);
}
}
Collections.sort(al);
boolean tf=false;
for (int k = 0; k < al.size(); k++) {
int x = al.get(k);
for (int j = 2; j*j <= x; j++) {
if (x % j == 0) {
if (x / j != j&& x / j != al.get(al.size()-k-1)&&al.get(al.size()-k-1)!=j) {
out1.append((x/j)+" "+ j +" "+ al.get(al.size()-k-1));
tf=true;break;}
}
}if(tf){break;}
}
if(!tf){out.append("NO\n");}
else{out.append("YES\n"+out1+"\n");}
}
System.out.print(out);
}
}
| JAVA |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int main() {
int t;
cin >> t;
while (t--) {
long long n, i, x = 0, y = 0, j;
cin >> n;
for (i = 2; i <= sqrt(n); i++) {
if (n % i == 0) {
x = i;
n /= i;
for (j = i + 1; j <= sqrt(n); j++) {
if (n % j == 0) {
y = j;
n /= j;
break;
}
}
if (x != 0 && y != 0) break;
}
}
if (x != 0 && y != 0 && x != y && y != n && n != x)
cout << "YES" << endl << x << " " << y << " " << n << endl;
else
cout << "NO" << endl;
}
return 0;
}
| CPP |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | #include <bits/stdc++.h>
const long long int mod = 1e9 + 7;
const long long int INF = 10000000000000;
using namespace std;
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
long long int t = 0;
cin >> t;
while (t--) {
long long int n = 0;
cin >> n;
long long int tn = n;
vector<long long int> fac;
for (long long int i = 2; i * i <= n; ++i) {
if (n % i == 0) {
fac.push_back(i);
n /= i;
}
}
fac.push_back(n);
long long int p = 1;
if (fac.size() < 3)
cout << "NO\n";
else {
for (long long int i = 3; i < fac.size(); ++i) {
fac[2] *= fac[i];
}
if ((fac[0] == fac[1]) || (fac[1] == fac[2]) || fac[2] == fac[0])
cout << "NO\n";
else {
cout << "YES\n";
cout << fac[0] << " " << fac[1] << " " << fac[2] << "\n";
}
}
}
return 0;
}
| CPP |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import sys
from collections import defaultdict
from math import sqrt
input=sys.stdin.readline
for _ in range(int(input())):
n=int(input())
nn=n
d=defaultdict(int)
while n%2==0:
d[2]+=1
n//=2
for i in range(3,int(sqrt(n))+1,2):
while n%i==0:
d[i]+=1
n//=i
if n>1:
d[n]=1
if len(d)==1 and sum(d.values())<6:
print("NO")
continue
if len(d)==1:
a=list(d.keys())[0]
b=a*2
c=nn//(a*b)
print("YES")
print(a,b,c)
continue
if len(d)==2 and sum(d.values())<4:
print("NO")
continue
x=list(d.keys())
a=x[0]
b=x[1]
c=nn//(a*b)
print("YES")
print(a,b,c)
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | # Why do we fall ? So we can learn to pick ourselves up.
from functools import reduce
def factors(n):
k = sorted(list(set(reduce(list.__add__,
([i,n//i] for i in range(1,int(n**0.5+1)) if n%i == 0)))))
return k[1:]
def solve():
n = int(input())
f = factors(n)
if len(f) < 3:
print('NO')
else:
sss = 'NO'
a, b, c = -1, -1, -1
for i in range(0, len(f)):
for j in range(i + 1, len(f)):
temp = n // f[i]
temp //= f[j]
if temp in f and temp != f[i] and temp != f[j]:
a, b, c = f[i], f[j], temp
sss = 'YES'
break
if sss == 'YES':
break
if sss == 'YES':
print(sss)
print(a, b, c)
else:
print(sss)
t = int(input())
for _ in range(0,t):
solve()
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | from collections import Counter
def prime_factorize(n):
a = []
while n % 2 == 0:
a.append(2)
n //= 2
f = 3
while f * f <= n:
if n % f == 0:
a.append(f)
n //= f
else:
f += 2
if n != 1:
a.append(n)
a = Counter(a)
return a
t = int(input())
for i in range(t):
n = int(input())
a = prime_factorize(n)
flag = True
ans = ""
if len(a.keys()) == 1:
for k, v in a.items():
if v <= 5:
flag = False
else:
ans = "{} {} {}".format(k, k**2, k**(v-3))
elif len(a.keys()) == 2:
v_pair = []
vk_pair = []
for k, v in a.items():
v_pair.append(v)
vk_pair.append((v, k))
if max(v_pair) >= 3 or (v_pair[0] >= 2 and v_pair[1] >= 2):
ans = "{} {} {}".format(
vk_pair[0][1], vk_pair[1][1], vk_pair[0][1] ** (vk_pair[0][0] - 1) * vk_pair[1][1] ** (vk_pair[1][0] - 1))
else:
flag = False
else:
b = 1
for j, (k, v) in enumerate(a.items()):
if j <= 1:
ans += "{} ".format(str(k ** v))
else:
b *= k ** v
ans += "{}".format(b)
if flag:
print("YES")
print(ans)
else:
print("NO")
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import math
def printDivisors(n) :
i = 2
factors = {}
while i <= math.sqrt(n):
if (n % i == 0) :
if (n // i == i) :
factors[i] = 1
else :
factors[n//i] = 1
factors[i] = 1
i = i + 1
return factors
for t in range(int(input())):
n = int(input())
f = printDivisors(n)
factors = list(f.keys())
factors.sort()
if len(factors) < 3:
print('NO')
else:
i = 1
flag = False
while i < len(factors)//2:
temp = factors[0]*factors[i]
try:
x = f[temp]
if n//temp != factors[0] and n//temp != factors[i]:
print('YES')
print(factors[0],factors[i],n//temp)
flag = True
break
else:
i += 1
except KeyError:
i += 1
if not(flag):
print('NO') | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import math
t = int(input())
for _ in range(t):
n = int(input())
a = []
count = 0
for i in range(2, int(math.sqrt(n))):
if n%i==0:
n = n//i
a.append(i)
count += 1
if count == 2:
break
if count == 2 and a[1]!=n and a[0]!=n and n!=1:
print("YES")
print(a[0], a[1], n)
else:
print("NO")
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import math
n = int(input())
Resulting_bool_list = []
Resulting_factor_list = []
for i in range(0,n):
factor_list = []
temp = []
num = int(input())
lim = math.floor(math.sqrt(num))
for i in range(2,lim+1):
if num % i == 0:
factor_list.append(i)
len_ = len(factor_list)
if len_ < 3:
Resulting_bool_list.append('NO')
Resulting_factor_list.append([])
else:
Resulting_bool_list.append('YES')
temp.append(factor_list[0])
num = num / factor_list[0]
for j in range(1,len_):
if num % factor_list[j] == 0:
temp.append(factor_list[j])
temp.append(int(num/factor_list[j]))
break
Resulting_factor_list.append(temp)
for i in range(0,n):
if Resulting_bool_list[i] == 'NO':
print(Resulting_bool_list[i])
else:
print(Resulting_bool_list[i])
for j in Resulting_factor_list[i]:
print(j,end=' ')
print('') | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | for i in range(int(input())):
