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| description
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| difficulty
int64 0
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| solution
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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 | 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**2 <= 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**2 <= 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 | import math
def primefactorisation(n):
p=[]
while(n%2==0):
n//=2
p.append(2)
i=3
while(i<=math.sqrt(n)):
while(n%i==0):
n//=i
p.append(i)
i+=2
if n>2:
p.append(n)
return p
for _ in range(int(input())):
n=int(input())
A=primefactorisation(n)
if len(A)>=3:
a=A[0]
b=A[1]
if b==a:
b=A[1]*A[2]
c=n//(a*b)
if c==a or c==b or c==1:
print("NO")
else:
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;
signed main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
long long int t = 1;
cin >> t;
while (t--) {
long long int n;
cin >> n;
vector<long long int> v;
long long int x = n;
while (n % 2 == 0) {
v.push_back(2);
n /= 2;
}
for (long long int i = 3; i * i <= n; i += 2) {
while (n % i == 0) {
v.push_back(i);
n /= i;
}
}
if (n > 2) v.push_back(n);
map<long long int, long long int> mp;
for (long long int i = 0; i < v.size(); i++) mp[v[i]]++;
vector<long long int> v1;
if (mp.size() > 2) {
cout << "YES\n";
map<long long int, long long int>::iterator it;
long long int cnt = 0;
for (it = mp.begin(); it != mp.end(); it++) {
if (cnt == 2) break;
cout << (long long int)pow(it->first, it->second) << " ";
cnt++;
}
long long int x = 1;
for (; it != mp.end(); it++) x = (x * pow(it->first, it->second));
cout << x << '\n';
} else if (mp.size() == 2) {
map<long long int, long long int>::iterator it = mp.begin();
long long int cnt = it->second;
it++;
cnt += it->second;
if (cnt < 4)
cout << "NO\n";
else {
cout << "YES\n";
it = mp.begin();
long long int x, y, z;
x = it->first;
y = pow(it->first, it->second - 1);
it++;
z = it->first;
y *= pow(it->first, it->second - 1);
cout << x << " " << y << " " << z << '\n';
}
} else {
map<long long int, long long int>::iterator it = mp.begin();
if (it->second >= 6) {
cout << "YES\n";
cout << it->first << " " << (long long int)pow(it->first, 2) << " "
<< (long long int)pow(it->first, it->second - 3) << '\n';
} 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 | from collections import Counter, defaultdict
from sys import stdin, stdout
raw_input = stdin.readline
for t in xrange(input()):
n=input()
l=[]
f=0
for i in xrange(2,min(n+1,10001)):
if n%i==0:
l.append(i)
n/=i
if len(l)==2 and n not in l and n>1:
f=1
break
elif len(l)==2:
break
if f:
stdout.write('YES\n')
for j in l:
stdout.write(str(j)+' ')
stdout.write(str(n)+'\n')
else:
stdout.write('NO\n')
| 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 itertools import permutations
def divisores(n):
large_divisors = []
for k in range(1, int(math.sqrt(n) + 1)):
if(n % k == 0):
yield k
if(k*k != n):
large_divisors.append(n//k)
for d in reversed(large_divisors):
yield d
t = int(input())
for k in range(t):
q = int(input())
divisoresL = (list(divisores(q)))
if(len(divisoresL) < 5):
print("NO")
else:
flag = 0
divisoresL.remove(divisoresL[0])
divisoresL.remove(divisoresL[len(divisoresL)-1])
for subset in permutations(divisoresL, 3):
if(subset[0]*subset[1]*subset[2] == q):
print("YES")
print("{} {} {}".format(subset[0], subset[1], subset[2]))
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 |
import collections
class Solution:
def solve(self, n):
used = set()
a = 2
while a*a <= n:
if n % a == 0 and a not in used:
used.add(a)
n //= a
break
a += 1
b = 2
while b*b <= n:
if n % b == 0 and b not in used:
used.add(b)
n //= b
break
b += 1
# print(used)
if len(used) < 2 or n in used or n == 1:
return (False, None)
else:
return (True, list(used) + [n])
# Gets Time limit exceeded
# for a in range(2,n//(2**2)+1):
# for b in range(2, a):
# for c in range(2, b):
# res = a*b*c
# if res == n:
# return (True, sorted([c,b,a]))
# elif res > n:
# break
# if a*b > n//2:
# break
# return (False, None)
sol = Solution()
t = int(input().strip())
for i in range(t):
n = int(input().strip())
answer = sol.solve(n)
if answer[0]:
print("YES")
print(" ".join(map(str, answer[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 math
for _ in range(int(input())):
n = int(input())
flag = False
if n < 24:
print('NO')
elif n == 24:
print('YES')
print(2, 3, 4)
else:
for i in range(2, int(math.sqrt(n)) + 1):
if n % i == 0:
a = i
n = n / i
for j in range(a + 1, int(math.sqrt(n)) + 1):
if n % j == 0:
b = j
c = int(n / j)
if c != b and a != c:
print('YES')
print(a, b, c)
flag = True
break
if flag is True:
break
if flag is False:
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;
bool test(void) {
long long N;
scanf("%lld", &N);
long long u = N;
vector<pair<long long, int>> A;
for (long long i = 2; i * i <= u; ++i) {
int q = 0;
while (u % i == 0) {
++q;
u /= i;
}
if (q) {
A.push_back({i, q});
}
}
if (u > 1) A.push_back({u, 1});
vector<long long> D{1};
for (auto [p, q] : A) {
int s = D.size();
for (int i = 0; i < s; ++i) {
long long u = D[i];
for (int k = 0; k < q; ++k) {
u *= p;
D.push_back(u);
}
}
}
int s = D.size();
for (int i = 0; i < s; ++i) {
long long a = D[i];
if (a == 1) continue;
for (int j = 0; j < i - 1; ++j) {
long long b = D[j];
if (b == 1) continue;
if (N % (a * b) == 0) {
long long c = N / (a * b);
if (c == 1) continue;
if (a != c && b != c) {
printf("YES\n");
printf("%lld %lld %lld\n", a, b, c);
return true;
}
}
}
}
return false;
}
int main(void) {
int T;
scanf("%d", &T);
while (T--) {
if (!test()) 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 | import math
for _ in range(int(input())):
n = int(input())
t = n
a1 = []
while n % 2 == 0:
a1.append(2),
n = n / 2
for i in range(3,int(math.sqrt(n))+1,2):
while n % i== 0:
a1.append(int(i)),
n = n / i
if n > 2:
a1.append(int(n))
l = len(a1)
a = list(set(a1))
p = len(a)
if p == 1 and l >= 6:
if a[0] != t//(a[0]**3) and a[0]**2 != t//(a[0]**3):
print("YES")
print(a[0],a[0]**2,t//(a[0]**3))
else:
print("NO")
elif (p >= 2 )and (l>= 3):
if a[0] != t//(a[0]*a[1]) and a[1] != t//(a[0]*a[1]):
print("YES")
print(a[0],a[1],t//(a[0]*a[1]))
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 | t = int(input())
for qwe in range(t):
# n1 = int(f.readline())
n1 = int(input())
flag = False
for i1 in range(2, int(n1**0.5) + 1):
if n1 % i1 == 0:
a = i1
n2 = n1 // i1
for i2 in range(2, int(n2**0.5) + 1):
if n2 % i2 == 0 and i2 != a:
b = i2
c = n2 // i2
if c != a and c != b:
flag = True
if flag:
break
if flag:
break
if flag:
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() {
int t;
cin >> t;
while (t--) {
int n, arr[10];
cin >> n;
int cnt = 0;
for (int i = 2; i < sqrt(n); i++) {
if (n % i == 0 && ((n / i) != i)) {
arr[0] = i;
n = n / i;
cnt = 5;
break;
}
}
if (cnt != 5)
cout << "NO" << endl;
else {
for (int i = 2; i < sqrt(n); i++) {
if (n % i == 0 && ((n / i) != i) && (arr[0] != i) &&
(arr[0] != n / i)) {
arr[1] = i;
arr[2] = n / i;
cnt = 10;
break;
}
}
if (cnt != 10)
cout << "NO" << endl;
else {
cout << "YES" << endl;
cout << arr[0] << " " << arr[1] << " " << arr[2] << 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 | #Codeforces - Product of Three Numbers
t = int(input())
while t != 0:
n = int(input())
i = 2
result = 'NO'
while i * i * i < n:
if n % i == 0:
j = i + 1
while j * j < int(n/i):
if n % (i * j) == 0:
result = '{} {} {}'.format(i, j, int(n/(i*j)))
break
j += 1
break
i += 1
if result != 'NO':
print('YES')
print(result)
t -= 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 | ans=""
for _ in range(int(input())):
n=nn=int(input())
l1=[]; i=2
while len(l1)<2 and i*i<=nn:
if n%i==0: l1.append(i); n//=i
i+=1
if len(l1)<2 or l1[1]==n or l1[0]==n: ans+="NO\n"
else: ans+="YES\n"; ans+=f"{l1[0]} {l1[1]} {n}\n"
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())
l=[]
x=n
for i in range(2,int(math.sqrt(n))+1):
if(len(l)<2):
if(x%i==0):
l.append(int(i))
x/=i
else:
break
if((x not in l) and (len(l)==2)):
print('YES')
print(str(l[0])+' '+str(l[1])+' '+str(int(x)))
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
sys.setrecursionlimit(10**9)
import atexit
import io
from collections import defaultdict, Counter
_INPUT_LINES = sys.stdin.read().splitlines()
input = iter(_INPUT_LINES).__next__
_OUTPUT_BUFFER = io.StringIO()
sys.stdout = _OUTPUT_BUFFER
@atexit.register
def write():
sys.__stdout__.write(_OUTPUT_BUFFER.getvalue())
IA =lambda: map(int,input().split())
ans=[0,0,0,0]
def solve(n):
num=int(0)
i=int(2)
tmp=1
m=n
while i*i<=n:
if n%i==0:
n=n//i
tmp*=i
if tmp not in ans:
num+=1
ans[num]=tmp
tmp=1
if num>=3:
ans[num]*=n
return 1
i+=1
if n>1:
if tmp*n not in ans:
num+=1
ans[num]=n*tmp
if num>=3: return 1
else:return -1
T=int(input())
for t in range(0,T):
ans=[0,0,0,0]
n=int(input())
if solve(n)==1:
print("YES")
for i in range(1,4):