n = int(input())
a = []
i=2
while(len(a)<2 and i*i <n):
if n%i ==0:
n=n//i
a.append(i)
i+=1
if len(a)==2 and n not in a:print("YES");print(n,*a)
else:print("NO")
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import sys,math,itertools
from collections import Counter,deque,defaultdict
from bisect import bisect_left,bisect_right
from heapq import heappop,heappush,heapify, nlargest
from copy import deepcopy
mod = 10**9+7
INF = float('inf')
def inp(): return int(sys.stdin.readline())
def inpl(): return list(map(int, sys.stdin.readline().split()))
def inpl_1(): return list(map(lambda x:int(x)-1, sys.stdin.readline().split()))
def inps(): return sys.stdin.readline()
def inpsl(x): tmp = sys.stdin.readline(); return list(tmp[:x])
def err(x): print(x); exit()
def prime_factorize(n):
if n == 1:
return [1]
a = []
while n % 2 == 0:
a.append(2)
n //= 2
f = 3
while f * f <= n:
if n % f == 0:
a.append(f)
n //= f
else:
f += 2
if n != 1:
a.append(n)
return a
for _ in range(inp()):
n = inp()
pp = prime_factorize(n)
c = Counter(pp)
bad = False
if (len(c.keys()) == 1 and len(pp)<6) or (len(c.keys())==2 and len(pp)<4):
print('NO')
else:
keys = list(c.keys())
if len(c.keys()) > 1:
a = keys[0]; c[keys[0]] -= 1
b = keys[1]; c[keys[1]] -= 1
C = 1
for key in keys:
if c[key] == 0: continue
C *= key**c[key]
else:
a = keys[0]*1; b = keys[0]**2
C = keys[0]**(c[keys[0]]-3)
print('YES')
print(a,b,C) | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | from math import sqrt
for i in range(int(input())):
a=int(input())
for j in range (2,int(sqrt(a))+1):
if a%j==0:
a=a//j
for k in range(j+1,int(sqrt(a))+1):
if a%k==0:
a=a//k
if a>k:
print("YES")
print(j,k,a)
break
else:
print("NO")
break
else:
print("NO")
break
break
else:
print("NO") | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayDeque;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.HashMap;
import java.util.HashSet;
import java.util.List;
import java.util.Map;
import java.util.Set;
import java.util.StringTokenizer;
public class CSES {
static long mod = 1000000007;
public static void main(String[] args) throws IOException {
FastScanner s=new FastScanner();
int t1 = s.nextInt();
// int t1 =1;
while(t1-->0){
long n = s.nextLong();
long a=0,b=0,c=0;
for(int i=2;i*i<=n;i++){
if(n%i==0) {
if (a == 0 ){
a=i;
n/=i;
break;
}
}
}
for(int i=2;i*i<=n;i++){
if(n%i==0) {
if(b==0 && i!=a){
b=i;
n/=i;
}
}
}
if(a==0 || b==0 || n==1 || n==a || n==b) System.out.println("NO");
else{
System.out.println("YES");
System.out.println(a + " " + b + " " + n);
}
}
}
static long gcd(long a, long b)throws IOException{return (b==0)?a:gcd(b,a%b);}
static int gcd(int a, int b)throws IOException{return (b==0)?a:gcd(b,a%b);}
static void sort(int[] a) {
ArrayList<Integer> l=new ArrayList<>();
for (int i:a) l.add(i);
Collections.sort(l);
for (int i=0; i<a.length; i++) a[i]=l.get(i);
}
static void sort(long[] a) {
ArrayList<Long> l=new ArrayList<>();
for (long i:a) l.add(i);
Collections.sort(l);
for (int i=0; i<a.length; i++) a[i]=l.get(i);
}
static class FastScanner {
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st=new StringTokenizer("");
String next() {
while (!st.hasMoreTokens())
try {
st=new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
int[] readArray(int n) {
int[] a=new int[n];
for (int i=0; i<n; i++) a[i]=nextInt();
return a;
}
long nextLong() {
return Long.parseLong(next());
}
}
} | JAVA |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const int N = (int)998244353;
const int M = (int)1000000007;
long long int gcd(long long int a, long long int b) {
if (b == 0) return a;
return gcd(b, a % b);
}
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
long long int t;
cin >> t;
for (int j = 0; j < t; j++) {
int n, count = 0, b = 0, c = 0, flag = 0;
cin >> n;
for (int i = 2; i <= sqrt(n); i++) {
if (n % i == 0) {
c = i;
n = n / i;
flag = 1;
break;
}
}
for (int i = c + 1; i < sqrt(n) + 1 && flag; ++i) {
if (n % i == 0) {
b = i;
n = n / i;
break;
}
}
if (n == 0 || b == 0 || c == 0 || b == n || c == n)
cout << "NO"
<< "\n";
else {
cout << "YES"
<< "\n";
cout << c << " " << b << " " << n << "\n";
}
}
return 0;
}
| CPP |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | nn = int(input())
import math
def isprime(n):
if n <= 7:
return False, 1
sqrt = int( math.sqrt(n) ) + 1
for i in range(2, sqrt):
if n % i == 0:
return True, i
return False, 1
def isprime2(n):
if n <= 3:
return False, 1
sqrt = int( math.sqrt(n) ) + 1
for i in range(2, sqrt):
if n % i == 0:
return True, i
return False, 1
for dfdfdi in range(nn):
v = int(input())
if v <= 7:
print("NO")
continue
firstTrue, firstPrime = isprime(v)
if not firstTrue:
print("NO")
continue
v = v // firstPrime
secondTrue, secondPrime = isprime2(v)
if not secondTrue:
print("NO")
continue
final = v // secondPrime
if firstPrime == secondPrime or secondPrime == final:
thirdTrue, thirdPrime = isprime2(final)
if not thirdTrue:
print("NO")
else:
secondPrime *= thirdPrime
final = final // thirdPrime
#if secondPrime == final:
if secondPrime == final or secondPrime == firstPrime or final == firstPrime:
print("NO")
continue
else:
print("YES")
print(str(firstPrime) + " " + str(secondPrime) + " " + str(final))
else:
print("YES")
print(str(firstPrime) + " " + str(secondPrime) + " " + str(final))
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 |
# -*- coding: utf-8 -*-
# @Date : 2020-01-23 16:32:49
# @Author : raj lath ([email protected])
# @Link : link
# @Version : 1.0.0
import sys
sys.setrecursionlimit(10**5+1)
inf = int(10 ** 20)
max_val = inf
min_val = -inf
RW = lambda : sys.stdin.readline().strip()
RI = lambda : int(RW())
RMI = lambda : [int(x) for x in sys.stdin.readline().strip().split()]
RWI = lambda : [x for x in sys.stdin.readline().strip().split()]
for _ in range(RI()):
inp = RI()
answer = []
i = 2
while len(answer) < 2 and i * i < inp:
if inp % i == 0:
answer.append(i)
inp //= i
i += 1
if len(answer) == 2 and inp not in answer:
print("YES")