print(ans[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 | def solve():
n = int(input())
divs = []
for i in range(2, int(n**0.5) + 2):
if n % i == 0:
n //= i
divs.append(i)
if len(divs) == 2:
if n > divs[1]:
print ("YES")
print (*divs, n)
return
else:
print ("NO")
return
print ("NO")
t = int(input())
for _ in range(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 | #include <bits/stdc++.h>
using namespace std;
int main() {
int t, n;
cin >> t;
while (t--) {
cin >> n;
set<int> values;
for (int i = 2; i * i <= n; i++) {
if (n % i == 0) {
values.insert(i);
n /= i;
break;
}
}
for (int i = 2; i * i <= n; i++) {
if (n % i == 0 && !values.count(i)) {
values.insert(i);
n /= i;
break;
}
}
if ((int)values.size() < 2 || n == 1 || values.count(n)) {
cout << "NO" << endl;
} else {
values.insert(n);
cout << "YES" << endl;
for (auto it : values) cout << it << " ";
cout << 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 math
for _ in range(int(input())):
n = int(input())
dic = {}
x = n
if x%2==0:
dic[2] = 0
while x%2==0:
x//=2
dic[2] += 1
for i in range(3,int(math.sqrt(x))+1,2):
if x%i==0:
dic[i] = 0
while x%i==0:
x//=i
dic[i] += 1
if x>2:
dic[x] = 1
# print(dic)
if len(dic)==3:
print('YES')
ans = []
for i in dic:
ans += [pow(i,dic[i])]
print(*ans)
elif len(dic)>3:
print('YES')
ans = []
c = 0
for i in dic:
ans += [pow(i,dic[i])]
c+=1
dic[i] = 0
if c==3:
break
t = 1
for i in dic:
t*= pow(i,dic[i])
ans[-1]*=t
print(*ans)
elif len(dic)==2:
s = 0
for i in dic:
s+=dic[i]
if s>=4:
print('YES')
ans = []
b = 1
for i in dic:
ans += [i]
b*= pow(i,dic[i]-1)
ans+=[b]
print(*ans)
else:
print('NO')
else:
for i in dic:
if dic[i]>=6:
print('YES')
print(i,pow(i,2),pow(i,dic[i]-3))
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 ll = long long;
using namespace std;
const int N = 1e6 + 1, mod = 1e9 + 7;
int t;
int main() {
cin.tie(0)->sync_with_stdio(0);
cin >> t;
while (t--) {
int n;
cin >> n;
vector<int> divisors;
int d = n, s = sqrt(n);
for (int i = 2; i <= s; i++) {
if (d % i == 0) {
divisors.push_back(i);
d /= i;
}
}
if (d > 1) {
divisors.push_back(d);
}
if (divisors.size() == 1) {
int a = divisors[0], cnt = 0;
while (n % a == 0) {
n /= a;
cnt++;
}
if (cnt >= 6) {
cout << "YES\n";
cout << a << ' ' << a * a << ' ' << n / (a * a * a) << '\n';
} else {
cout << "NO\n";
}
} else if (divisors.size() >= 3) {
int a = divisors[0], b = divisors[1];
int c = n / divisors[0] / divisors[1];
if (a == c || b == c) {
cout << "NO\n";
} else {
cout << "YES\n";
cout << a << ' ' << b << ' ' << c << '\n';
}
} 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 |
for _ in range(int(input())):
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):
if(i!=a):
b=i
n//=i
break
if(n!=1 and n!=a and n!=b):
c=n
if(a==1 or b==1 or c==1):
print("NO")
else:
print("YES")
print(a,b,c,sep=" " ) | 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.*;
import java.util.*;
import static java.lang.Double.parseDouble;
import static java.lang.Integer.parseInt;
import static java.lang.Long.parseLong;
import static java.lang.Math.abs;
import static java.lang.System.exit;
public class Main {
static BufferedReader in;
static PrintWriter out;
static StringTokenizer tok;
void Case() throws IOException {
int n = nextInt();
List<Integer> divs = new ArrayList<>();
for (int i = 2; i * i <= n; i++)
if (n % i == 0) {
divs.add(i);
divs.add(n / i);
}
for (int yyy : divs) for (int xxx : divs) if(xxx != 1 && yyy != 1)
for (int x : divs)
for (int y : divs)
for (int z : divs)
if (x != y && x != z && y != z && x * y * z == n) {
out.println("Yes\n" + x + " " + y + " " + z);
return;
}
out.println("NO");
}
void solve() throws Exception {
int t = nextInt();
while (t-- > 0) Case();
}
// call it like this: lower_bound(a, x + 1) ( /!\ + 1 )
public static int lower_bound(Integer[] a, int v) {
int low = -1, high = a.length;
while (high - low > 1) {
int h = high + low >>> 1;
if (a[h] >= v) {
high = h;
} else {
low = h;
}
}
return high;
}
private String getFraction(int a, int b) {
assert b != 0;
String sign = (a > 0 && b > 0) || (a < 0) && (b < 0) ? "+" : "-";
a = abs(a);
b = abs(b);
int gcd = gcd(a, b);
return sign + (a / gcd) + "/" + (b / gcd);
}
private int gcd(int a, int b) {
while (b > 0) {
int temp = b;
b = a % b;
a = temp;
}
return a;
}
private int lcm(int a, int b) {
return a * (b / gcd(a, b));
}
public static int[] radixSort(int[] f) {
if (f.length < 100) {
Arrays.sort(f);
return f;
}
int[] to = new int[f.length];
{
int[] b = new int[65537];