print(*answer, inp)
else:
print("NO")
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import math
from collections import Counter
# def printDivisors(n) :
# i = 1
# l = []
# while i <= math.sqrt(n):
# if (n % i == 0) :
# if (n / i == i) :
# l.append(i)
# else :
# l.append(i)
# l.append(n//i)
# i = i + 1
# return l
def primeFactors(n):
l = []
while n % 2 == 0:
l.append(2)
n = n // 2
for i in range(3,int(math.sqrt(n))+1,2):
while n % i == 0:
l.append(i)
n = n // i
if n > 2:
l.append(n)
return l
for _ in range(int(input())):
n = int(input())
k = primeFactors(n)
# print(*k)
klen = len(set(k))
if len(k) < 3:
print("NO")
elif klen == 3 and len(k) == 3:
print("YES")
print(*k)
elif klen == 2 and len(k) >= 4:
print("YES")
a = k[0]
b = k[-1]
c = 1
for t in k[1:len(k)-1]:
c *= t
print(a, b, c)
elif klen >= 3:
print("YES")
kl = set(k)
jk = list(kl)
a = jk[0]
b = jk[1]
x = 1
for t in k:
x *= t
print(a, b, ((x//a)//b))
elif len(k) >= 6:
print("YES")
a = k[0]
b = k[1]*k[2]
c = 1
for t in range(3, len(k)):
c *= k[t]
print(a, b, c)
else:
a = k[0]
flag = 0
x = 1
for t in k[1::]:
x *= t
b = -1
c = -1
for t in range(1, len(k)):
b = k[t]
c = x//k[t]
x //= k[t]
if a != b and b != c and a != c and a*b*c == n:
print("YES")
flag = 1
print(a, b, c)
break
if flag == 0:
print("NO") | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import math
def div(n):
y=n
c=0
ans=[]
for i in range(2,int(math.sqrt(n))+1):
if y%i==0:
c+=1
y=y//i
ans.append(i)
if c>=2:
break
print("YES" if c>=2 and y not in ans else "NO")
if c>=2 and y not in ans:
print(*(ans+[y]))
for _ in range(int(input())):
n = int(input())
div(n) | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import sys
from math import sqrt
inp = sys.stdin.readline
for i in range(int(inp())):
n = int(inp())
arr = ""
c = 0
flag = 0
for i in range(2, round(sqrt(n))+1):
if n % i == 0:
c += 1
arr += str(i)+" "
n = n//i
if c == 2 and n > i:
flag = 1
arr += str(n)
break
if flag:
print("YES\n" + arr)
else:
print("NO") | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import math
t=int(input())
for i in range(t):
s=int(input())
s1=0
d=0
f=0
for j in range(2,int(math.sqrt(s))+1):
if s%j==0:
s1=j
d=s/j
for k in range(2,int(math.sqrt(d))+1):
if d%k==0:
s2=k
s3=d/k
if s2!=s3 and s3!=s1 and s2!=s1:
print("YES")
print(int(s1),int(s2),int(s3))
f=1
break
if f==1:
break
for k in range(2,int(math.sqrt(s1))+1):
if s1%k==0:
s2=k
s3=s1/k
if s2!=s3 and s3!=d and s2!=d:
print("YES")
print(int(s1),int(s2),int(s3))
f=1
break
if f==1:
break
if f==0:
print("NO")
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | for i in range(int(input())):
n = int(input())
m = []
i = 2
while i**2 < n and len(m) != 2:
if n % i == 0:
m.append(i)
n = n / i
i += 1
if len(m) != 2 or n == 1:
print('NO')
continue
m.append(int(n))
print('YES')
for i in m:
print(i, end = ' ')
print() | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
int t;
cin >> t;
while (t--) {
int n;
cin >> n;
vector<int> divisors;
for (int i = 2; i * i <= n; i++) {
if (n % i == 0) {
divisors.push_back(i);
if (i * i < n) divisors.push_back(n / i);
}
}
divisors.push_back(n);
vector<int> ans;
int val;
for (auto it1 : divisors) {
for (auto it2 : divisors) {
if (it1 == it2) continue;
val = n;
val /= it1;
val /= it2;
if (val >= 2 && ((val * it1) * it2) == n && val != it1 && val != it2) {
ans.push_back(it1);
ans.push_back(it2);
ans.push_back(val);
break;
}
}
if (!ans.empty()) break;
}
if (!ans.empty()) {
cout << "YES\n";
for (auto it : ans) cout << it << " ";
} else
cout << "NO";
cout << "\n";
}
}
| CPP |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int dx8[] = {0, 0, 1, 1, 1, -1, -1, -1};
int dy8[] = {1, -1, 1, -1, 0, 0, -1, 1};
const double PI = acos(-1.0);
const double EPS = 1e-6;
const int MOD = (int)1e9 + 7;
const int maxn = (int)2e5 + 5;
const int LOGN = 20;
bitset<10000005> isprime;
vector<long long> prime;
vector<long long> ans;
void sieve() {
isprime[0] = isprime[1] = 1;
prime.push_back(2);
for (long long i = 3; i * i <= 10000005; i += 2) {
if (!isprime[i]) {
for (long long j = i * i; j <= 10000005; j += i + i) {
isprime[j] = 1;
}
}
}
for (long long i = 3; i <= 10000005; i += 2) {
if (!isprime[i]) {
prime.push_back(i);
}
}
}
int main() {
long long a, b, cnt, n, s, t;
cin >> t;
sieve();
while (t--) {
int f = 0;
cin >> n;
b = n;
for (int i = 0; prime[i] * prime[i] <= n && i < prime.size(); i++) {
if (n % prime[i] == 0) {
while (n % prime[i] == 0) {
n /= prime[i];
ans.push_back(prime[i]);
}
}
}
if (n != 1) ans.push_back(n);
int x = ans[0], y = 1, z = 1, j;
for (int i = 1; i < ans.size(); i++) {
y *= ans[i];
if (y == x)
continue;
else {
j = i;
break;
}
}
for (j++; j < ans.size(); j++) {
z *= ans[j];
}
if (x != 1 && y != 1 && z != 1 && x != y && x != z && y != z) {
cout << "YES" << endl;
cout << x << " " << y << " " << z << endl;
} else
cout << "NO" << endl;
ans.clear();
}
return 0;
}
| CPP |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | def factor(n):
s=set()
for i in range(2,int(n**0.5)+1):
if n%i==0:
s.add(i)
s.add(n//i)
f=list(s)
f.sort()
return f
t=int(input())
while t:
t-=1
n=int(input())
f=factor(n)
# print(f)
b=0
if f:
for i in range(len(f)//2):
for j in range(i+1,len(f)//2):
if n%(f[i]*f[j])==0:
l=[f[i],f[j],n//(f[i]*f[j])]
if len(set(l))==3:
print('YES')
print(*l)
b=1
break
if b:break
if b==0:
print('NO')
else:
print('NO')
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | t = int(input())
import math
for i in range(t):
n = int(input())
res = []
c = 0
a = 0
q = int(math.sqrt(n)) + 2
for j in range(2,q):
if n % j == 0:
n = n // j
res.append(j)
a = j + 1
c += 1
break
if c == 1:
q = int(math.sqrt(n)) + 2
for j in range(a,q):
if n % j == 0:
n = n // j
res.append(j)
c += 1
break
if c < 2 or res[0] == n or res[1] == n:
print('No')
else:
print('Yes')
print(res[0],res[1],n) | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | /**
* BaZ :D
*/
import java.util.*;
import java.io.*;
import static java.lang.Math.*;
public class Main
{
static MyScanner scan;
static PrintWriter pw;
static long MOD = 1_000_000_007;
static long INF = 1_000_000_000_000_000_000L;
static long inf = 2_000_000_000;
public static void main(String[] args) {
new Thread(null, null, "BaZ", 1 << 27) {
public void run() {
try {
solve();
} catch (Exception e) {
e.printStackTrace();
System.exit(1);
}
}
}.start();
}
static void solve() throws IOException
{
//initIo(true);
initIo(false);
StringBuilder sb = new StringBuilder();
int t = ni();
while(t-->0) {
long n = ni();
long temp = n;
ArrayList<Long> primes = new ArrayList<>();
for(long i=2;i*i<=temp;++i) {
if(temp%i==0) {
primes.add(i);
while(temp%i==0) {
temp/=i;
}
}
}
if(temp>1) {
primes.add(temp);
}
if(primes.size()==1) {
long p = primes.get(0);
long curr = p;
for(int i=2;i<=6;++i) {
curr*=p;
if(curr>n) {
break;
}
}
//pl("curr : "+curr);
if(curr<=n) {
pl("YES");
pl(p+" "+(p*p)+" "+(n/(p*p*p)));
}
else {
pl("NO");
}
}
else {
long pro = primes.get(0) * primes.get(1);
if(n!=pro && n/pro != primes.get(0) && n/pro != primes.get(1)) {
pl("YES");
pl(primes.get(0)+" "+primes.get(1)+" "+(n/pro));
}
else {
pl("NO");
}
}
}
pw.flush();
pw.close();
}
static void initIo(boolean isFileIO) throws IOException {
scan = new MyScanner(isFileIO);
if(isFileIO) {
pw = new PrintWriter("/Users/amandeep/Desktop/output.txt");
}
else {
pw = new PrintWriter(System.out, true);
}
}
static int ni() throws IOException
{
return scan.nextInt();
}
static long nl() throws IOException
{
return scan.nextLong();
}
static double nd() throws IOException
{
return scan.nextDouble();
}
static String ne() throws IOException
{
return scan.next();
}
static String nel() throws IOException
{
return scan.nextLine();
}
static void pl()
{
pw.println();
}
static void p(Object o)
{
pw.print(o+" ");
}
static void pl(Object o)
{
pw.println(o);
}
static void psb(StringBuilder sb)
{
pw.print(sb);
}
static void pa(String arrayName, Object arr[])
{
pl(arrayName+" : ");
for(Object o : arr)
p(o);
pl();
}
static void pa(String arrayName, int arr[])
{
pl(arrayName+" : ");
for(int o : arr)
p(o);
pl();
}
static void pa(String arrayName, long arr[])
{
pl(arrayName+" : ");
for(long o : arr)
p(o);
pl();
}
static void pa(String arrayName, double arr[])
{
pl(arrayName+" : ");
for(double o : arr)
p(o);
pl();
}
static void pa(String arrayName, char arr[])
{
pl(arrayName+" : ");
for(char o : arr)
p(o);
pl();
}
static void pa(String listName, List list)
{
pl(listName+" : ");
for(Object o : list)
p(o);
pl();
}
static void pa(String arrayName, Object[][] arr) {
pl(arrayName+" : ");
for(int i=0;i<arr.length;++i) {
for(Object o : arr[i])
p(o);
pl();
}
}
static void pa(String arrayName, int[][] arr) {
pl(arrayName+" : ");
for(int i=0;i<arr.length;++i) {
for(int o : arr[i])
p(o);
pl();
}
}
static void pa(String arrayName, long[][] arr) {
pl(arrayName+" : ");
for(int i=0;i<arr.length;++i) {
for(long o : arr[i])
p(o);
pl();
}
}
static void pa(String arrayName, char[][] arr) {
pl(arrayName+" : ");
for(int i=0;i<arr.length;++i) {
for(char o : arr[i])
p(o);
pl();
}
}
static void pa(String arrayName, double[][] arr) {
pl(arrayName+" : ");
for(int i=0;i<arr.length;++i) {
for(double o : arr[i])
p(o);
pl();
}
}
static class MyScanner
{
BufferedReader br;
StringTokenizer st;
MyScanner(boolean readingFromFile) throws IOException
{
if(readingFromFile) {
br = new BufferedReader(new FileReader("/Users/amandeep/Desktop/input.txt"));
}
else {
br = new BufferedReader(new InputStreamReader(System.in));
}
}
String nextLine()throws IOException
{
return br.readLine();
}
String next() throws IOException
{
if(st==null || !st.hasMoreTokens())
st = new StringTokenizer(br.readLine());
return st.nextToken();
}
int nextInt() throws IOException
{
return Integer.parseInt(next());
}
long nextLong() throws IOException
{
return Long.parseLong(next());
}
double nextDouble() throws IOException
{
return Double.parseDouble(next());
}
}
} | JAVA |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | def product(n):
for i in range(2,int(n**(1/2))):
if n%i==0:
for j in range(2,int((n)**(1/2))):
if (n//i)%j==0 and j!=i and (n//i)/j!=i and j!=(n//i)**(1/2):
print("YES")
print(i,j,(n//i)//j)
return
print("NO")
return
t=int(input())
a=[]
for i in range(t):
a.append(int(input()))
for x in a:
product(x) | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | # imports
"""Fill in here"""
import math
# Set to False for input from standard in
debug = False
if debug:
f = open('test', 'r')
else:
f = None
def line(file):
if debug:
return file.__next__().strip()
return input()
# Solve
"""Fill in here"""
# array = [int(i) for i in line(f).split(' ')]
# Instead of using input() use line(f)
test_cases = int(line(f))
for _ in range(test_cases):
n = int(line(f))
def solve(n):
throot = int(math.pow(n, 1/3)) + 1
for a in range(2, throot):
if n % a == 0:
twroot = int(math.sqrt(n / a)) + 1
for b in range(a + 1, twroot):
if (n / a) % b == 0:
c = int(n / a / b)
if c != b:
return f'{a} {b} {c}'
raise EnvironmentError()
try:
string = solve(n)
print('YES')
print(string)
except EnvironmentError:
print('NO')
####
if debug:
f.close()
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | fast=lambda:stdin.readline().strip()
zzz=lambda:[int(i) for i in fast().split()]
z,zz=input,lambda:list(map(int,z().split()))
szz,graph,mod,szzz=lambda:sorted(zz()),{},10**9+7,lambda:sorted(zzz())
from re import *
from sys import *
from math import *
from heapq import *
from queue import *
from bisect import *
from string import *
from itertools import *
from collections import *
from math import factorial as f
from bisect import bisect as bs
from bisect import bisect_left as bsl
from collections import Counter as cc
from itertools import accumulate as ac
def lcd(xnum1,xnum2):return (xnum1*xnum2//gcd(xnum1,xnum2))
def output(answer):stdout.write(str(answer))
###########################---Test-Case---#################################
"""
If you Know me , Then you probably don't know me !