for (int i = 0; i < f.length; i++) b[1 + (f[i] & 0xffff)]++;
for (int i = 1; i <= 65536; i++) b[i] += b[i - 1];
for (int i = 0; i < f.length; i++) to[b[f[i] & 0xffff]++] = f[i];
int[] d = f;
f = to;
to = d;
}
{
int[] b = new int[65537];
for (int i = 0; i < f.length; i++) b[1 + (f[i] >>> 16)]++;
for (int i = 1; i <= 65536; i++) b[i] += b[i - 1];
for (int i = 0; i < f.length; i++) to[b[f[i] >>> 16]++] = f[i];
int[] d = f;
f = to;
to = d;
}
return f;
}
public static long pow(long a, long n, long mod) {
long ret = 1;
int x = 63 - Long.numberOfLeadingZeros(n);
for (; x >= 0; x--) {
ret = ret * ret % mod;
if (n << 63 - x < 0)
ret = ret * a % mod;
}
return ret;
}
public static boolean nextPermutation(Integer[] a) {
int n = a.length;
int i;
for (i = n - 2; i >= 0 && a[i] >= a[i + 1]; i--) ;
if (i == -1)
return false;
int j;
for (j = i + 1; j < n && a[i] < a[j]; j++) ;
int d = a[i];
a[i] = a[j - 1];
a[j - 1] = d;
for (int p = i + 1, q = n - 1; p < q; p++, q--) {
d = a[p];
a[p] = a[q];
a[q] = d;
}
return true;
}
void print(Object x) {
out.print(String.valueOf(x));
out.flush();
}
void println(Object x) {
out.println(String.valueOf(x));
out.flush();
}
// for Map with custom key/value, override toString in your custom class
void printMap(Map map) {
if (map.keySet().size() == 0) return;
Object firstValue = map.keySet().iterator().next();
if (map.get(firstValue) instanceof Queue || map.get(firstValue) instanceof List) {
for (Object key : map.keySet()) {
out.print(String.valueOf(key) + ": ");
Collection values = (Collection) map.get(key);
for (Object value : values) out.print(String.valueOf(value) + " ");
out.println();
}
} else if (map.get(firstValue).getClass().isArray()) {
for (Object key : map.keySet()) {
out.print(String.valueOf(key) + ": ");
Object[] values = (Object[]) map.get(key);
for (Object value : values) out.print(String.valueOf(value) + " ");
out.println();
}
} else {
for (Object key : map.keySet()) {
out.println(String.valueOf(key) + ": " + map.get(key));
}
}
}
private int[] na(int n) throws IOException {
int[] a = new int[n];
for (int i = 0; i < n; i++) a[i] = nextInt();
return a;
}
private long[] nal(int n) throws IOException {
long[] a = new long[n];
for (int i = 0; i < n; i++) a[i] = nextLong();
return a;
}
int nextInt() throws IOException {
return parseInt(next());
}
long nextLong() throws IOException {
return parseLong(next());
}
double nextDouble() throws IOException {
return parseDouble(next());
}
String next() throws IOException {
while (tok == null || !tok.hasMoreTokens()) {
tok = new StringTokenizer(in.readLine());
}
return tok.nextToken();
}
public static void main(String[] args) throws Exception {
try {
boolean isLocal = false;
if (isLocal) {
in = new BufferedReader(new FileReader("solution.in"));
out = new PrintWriter(new BufferedWriter(new FileWriter("solution.out")));
} else {
in = new BufferedReader(new InputStreamReader(System.in));
out = new PrintWriter(new OutputStreamWriter(System.out));
}
//long lStartTime = System.currentTimeMillis();
new Main().solve();
//long lEndTime = System.currentTimeMillis();
//out.println("Elapsed time in seconds: " + (double)(lEndTime - lStartTime) / 1000.0);
in.close();
out.close();
} catch (Throwable e) {
e.printStackTrace();
exit(1);
}
}
} | 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())
fin=[]
for i in range(2,40000):
if(n%i==0):
n=n//i
fin.append(i)
break
if(len(fin)==0):
print("NO")
else:
for i in range(2,40000):
if(n%i==0 and i!=fin[0] ):
n=n//i
fin.append(i)
break
if(len(fin)<=1 or n==fin[0] or n==1 or n==fin[1]):
print("NO")
else:
fin.append(n)
print("YES")
print(*fin) | 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 p3n(cur):
j = 2
while j * j <= cur:
if cur % j == 0:
# print(cur, j)
oth = cur // j
k = 2
while k * k <= oth:
if oth % k == 0 and j != k != oth // k != j:
print('YES')
print(j, k, oth // k)
return
k += 1
j += 1
print('NO')
t = int(input())
tc = []
for i in range(t):
tc.append(int(input()))
for i in range(t):
p3n(tc[i])
| 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())
a = -99
t = 0
if(n < 24):
print("NO")
else:
for i in range(2, (int)(math.sqrt(n)) + 1):
if(n%i == 0):
if(a > 0):
if(n%(i*a) == 0):
if(a != i) and (i != (n//(i*a))) and (n//(i*a) > 2) and (a != (n//(i*a))):
print("YES")
print(a,end=' ')
print(i, end=' ')
print((n//(i*a)))
t = 1
break
else:
a = i
if(n%(i*a) == 0):
if(a != i) and (i != (n//(i*a))) and (n//(i*a) > 2) and (a != (n//(i*a))):
print("YES")
print(a,end=' ')
print(i, end=' ')
print((n//(i*a)))
t = 1
break
if(t == 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;
signed main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
;
long long t;
cin >> t;
while (t--) {
long long n, f = 0;
cin >> n;
long long i = 2;
while (i * i < n) {
if (n % i == 0) {
f = 1;
break;
}
i++;
}
n = n / i;
if (n > i && f) {
f = 0;
long long j = i + 1;
while (j * j < n) {
if (n % j == 0) {
f = 1;
break;
}
j++;
}
n /= j;
if (f && n > j) cout << "YES\n" << i << " " << j << " " << n << "\n";
}
if (f == 0) 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>
using namespace std;
int main() {
int t;
cin >> t;
int n;
while (t--) {
cin >> n;
vector<int> v;
for (int i = 2; i * i <= n; i++) {
if (n % i == 0) {
v.push_back(i);
n /= i;
}
if (v.size() == 2) break;
}
if (n != 1) v.push_back(n);
if (v.size() == 3 && v[0] != v[1] && v[0] != v[2] && v[1] != v[2]) {
printf("YES\n");
for (auto x : v) cout << x << ' ';
cout << endl;
} else {
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 | import math
for i in range(int(input())):
q=int(input())
k=[]
t=0
for i in range(2,int(math.sqrt(q))+1):
if q%i==0:
q=q//i
t=1
k.append(str(i))
break
if t==1:
for i in range(2,int(math.sqrt(q))+1):
if q%i==0 and (str(i) not in k):
q=q//i
t=2
k.append(str(i))
break
else:
print('NO')
if t==2 and str(q) not in k:
k.append(str(q))
print('YES')
print(' '.join(k))
elif t!=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 _ in range(t):
n=int(input())
a=[]
append=a.append
for i in range(2,int(n**0.5)+1):
if(n%i==0):
append(i)
n=n//i
if(len(a)==2):
if n!=1:
append(n)
break
a=list(set(a))
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 java.io.*;
import java.util.*;
public class main {
public static void main(String[] args) throws NumberFormatException, IOException
{
BufferedReader br= new BufferedReader(new InputStreamReader(System.in));
PrintWriter pw= new PrintWriter(System.out);
int t= Integer.parseInt(br.readLine());
for (int i = 0; i < t; i++) {
int n = Integer.parseInt(br.readLine());
boolean possible=false;
o:for (int j = 2; j <= Math.sqrt(n); j++) {
if(n%j==0)
{
int temp=n/j;
for (int k = j+1; k <= Math.sqrt(temp); k++) {
if(temp%k==0)
{
if(j!= k && k!=temp/k && j!=temp/k)
{
pw.println("YES");
pw.println(temp/k + " " + k + " " + j);
possible=true;
}
break o;
}
}
}
}
if(!possible)pw.println("NO");
}
pw.close();
}
} | 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())
nn = n
fact = dict()
i = 2
while i <= math.sqrt(nn):
if n % i == 0:
n //= i
try:
fact[i] += 1
except:
fact[i] = 1
else:
i += 1
if n != 1:
fact[n] = 1
if len(fact) == 0:
print("NO")
continue
if len(fact) > 3:
k = list(fact.keys())
print("YES")
print(k[0]**fact[k[0]], k[1]**fact[k[1]], end=" ")
num = 1
for i in range(2, len(k)):
num *= k[i] ** fact[k[i]]
print(num)
if len(fact) == 3:
print("YES")
a, b, c = list(fact.keys())
print(a**fact[a], b**fact[b], c**fact[c])
if len(fact) == 1:
k = list(fact.keys())[0]
if fact[k] < 6:
print("NO")
else:
print("YES")
print(k, k*k, k**(fact[k]-3))
continue
if len(fact) == 2:
a, b = list(fact.keys())
if fact[a] >= 2 and fact[b] >= 2:
print("YES")
print(a, b, (a ** (fact[a] - 1)) * (b ** (fact[b] - 1)))
elif fact[a] == 1 and fact[b] >= 3:
print("YES")
print(a, b, b ** (fact[b]-1))
elif fact[b] == 1 and fact[a] >= 3:
print("YES")
print(a, b, a ** (fact[a]-1))
else:
print("NO")
continue | 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;
using vi = vector<long long int>;
using vvi = vector<vi>;
using vb = vector<bool>;
using vc = vector<char>;
using vs = vector<string>;
using vld = vector<long double>;
using pii = pair<long long int, long long int>;
using psi = pair<string, long long int>;
using pci = pair<char, long long int>;
using vpii = vector<pii>;
long long int mod = 1e9 + 7;
long long int const maxn = 1e5 + 5;
long long int const inf = 1e18;
long long int add(long long int a, long long int b) {
return ((a % mod) + (b % mod)) % mod;
}
long long int mul(long long int a, long long int b) {
return ((a % mod) * (b % mod)) % mod;
}
long long int powm(long long int x, long long int n, long long int M) {
long long int result = 1;
while (n > 0) {
if (n % 2 == 1) result = (result * x) % M;
x = (x * x) % M;
n = n / 2;
}
return result;
}
long long int modinverse(long long int a, long long int m) {
return powm(a, m - 2, m);
}
bool prime(long long int x) {
if (x <= 1) return false;
for (int i = 2; i <= sqrt(x); i++)
if (x % i == 0) return false;
return true;
}
long long int divisor(long long int x) {