"""
###########################---START-CODING---##############################
num=int(z())
for _ in range( num ):
n=int(fast())
x=int(n**.5)
lst=[]
for i in range(2,x+1):
if n%i==0:
lst.append(i)
n//=i
if len(lst)==2:
if lst[1]!=n and lst[0]!=n:
lst.append(n)
break
if len(lst)==3:
print('YES')
print(*lst)
else:
print('NO')
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
int i, j, t, n;
cin >> t;
while (t--) {
cin >> n;
if (n == 1) {
cout << "NO" << endl;
continue;
}
int m = n;
map<int, int> mp;
vector<int> v;
for (i = 2; i <= sqrt(n) + 1 && m > 1; i++) {
if (m % i == 0)
v.push_back(i);
else
continue;
while (m % i == 0) {
m /= i;
mp[i]++;
}
}
if (m != 1) {
v.push_back(m);
mp[m]++;
}
if (v.size() == 0) {
cout << "NO" << endl;
continue;
}
if (v.size() > 2) {
cout << "YES" << endl;
cout << v[0] << " ";
n /= v[0];
cout << v[1] << " ";
n /= v[1];
cout << n << endl;
continue;
}
if (v.size() == 1) {
int x = mp[v[0]];
if (x < 6) {
cout << "NO" << endl;
continue;
}
int y = v[0];
n /= pow(y, 3);
cout << "YES" << endl;
cout << y << " " << y * y << " " << n << endl;
continue;
} else {
int x = mp[v[0]], y = mp[v[1]];
if ((x < 2 && y < 2) || x + y < 4) {
cout << "NO" << endl;
continue;
}
cout << "YES" << endl;
n /= (v[0] * v[1]);
cout << v[1] << " " << v[0] << " " << n << endl;
continue;
}
}
return 0;
}
| CPP |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import sys
T = int(sys.stdin.readline())
for _ in range(T):
n = int(sys.stdin.readline())
res = set()
start_i = 2
for _ in range(2):
for i in range(start_i, int(n**0.5)+1):
if not n%i:
n//=i
start_i=i+1
res.add(i)
break
res.add(n)
if len(res)==3:
print('YES')
[print(a, end=' ') for a in res]
print()
else:
print('NO') | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import sys
import math
import itertools
import functools
import collections
import operator
import fileinput
import copy
ORDA = 97 # a
def ii(): return int(input())
def mi(): return map(int, input().split())
def li(): return [int(i) for i in input().split()]
def lcm(a, b): return abs(a * b) // math.gcd(a, b)
def revn(n): return str(n)[::-1]
def dd(): return collections.defaultdict(int)
def ddl(): return collections.defaultdict(list)
def sieve(n):
if n < 2: return list()
prime = [True for _ in range(n + 1)]
p = 3
while p * p <= n:
if prime[p]:
for i in range(p * 2, n + 1, p):
prime[i] = False
p += 2
r = [2]
for p in range(3, n + 1, 2):
if prime[p]:
r.append(p)
return r
def divs(n, start=2):
r = []
for i in range(start, int(math.sqrt(n) + 1)):
if (n % i == 0):
if (n / i == i):
r.append(i)
else:
r.extend([i, n // i])
return r
def divn(n, primes):
divs_number = 1
for i in primes:
if n == 1:
return divs_number
t = 1
while n % i == 0:
t += 1
n //= i
divs_number *= t
def prime(n):
if n == 2: return True
if n % 2 == 0 or n <= 1: return False
sqr = int(math.sqrt(n)) + 1
for d in range(3, sqr, 2):
if n % d == 0: return False
return True
def convn(number, base):
new_number = 0
while number > 0:
new_number += number % base
number //= base
return new_number
def cdiv(n, k): return n // k + (n % k != 0)
def ispal(s): # Palindrome
for i in range(len(s) // 2 + 1):
if s[i] != s[-i - 1]:
return False
return True
# a = [1,2,3,4,5] -----> print(*a) ----> list print krne ka new way
for i in range(ii()):
n = ii()
r = sieve(int(n**0.5))
#n = ii()
if prime(n):
print('NO')
continue
else:
a = []
for i in r:
if n% i == 0:
a.append(i)
n = n//i
break
i= 2
while i < n**0.5:
if n %i == 0 and i not in a:
if n//i != i:
a.append(i)
a.append(n//i)
break
i += 1
if len(a) == 3:
print('YES')
print(a[0],a[1],a[2])
else:
print('NO')
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import math
for p in range(int(input())):
num=int(input())
b=[]
a=num
for i in range(2,int(math.sqrt(num))+1):
if(len(b)<2):
if(a%i==0):
b.append(int(i))
a/=i
else:
break
if((a not in b) and (len(b)==2)):
print('YES')
print(str(b[0])+' '+str(b[1])+' '+str(int(a)))
else:
print('NO') | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import math
def pr(a,m):
n=len(a)
d={}
l=[]
for i in range(n):
d[a[i]]=i
c=0
for i in range(n-1):
for j in range(i+1,n):
if ((a[i]*a[j]<=m) and (a[i]*a[j]!=0) and (m%(a[i]*a[j])==0)):
check=m//(a[i]*a[j])
if( check!=1 and check!=a[i] and check!=a[j] and check in d and d[check]>i and d[check]>j):
l=[a[i],a[j],check]
return l
def p(n) :
l=[]
i = 1
while i <= math.sqrt(n):
if (n % i == 0) :
if (n / i == i) :
if(i!=1):
l.append(i)
else :
if(i!=1):
l.append(i)
if(n//i!=1):
l.append(n//i)
i = i + 1
return l
for _ in range(int(input())):
x=int(input())
l=p(x)
l.sort()
#print(l,"d")
aa=pr(l,x)
if(aa==None):
print("NO")
else:
print("YES")
aa.sort()
print(*aa)
#print(aa) | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | for _ in range(int(input())):
num = int(input())
done = False
for i in range(2, int(num ** (1/3)) + 8):
if num % i != 0:
continue
new_num = num // i
for j in range(i + 1, int(num ** .5) + 8):
if new_num % j == 0 and new_num // j not in [i, j, 1]:
print('YES')
print(f'{i} {j} {new_num // j}')
done = True
break
if done:
break
if not done:
print('NO')
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
vector<long long> vec;
void primeFactors(long long n) {
while (n % 2 == 0) {
vec.push_back(2);
n = n / 2;
}
for (long long i = 3; i <= sqrt(n); i = i + 2) {
while (n % i == 0) {
vec.push_back(i);
n = n / i;
}
}
if (n > 2) vec.push_back(n);
}
int main() {
long long t;
cin >> t;
while (t--) {
long long n, i;
cin >> n;
primeFactors(n);
if (vec.size() < 3) {
cout << "NO" << endl;
} else {
set<long long> s;
s.insert(vec[0]);
long long x = 1;
for (i = 1; i < vec.size(); i++) {
x *= vec[i];
if (x != vec[0]) break;
}
s.insert(x);
x = (n / x) / vec[0];
if (x != 1) s.insert(x);
if (s.size() == 3) {
cout << "YES" << endl;
for (auto it = s.begin(); it != s.end(); it++) {
cout << *it << " ";
}
cout << endl;
} else {
cout << "NO" << endl;
}
}
vec.clear();
}
return 0;
}
| CPP |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | t = int(input())
for _ in range(t):
n = int(input())
rt = int(n**.5)
st = set()
flag=0
for i in range(2,rt+1):
if(n%i==0):
k = n//i
v = int(k**.5)
for j in range(2,v+1):
if(k%j==0 and i!=j):
if(k//j!=i and k//j !=j):
print("YES")
print(i,j,k//j)
flag=1
break
if(flag==1):
break
if(flag==0):
print("NO")
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | t = int(input())
for queries in range(t):
n = int(input())
m = n
r = 2
array = []
while r <= 25000:
if n % r == 0:
array.append(r)
n //= r
r += 1
if len(array) == 2:
break
if len(array) == 2 and n > array[1]:
print("YES")
print(array[0],array[1],m//array[0]//array[1])