long long int cnt = 0;
for (int i = 1; i <= sqrt(x); i++) {
if (x % i == 0) {
if (i != x / i)
cnt += 2;
else
cnt += 1;
}
}
return cnt;
}
vector<long long int> sieve(long long int n) {
bool prim[n + 1];
memset(prim, true, sizeof(prim));
for (long long int p = 2; p * p <= n; p++) {
if (prim[p] == true) {
for (int i = p * p; i <= n; i += p) prim[i] = false;
}
}
vector<long long int> v;
for (int i = 2; i <= n; i++)
if (prim[i]) v.push_back(i);
return v;
}
void solve() {
set<long long int> st;
long long int x = 0;
long long int n;
cin >> n;
long long int m = n;
for (int i = 2; i * i <= m and x < 2; i++) {
if (n % i == 0) {
st.insert(i);
x++;
n /= i;
}
}
st.insert(n);
if ((long long int)st.size() == 3) {
cout << "YES\n";
for (auto it = st.begin(); it != st.end(); it++) cout << *it << " ";
cout << endl;
} else
cout << "NO\n";
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
;
long long int t;
t = 1;
cin >> t;
while (t--) {
solve();
}
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 math
from functools import reduce
t = int(input())
for cas in range(t):
n = int(input())
ans = []
m = n
sqrt_n = int(math.sqrt(m))
for i in range(2, sqrt_n):
cal = 0
if n == 1:
break
while n % i == 0 and n > 1:
n //= i
ans.append(i)
if n > 1:
ans.append(n)
if len(ans) < 3:
print("NO")
elif len(ans) == 3:
if len(set(ans)) < 3:
print("NO")
else:
print("YES")
print(ans[0], ans[1], ans[2])
elif len(ans) == 4:
if len(set(ans)) == 1:
print("NO")
else:
print("YES")
print(ans[0], ans[1] * ans[2], ans[3])
elif len(ans) == 5:
if len(set(ans)) == 1:
print("NO")
else:
print("YES")
print(ans[0], ans[1] * ans[2] * ans[3], ans[4])
else:
print("YES")
print(ans[0], ans[1] * ans[2], m//(ans[0]*ans[1]*ans[2]))
#reduce(lambda x, y: x * y, ans[3:]) | 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 | #!/usr/bin/env pypy
rr= lambda: input().strip()
rri= lambda: int(rr())
rrm= lambda: [int(x) for x in rr().split()]
def fact(n):
res=[]
for i in range(2, int(n**.5)+1):
if n%i==0:
res.append(i)
res.append(n//i)
if (int(n**.5)**2==n):
res.pop()
return res
def sol():
n=rri()
f=fact(n)
#print(f)
if (len(f)<3):
print("NO")
return
f.sort()
a=f[0]
d=n//a
for i in range(1, len(f)):
for j in range(i+1, len(f)):
if f[i]*f[j]==d:
print("YES")
print (a, f[i], f[j])
return
print("NO")
return
t=rri()
for _ in range(t):
#a,b,c=rrm()
sol()
#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 sys import stdin
input = lambda: stdin.readline().strip()
for _ in range(int(input())):
n = int(input())
a = []
for i in range(2,n//2):
if(n%i==0):
n //= i
a.append(i)
if(len(a)==2 or n<i*i):
break
if(len(a)==2 and n not in a and n>=2):
print("YES")
print(*a,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 |
tests = int(input())
for t in range(tests):
n = int(input())
initN = n
i = 2
a,b = 0,0
while True:
if i**2 > initN:
print("NO")
break
if n%i == 0:
n=n/i
if a == 0:
a=i
elif b==0:
b=i
if a > 0 and b > 0 and n > 1 and a != b and b !=n and a!=n:
print("YES")
print(str(a)+" "+str(b)+" "+str(int(n)))
break
i+=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 | import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = 'x' in file.mode or 'r' not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b'\n') + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode('ascii'))
self.read = lambda: self.buffer.read().decode('ascii')
self.readline = lambda: self.buffer.readline().decode('ascii')
def inp():
return sys.stdin.readline().rstrip()
def mpint():
return map(int, inp().split(' '))
def itg():
return int(inp())
# ############################## import
# ############################## main
def solve():
n = itg()
ans = []
a = 2
while a * a <= n:
if not n % a:
ans.append(a)
n //= a
if len(ans) == 2:
ans.append(n)
break
a += 1
if len(ans) == 3 and ans[0] < ans[1] < ans[2]:
print("YES")
print(*ans)
else:
print("NO")
def main():
# solve()
# print(solve())
for _ in range(itg()):
# print(solve())
solve()
# print("yes" if solve() else "no")
# print("YES" if solve() else "NO")
DEBUG = 0
URL = ''
if __name__ == '__main__':
# 0: normal, 1: runner, 2: interactive, 3: debug
if DEBUG == 1:
import requests
from ACgenerator.Y_Test_Case_Runner import TestCaseRunner
runner = TestCaseRunner(main, URL)
inp = runner.input_stream
print = runner.output_stream
runner.checking()
else:
if DEBUG != 3:
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
if DEBUG:
_print = print
def print(*args, **kwargs):
_print(*args, **kwargs)
sys.stdout.flush()
main()