else:
print("NO")
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import java.io.ByteArrayInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.InputMismatchException;
public class Main {
InputStream is;
PrintWriter out;
String INPUT = "";
void solve() {
outer:
for(int T = ni(); T > 0; T--) {
long n = nl();
long[][] f = factorX(n);
for(int i = 0; i < f.length; i++) {
if(f[i][0] < 2 || f[i][1] < 2) continue;
long[] ret = ok(f[i][0], f[i][1], n);
if(ret != null) {
if(ret[0] >= 2 && ret[1] >= 2 && ret[2] >= 2) {
out.println("YES");
for(int x = 0; x < ret.length; x++) {
out.print(ret[x]+" ");
}
out.println();
continue outer;
}
}
ret = ok(f[i][1], f[i][0], n);
if(ret != null) {
if(ret[0] >= 2 && ret[1] >= 2 && ret[2] >= 2) {
out.println("YES");
for(int x = 0; x < ret.length; x++) {
out.print(ret[x]+" ");
}
out.println();
continue outer;
}
}
}
out.println("NO");
}
}
public long[] ok(long a, long b, long n) {
for(long i = 2; i * i <= a; i++) {
if(n % i == 0) {
if(i * (a / i) * b == n) {
if(i != a / i && i != b && a / i != b) return new long[] {i, a / i, b};
}
}
}
return null;
}
public long[][] factorX(long n) {
long[][] f = new long[1000][2];
int p = 0;
for(int i = 2; i * i <= n; i++) {
if(n % i == 0) {
f[p][0] = i;
f[p++][1] = n / i;
}
}
return Arrays.copyOf(f, p);
}
void run() throws Exception {
is = INPUT.isEmpty() ? System.in : new ByteArrayInputStream(INPUT.getBytes());
out = new PrintWriter(System.out);
long s = System.currentTimeMillis();
solve();
out.flush();
if (!INPUT.isEmpty())
tr(System.currentTimeMillis() - s + "ms");
}
public static void main(String[] args) throws Exception {
new Main().run();
}
private byte[] inbuf = new byte[1024];
public int lenbuf = 0, ptrbuf = 0;
private int readByte() {
if (lenbuf == -1)
throw new InputMismatchException();
if (ptrbuf >= lenbuf) {
ptrbuf = 0;
try {
lenbuf = is.read(inbuf);
} catch (IOException e) {
throw new InputMismatchException();
}
if (lenbuf <= 0)
return -1;
}
return inbuf[ptrbuf++];
}
private boolean isSpaceChar(int c) {
return !(c >= 33 && c <= 126);
}
private int skip() {
int b;
while ((b = readByte()) != -1 && isSpaceChar(b))
;
return b;
}
private double nd() {
return Double.parseDouble(ns());
}
private char nc() {
return (char) skip();
}
private String ns() {
int b = skip();
StringBuilder sb = new StringBuilder();
while (!(isSpaceChar(b))) { // when nextLine, (isSpaceChar(b) && b != ' ')
sb.appendCodePoint(b);
b = readByte();
}
return sb.toString();
}
private char[] ns(int n) {
char[] buf = new char[n];
int b = skip(), p = 0;
while (p < n && !(isSpaceChar(b))) {
buf[p++] = (char) b;
b = readByte();
}
return n == p ? buf : Arrays.copyOf(buf, p);
}
private char[][] nm(int n, int m) {
char[][] map = new char[n][];
for (int i = 0; i < n; i++)
map[i] = ns(m);
return map;
}
private int[] na(int n) {
int[] a = new int[n];
for (int i = 0; i < n; i++)
a[i] = ni();
return a;
}
private int ni() {
int num = 0, b;
boolean minus = false;
while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'))
;
if (b == '-') {
minus = true;
b = readByte();
}
while (true) {
if (b >= '0' && b <= '9') {
num = num * 10 + (b - '0');
} else {
return minus ? -num : num;
}
b = readByte();
}
}
private long nl() {
long num = 0;
int b;
boolean minus = false;
while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'))
;
if (b == '-') {
minus = true;
b = readByte();
}
while (true) {
if (b >= '0' && b <= '9') {
num = num * 10 + (b - '0');
} else {
return minus ? -num : num;
}
b = readByte();
}
}
private static void tr(Object... o) {
System.out.println(Arrays.deepToString(o));
}
} | JAVA |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import java.util.Scanner;
public class Main {
/**
* @param args
*/
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
while(t-->0){
int n = sc.nextInt();
doTheTask(n);
}
}
static void doTheTask (int n)
{
for (int i=2; i*i<=n; i++)
{
if (n % i == 0)
{
int d = n/i;
for (int j=i+1; j*j<d; j++)
{
if (d % j == 0)
{
System.out.printf ("YES\n%d %d %d\n",i,j,d/j);
return;
}
}
}
}
System.out.println ("NO\n");
}
}
| JAVA |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import math
t = int(input())
for _ in range(t):
n = int(input())
ul = math.sqrt(n)
ul = int(ul)
ul += 1
l = []
count = 0
flag = 0
for i in range(2,ul+1):
if count == 2:
break
if n%i==0:
l.append(i)
count += 1
n = n//i
if count == 2 and n not in l and n!=1:
print("YES")
for i in range(2):
print(l[i],end=" ")
print(n)
else:
print("NO")
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | from math import *
t=int(input())
for _ in range(t):
n=int(input())
c=0
l=[]
r=floor(sqrt(n))+1
for i in range(2,r):
if n%i==0:
l.append(i)
c=c+1
n=n//i
if c==2:
break
if len(l)==2:
if n==l[0] or n==l[1]:
print('NO')
else:
print('YES')
print(l[0],l[1],n)
else:
print('NO')
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | for x in range(int(input())):
n = int(input())
k=0
a=set()
for i in range(2,int(n**0.5)+1):
if(n%i==0 and k<2):
n//=i
k+=1
a.add(i)
#print(i)
if(k==2 and n>1):
a.add(n)
if(len(a)==3 ):
#a.append(n)
a=sorted(a)
print('YES')
print(*a)
else:
print('NO')
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | def solve(n):
a, b, c = None, None, n
i = 2
while c >= 2 and i * i <= c:
if c % i == 0:
c //= i
if a is None:
a = i
elif c >= 2 and len(set((a, i, c))) == 3:
return a, i, c
i += 1
raise ValueError()
for T in range(int(input())):
try:
a, b, c = solve(int(input()))
print('YES')
print(a, b, c)
except ValueError:
print('NO')
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import java.io.IOException;
import java.util.*;
import javax.lang.model.util.ElementScanner6;
import java.lang.*;
import java.io.*;
public class Solution
{
public static void main(String args[])throws IOException
{/*
InputStreamReader isr=new InputStreamReader(System.in);
BufferedReader br=new BufferedReader(isr);
int a=Integer.parseInt(br.readLine());*/
FastReader in=new FastReader(System.in);
long n,d,d2,a,b,c;
long i,j,k=0;
int t=in.nextInt();
for(i=0;i<t;i++)
{
n=in.nextLong();
d=n;
d2=(long)Math.sqrt(n)+1;
int fg=0;
a=0;
b=0;
c=0;
if(n==3 || n==2 || n==1 || n==0)
{
System.out.println("NO");
}
for(j=2;j<d2;j++)
{
if(d%j==0 && fg==0)
{
a=j;
fg=1;
d=d/a;
}
else if(d%j==0 && fg==1)
{
b=j;
c=d/b;
if(a==c || b==c)
{
j=a+1;
a=0;b=0;c=0;
d=n;
fg=0;
}
else
{
System.out.println("YES");
System.out.println(a+" "+b+" "+c);
break;
}
}
else
{
;
}
if(j>=d)
{
System.out.println("NO");
break;
}
if(j==d2-1)
{
if(c==0)
System.out.println("NO");
}
}
}
}
}
class FastReader {
byte[] buf = new byte[2048];
int index, total;
InputStream in;