# Please check!
| 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.util.Map.Entry;
import java.lang.*;
import java.io.*;
import java.math.BigInteger;
public class CF {
private static FS sc = new FS();
private static class FS {
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) {}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
}
private static class extra {
static int[] intArr(int size) {
int[] a = new int[size];
for(int i = 0; i < size; i++) a[i] = sc.nextInt();
return a;
}
static long[] longArr(int size) {
long[] a = new long[size];
for(int i = 0; i < size; i++) a[i] = sc.nextLong();
return a;
}
static long intSum(int[] a) {
long sum = 0;
for(int i = 0; i < a.length; i++) {
sum += a[i];
}
return sum;
}
static long longSum(long[] a) {
long sum = 0;
for(int i = 0; i < a.length; i++) {
sum += a[i];
}
return sum;
}
static LinkedList[] graphD(int vertices, int edges) {
LinkedList[] temp = new LinkedList[vertices+1];
for(int i = 0; i <= vertices; i++) temp[i] = new LinkedList<>();
for(int i = 0; i < edges; i++) {
int x = sc.nextInt();
int y = sc.nextInt();
temp[x].add(y);
}
return temp;
}
static LinkedList[] graphUD(int vertices, int edges) {
LinkedList[] temp = new LinkedList[vertices+1];
for(int i = 0; i <= vertices; i++) temp[i] = new LinkedList<>();
for(int i = 0; i < edges; i++) {
int x = sc.nextInt();
int y = sc.nextInt();
temp[x].add(y);
temp[y].add(x);
}
return temp;
}
static void printG(LinkedList[] temp) {
for(LinkedList<Integer> aa:temp) System.out.println(aa);
}
}
static LinkedList[] temp;
static int mod = (int)Math.pow(10, 9) + 7;
public static void main(String[] args) {
int t = sc.nextInt();
// int t = 1;
StringBuilder ret = new StringBuilder();
while(t-- > 0) {
int n = sc.nextInt();
int a = 0, b = 0;
for(int i = 2; i*i <= n; i++) {
if(n%i == 0) {
n /= i;
a = i;
break;
}
}
for(int i = 2; i*i <= n; i++) {
if(n%i == 0 && i != a) {
n /= i;
b = i;
break;
}
}
if(a == 0 || b == 0 || a == n || b == n || n == 1) ret.append("NO\n");
else {
ret.append("YES\n");
ret.append(a + " " + b + " " + n + "\n");
}
}
System.out.println(ret);
}
}
| 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>
const long long int mod = 1e9 + 7;
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;
if (n < 24) {
cout << "NO\n";
continue;
} else {
vector<long long int> f;
for (long long int i = 2; i * i <= n; ++i) {
if (n % i == 0) {
f.push_back(i);
n /= i;
}
}
f.push_back(n);
sort(f.begin(), f.end());
if (f.size() > 3) {
for (long long int i = 3; i < f.size(); ++i) {
f[2] *= f[i];
}
}
if (f.size() < 3)
cout << "NO\n";
else if (f[0] != f[1] && f[0] != f[2] && f[1] != f[2]) {
cout << "YES\n";
cout << f[0] << " " << f[1] << " " << f[2] << "\n";
} else
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>
using namespace std;
using lli = long long int;
using ulli = unsigned long long int;
using db = double;
using vint = vector<int>;
using vlli = vector<vint>;
using vvint = vector<vint>;
using vvlli = vector<vlli>;
using cd = complex<double>;
const int inf = 1e9 + 5;
const lli infl = 1e18 + 5;
void solve() {
int n;
cin >> n;
int sqn = sqrt(n);
for (int a = 2; a <= sqn; ++a) {
if (n % a == 0) {
int sqna = sqrt(n / a);
for (int b = 2; b <= sqna; ++b) {
if ((n / a) % b == 0 and (a != b and b != (n / a) / b and
a != (n / a) / b and (n / a) / b >= 2)) {
cout << "YES"
<< "\n";
cout << a << " " << b << " " << (n / a) / b << "\n";
return;
}
}
}
}
cout << "NO"
<< "\n";
return;
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int t;
cin >> t;
while (t--) solve();
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())
j=0
while j<t:
n=int(input())
i=1
var1='NO'
var2='NO'
while i<(n**(1/3)) and var1=='NO':
i+=1
if n%i==0:
var1='YES'
a=i
if var1=='YES':
product=(n/a)
i=(a+1)
while i<(product**(0.5)) and var2=='NO':
if product%i==0:
var2='YES'
b=i
c=product/b
i+=1
if var1=='YES' and var2=='YES':
print('YES')
print(a,b,int(c))
else:
print('NO')
j+=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 | import math