FastReader(InputStream is) {
in = is;
}
int scan() throws IOException {
if (index >= total) {
index = 0;
total = in.read(buf);
if (total <= 0) {
return -1;
}
}
return buf[index++];
}
String next() throws IOException {
int c;
for (c = scan(); c <= 32; c = scan()) ;
StringBuilder sb = new StringBuilder();
for (; c > 32; c = scan()) {
sb.append((char) c);
}
return sb.toString();
}
String nextLine() throws IOException {
int c;
for (c = scan(); c <= 32; c = scan()) ;
StringBuilder sb = new StringBuilder();
for (; c != 10 && c != 13; c = scan()) {
sb.append((char) c);
}
return sb.toString();
}
char nextChar() throws IOException {
int c;
for (c = scan(); c <= 32; c = scan()) ;
return (char) c;
}
int nextInt() throws IOException {
int c, val = 0;
for (c = scan(); c <= 32; c = scan()) ;
boolean neg = c == '-';
if (c == '-' || c == '+') {
c = scan();
}
for (; c >= '0' && c <= '9'; c = scan()) {
val = (val << 3) + (val << 1) + (c & 15);
}
return neg ? -val : val;
}
long nextLong() throws IOException {
int c;
long val = 0;
for (c = scan(); c <= 32; c = scan()) ;
boolean neg = c == '-';
if (c == '-' || c == '+') {
c = scan();
}
for (; c >= '0' && c <= '9'; c = scan()) {
val = (val << 3) + (val << 1) + (c & 15);
}
return neg ? -val : val;
}
} | JAVA |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | def find(num):
a=0
b=0
i=2
while (i*i)<=num:
if num%i==0:
if a==0:a=i
elif b==0:b=i
num=num//i
if a>0 and b>0:
return [a,b]
i+=1
return None
for i in range(int(input())):
n=int(input())
A=find(n)
if not A or n<=2:print("NO")
else:
a=A[0]
b=A[1]
c=a*b
if n%a==0 and n//a!=a and n//a!=b and (a*b*(n//a))==n:
print("YES")
print(a,b,n//a)
if n%b==0 and n//b!=a and n//b!=b and (a*b*(n//b))==n:
print("YES")
print(a,b,n//b)
elif n%c==0 and n//c!=a and n//c!=b and (a*b*(n//c))==n:
print("YES")
print(a,b,n//c)
else:print("NO")
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | # -*- coding: utf-8 -*-
import sys
def input(): return sys.stdin.readline().strip()
def list2d(a, b, c): return [[c] * b for i in range(a)]
def list3d(a, b, c, d): return [[[d] * c for j in range(b)] for i in range(a)]
def list4d(a, b, c, d, e): return [[[[e] * d for j in range(c)] for j in range(b)] for i in range(a)]
def ceil(x, y=1): return int(-(-x // y))
def INT(): return int(input())
def MAP(): return map(int, input().split())
def LIST(N=None): return list(MAP()) if N is None else [INT() for i in range(N)]
def Yes(): print('Yes')
def No(): print('No')
def YES(): print('YES')
def NO(): print('NO')
# sys.setrecursionlimit(10 ** 9)
INF = 10 ** 18
MOD = 10 ** 9 + 7
def factorize(num: int) -> dict:
from math import sqrt
from collections import Counter
d = Counter()
for i in range(2, int(sqrt(num))+1):
while num % i == 0:
num //= i
d[i] += 1
if num == 1:
break
if num != 1:
d[num] += 1
return d
for _ in range(INT()):
N = INT()
primes = sorted(factorize(N).items())
ans = [1] * 3
if len(primes) == 1:
k, v = primes[0]
if v < 6:
NO()
continue
ans[0] = k
ans[1] = k ** 2
ans[2] = k ** (v-3)
elif len(primes) == 2:
k1, v1 = primes[0]
k2, v2 = primes[1]
if v1 == 1 and v2 <= 2 or v1 <= 2 and v2 == 1:
NO()
continue
ans[0] = k1
ans[1] = k2
ans[2] = k1**(v1-1) * k2**(v2-1)
else:
for i, (k, v) in enumerate(primes):
if i < 2:
ans[i] = k ** v
else:
ans[2] *= k ** v
YES()
print(*ans)
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | from collections import defaultdict, deque
from itertools import permutations
from sys import stdin,stdout
from bisect import bisect_left, bisect_right
from copy import deepcopy
#from random import randint
int_input=lambda : int(stdin.readline())
string_input=lambda : stdin.readline()
multi_int_input =lambda : map(int, stdin.readline().split())
multi_input = lambda : stdin.readline().split()
list_input=lambda : list(map(int,stdin.readline().split()))
string_list_input=lambda: list(string_input())
MOD = pow(10,9)+7
import math
test = int_input()
def solve(n):
for i in range(2, int(math.sqrt(n))+1):
if(n%i==0):
a=i
tmp=n//i
for j in range(i+1, int(math.sqrt(tmp))+1):
if(tmp%j==0 and (tmp//j!=a and tmp//j!=j)):
print("YES")
print(a,j,tmp//j)
return
print("NO")
return
for _ in range(test):
n = int_input()
solve(n)
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int pri[100003], top, n;
bool ntp[100002];
struct node {
int a, b;
bool operator<(const node &rhs) const { return b > rhs.b; }
};
void euler() {
for (int i = 2; i <= 100000; ++i) {
if (!ntp[i]) pri[top++] = i;
for (int j = 0; j < top && pri[j] * i <= 100000; ++j) {
ntp[pri[j] * i] = 1;
if (i % pri[j] == 0) break;
}
}
}
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
euler();
int t;
cin >> t;
while (t--) {
cin >> n;
int bn = n;
vector<node> v;
int sz = 1;
for (int i = 0; i < top && n != 1; ++i) {
if (n % pri[i] == 0) {
node x;
x.a = pri[i];
x.b = 0;
while (n % pri[i] == 0) ++x.b, n /= pri[i];
sz *= (x.b + 1);
v.push_back(x);
}
}
if (n != 1) {
node x;
x.a = n;
x.b = 1;
v.push_back(x);
}
if (v.size() == 0) {
cout << "NO\n";
} else if (v.size() == 1) {
if (v[0].b < 6) {
cout << "NO\n";
} else {
cout << "YES\n";
const int &t = v[0].a;
cout << t << ' ' << t * t << ' ' << bn / (t * t * t) << '\n';
}
} else if (v.size() == 2) {
if (v[0].b + v[1].b < 4) {
cout << "NO\n";
} else {
cout << "YES\n";
cout << v[0].a << ' ' << v[1].a << ' ' << bn / v[0].a / v[1].a << '\n';
}
} else {
cout << "YES\n";
cout << v[0].a << ' ' << v[1].a << ' ' << bn / v[0].a / v[1].a << '\n';
}
}
return 0;
}
| CPP |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | /*package whatever //do not write package name here */
import java.io.*;
import java.util.*;
public class GFG {
static String solve(ArrayList<Integer> div,int n)
{
HashSet<Integer> set=new HashSet<>();
int sz=div.size();
for(int i=0;i<sz;i++)
set.add(div.get(i));
for(int i=0;i<sz;i++)
{
for(int j=i+1;j<sz;j++)
{
int a=div.get(i);
int b=div.get(j);
if(a!=b){
int ab=a*b;
if(n%ab==0)
{
int c=n/ab;
if(set.contains(c)&&(a!=c)&&(b!=c))
{
return "YES\n"+c+" "+a+" "+" "+b;
}
}
}
}
}
return "NO";
}
static void findDiv(int n,ArrayList<Integer> div)
{
int p=(int)Math.sqrt(n);
for(int i=2;i<=p;i++)
{
if(n%i==0)
{
if(n/i==i)
{
div.add(i);
}
else
{
div.add(n/i);
div.add(i);
}
}
}
}
public static void main (String[] args) throws Exception{
BufferedReader bf=new BufferedReader(new InputStreamReader(System.in));
int t=Integer.parseInt(bf.readLine());
StringBuffer str=new StringBuffer("");
while(t-->0)
{
int n=Integer.parseInt(bf.readLine());
ArrayList<Integer> div=new ArrayList<>();
findDiv(n,div);
Collections.sort(div);
str.append(solve(div,n)+"\n");
}
System.out.println(str);
}
} | JAVA |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | t=int(input())
for _ in range(t):
n=int(input())
a=b=c=-1
for i in range(2,int(n**0.5)+1):
if n%i==0:
a=i
n//=i
break
for i in range(2,int(n**0.5)+1):
if n%i==0 and i!=a:
b=i
n//=i
break
if b!=-1 and n!=a and n!=b:
c=n
if c==-1:
print('NO')
else:
print('YES')
print('{} {} {}'.format(a,b,c)) | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | #include <bits/stdc++.h>
using Int = long long;
using namespace std;
vector<pair<Int, Int>> primeFactors(Int n) {
vector<pair<Int, Int>> result;
Int factor = 0;
Int factorCount = 0;
Int q = n;
if (n >= 2) {
for (Int i = 2; i * i <= n;) {
if (q % i == 0) {
if (i != factor) {
factor = i;
factorCount = 1;
} else {
factorCount++;
}
q /= i;
} else {
if (factor > 0) {
result.push_back(make_pair(factor, factorCount));
factor = 0;
}
i++;
}
}
}
if (q > 1) {
result.push_back(make_pair(q, 1));
}
return result;
}
int main() {
int t;
cin >> t;
for (int i = 0, i_len = (int)(t); i < i_len; i++) {
Int n;
cin >> n;
auto factors = primeFactors(n);
if (((int)(factors).size()) == 1) {
auto p = factors[0].first;
auto e = factors[0].second;
if (e >= 6) {
puts("YES");
printf("%lld %lld %lld \n", p, p * p, n / (p * p * p));
} else {
puts("NO");
}
} else if (((int)(factors).size()) == 2) {
auto p0 = factors[0].first;
auto e0 = factors[0].second;
auto p1 = factors[1].first;
auto e1 = factors[1].second;
if (e0 + e1 <= 3) {
puts("NO");
} else {
puts("YES");
printf("%lld %lld %lld \n", p0, p1, n / (p0 * p1));
}
} else {
auto p = factors[0].first;
auto q = factors[1].first;
puts("YES");
printf("%lld %lld %lld \n", p, q, n / (p * q));
}
}
}
| CPP |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
void helper(long long int n) {
set<int> nums;
for (int i = 2; i * i <= n; i++) {
if (n % i == 0 && !nums.count(i)) {
nums.insert(i);
n /= i;
break;
}
}
for (int i = 2; i * i <= n; i++) {
if (n % i == 0 && !nums.count(i)) {
nums.insert(i);
n /= i;
break;
}
}
if (nums.size() < 2 || nums.count(n) || n <= 1) {
cout << "NO" << endl;
} else {
cout << "YES" << endl;
nums.insert(n);
for (auto i : nums) {
cout << i << " ";
}
cout << endl;
}
}
int main() {
int t;
cin >> t;
while (t--) {
int n;
cin >> n;
helper(n);
}
return 0;
}
| CPP |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const long long INF = 1e9;
const long long N = 1e3 + 500;
const long long mod = 1e9 + 7;
const long double eps = 1E-7;
long long n, mx, mn, cnt, m, ans;
long long a, b, c;
string s;
long long used[N];
char z;
int main() {
ios_base::sync_with_stdio(0);
long long T;
cin >> T;
while (T--) {
cin >> n;
vector<long long> v;
for (long long i = 2; i <= 1000000; i++) {
if (n % i == 0 and v.size() < 2) {
v.push_back(i);
n /= i;
} else if (v.size() == 2)
break;
}
if (v.size() == 2 and n != v[0] and n != v[1] and n > 1) {
v.push_back(n);
}
if (v.size() == 3)
cout << "YES\n" << v[0] << " " << v[1] << " " << v[2] << endl;
else
cout << "NO\n";
v.clear();
}
}
| CPP |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | for _ in range(int(input())):
n = int(input())
l = []
til = int(n**0.5)+1
i = 2
while i < til:
if n%i==0:
til = int((n//i)**0.5)+1
n= (n//i)
l.append(i)
if len(l)==2:
break
i+=1
if n > 1 and n not in l and len(l)==2:
print("YES")
print(l[0],l[1], n)
else:
print("NO")
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | t = int(input())
for _ in range(t):
n = int(input())
i = 2
arr = [0,0,0]
k = 0
while(i * i <= n):
if (n % i == 0):
arr[k] = i
n = int(n/i)
k+=1
if (k == 2):
break
i+=1
arr[2] = n
if(arr[1] >1 and arr[0] > 1 and arr[2] > 1):
if (arr[0] == arr[1] or arr[1] == arr[2] or arr[0] == arr[1]):
print ("NO")
else:
print("YES")
print(str(arr[0]) + " " + str(arr[1]) + " " + str(arr[2]))
else:
print("NO")
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | for _ in range(int(input())):
num=int(input())
list1=[]
j=2
while(j*j<num and len(list1)<2):
if(num%j==0):
num=num//j
list1.append(j)
j+=1
if(len(list1)==2 and num not in list1):
print("YES")
for k in list1:
print(k,end=" ")
print(num)
else:
print("NO") | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import math
b= int(input())
t=[]
for j in range(b):
t.append(int(input()))
for a in t :
c=3
i = 2
s = ''
while a > 1 and a // i >= 1 and i <=math.pow(a,1/c):
if (a % i == 0):
a = a // i
s = s + i.__str__() + ' '
c=c-1
if (s.split(' ').__len__()>2 and a//(i+1)>0):
s = s + a.__str__() + ' '
a=0
elif s.split(' ').__len__()>2:
a=0
i = i + 1
if s.split(' ').__len__()>3:
print("YES \n", s, sep='')
else:
print("NO") | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | t = int(input())
for i in range(t):
n, s = int(input()), []
for i in range(2, int(n**(1/2)+1)):
if n % i == 0:
s.append(i)
n = n // i
if len(s) == 2 and s[0]!=n and s[1]!=n:
break
if len(s) == 2 and s[0]!=n and s[1]!=n:
print('YES')
print(str(s[0])+' '+str(s[1])+' '+str(n))
else:
print('NO')
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import math
n = int(input())
def sd(n, notin):
i = 2
while i * i < n:
if n % i == 0 and i not in notin:
return i
i += 1
return n
for i in range(n):
a = int(input())
aa = a
sq = int(math.sqrt(a)) + 1
divisor = []
while True:
s = sd(a, divisor)
if s == a:
print("NO")
break
a = a // s
divisor.append(s)
if len(divisor) == 2:
if a != divisor[0] and a != divisor[1]:
print("YES")
print(divisor[0], divisor[1], a)
else:
print("NO")
break
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | def product_of_three_numbers():
num_test = int(input())
count = 0
n_list = []
while count < num_test:
n_list += [int(input())]
count += 1
answer_list = []
for idx, n in enumerate(n_list):
if n <= 5:
answer_list += ["NO"]
continue
breakit = False
for a in range(2, n+1):
if (a-1)*(a-1) >= n:
answer_list += ["NO"]
break
if n%a == 0:
divbya = int(n/a)
if divbya <= a:
answer_list += ["NO"]
break
for b in range(a, divbya):
if (b-1)*(b-1) >= divbya:
answer_list += ["NO"]
breakit = True
break
if divbya%b == 0:
if b==a:
continue
divbyb = int(divbya/b)
if divbyb <= b:
answer_list += ["NO"]
breakit = True
break
answer_list += ["YES"]
answer_list += [str(a)+" "+str(b)+" "+str(divbyb)]
breakit = True
break
if breakit:
break
for answer in answer_list:
print(answer)
product_of_three_numbers() | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | v = int(input())
for i in range(v):
b = list(map(int,input().split()))[:v]
c = 0
a = []
for j in b:
for h in range(2,int(j**(0.5))+1):
if j%h==0:
c+=1
j = j//h
a.append(h)
if len(a)==2:
break
if len(a)==2 and j not in a:
print("YES")
print(*a,j)
else:
print("NO")
| PYTHON3 |
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