def solve(n):
for a in range(2, math.ceil(n ** (1 / 3))):
if n % a == 0:
for b in range(a + 1, math.ceil(math.sqrt(n // a))):
c = n // a // b
if c * a * b == n:
print('YES')
print('%d %d %d' % (a, b, c))
return False
return True
t = int(input())
for i in range(t):
n = int(input())
if solve(n):
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 | # lines = [int(line) for line in open('CodeForces_1294C.in', 'r').readlines()[1:]]
lines = [int(line) for line in [*open(0)][1:]]
for n in lines:
sol = []
d = 2
done = False
while d * d <= n:
if n % d == 0:
n //= d
sol.append(d)
if len(sol) == 2:
done = True
if n not in sol:
sol.append(n)
print("YES")
print(' '.join(map(str, sol)))
else:
print("NO")
break
d += 1
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 | from collections import Counter
from collections import defaultdict
import math
# method to print the divisors
def factors(n) :
l=list()
# Note that this loop runs till square root
i = 1
while i <= math.sqrt(n):
if (n % i == 0) :
# If divisors are equal, print only one
if (n / i == i) :
l.append(i)
else :
# Otherwise print both
l.append(i)
l.append(n//i)
i = i + 1
return l
t=int(input())
for _ in range(0,t):
n=int(input())
l=factors(n)
#print(l,t)
if(len(l)==2):
print("NO")
continue
else:
a=l[2]
b=n//a
l2=factors(b)
f=0
#print(a,l2)
if(len(l2)==2):
print("NO")
continue
for i in range(2,len(l2)):
if(a!=l2[i] and l2[i]!=b//l2[i] and b//l2[i]!=a and a>=2 and l2[i]>=2 and b//l2[i]>=2):
f=1
print("YES")
print(a,l2[i],b//l2[i])
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 | #right preference
q = int(input())
for rwre in range(q):
n = int(input())
dz = []
nn = n
kan = 2
while nn > 1 and kan**2 <= n:
if nn%kan == 0:
nn//= kan
dz.append(kan)
else:
kan += 1
if nn > 1:
dz.append(nn)
dz.append(52525425542542)
roz = []
for i in range(1,len(dz)):
if dz[i] != dz[i-1]:
roz.append(dz[i-1])
dz.pop(-1)
if len(roz) >= 3:
print("YES")
print(roz[0],roz[1],n//(roz[0]*roz[1]))
else:
if len(roz) == 1:
a = roz[0]
b = roz[0]**2
c = n//(roz[0]**3)
if c > 1 and c != b and c != a:
print("YES")
print(a,b,c)
else:
print("NO")
if len(roz) == 2:
c = n//(roz[1]*roz[0])
if c != roz[1] and c!= roz[0] and c > 1:
print("YES")
print(roz[0],roz[1],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 | from math import sqrt
t=int(input())
while t>0:
d=list()
n=int(input())
imp=int(n)
ctr=0
for i in range (2,int(sqrt(n))):
while(n%i==0):
n//=i
d.append(i)
if(n>1):
d.append(n)
if(len(set(d))==2 and len(d)>=4):
print("YES")
print(d[0],d[-1],int(imp/(d[0]*d[-1])))
ctr+=1
if(len(set(d))>=3 or len(d)>=6 and ctr!=1 ):
print("YES")
if(len(set(d))>=3):
print(d[0],d[-1],int(imp/(d[0]*d[-1])))
ctr+=1
else:
print(d[0],d[0]**2,int(imp/(d[0]**3)))
ctr+=1
if(ctr==0):
print("NO")
t-=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 | from collections import deque
from math import sqrt
def factors(n):
d=deque()
for x in range(2,int(sqrt(n)+1)):
if n%x==0 :
d.append(x)
if n//x!=x :
d.append(n//x)
return d,set(d)
t=int(input())
while(t):
n=int(input())
f,m=factors(n)
flag=a=b=c=0
for x in range(len(f)):
fa,ma=factors(f[x])
for y in range(len(fa)):
if f[x]//fa[y]!=fa[y] and fa[y]!=n//f[x] and f[x]//fa[y]!=n//f[x]:
# print(y,x,f[x],fa[y])
flag=1
a,b,c=fa[y],n//f[x],f[x]//fa[y]
break
if flag :
break
if flag:
print("YES")
print(a,b,c)
else:
print("NO")
t-=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 | import java.util.Scanner;
public class C1294 {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int T = in.nextInt();
main: for (int t=0; t<T; t++) {
long N = in.nextLong();
for (long a=2; a*a*a<N; a++) {
if (N%a == 0) {
for (long b=a+1; b*b<N/a; b++) {
if (N%(a*b) == 0) {
System.out.println("YES");
System.out.println(a + " " + b + " " + N/(a*b));
continue main;
}
}
}
}
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 math import sqrt, ceil
def primesfrom2to(n):
""" Returns a list of primes < n """
sieve = [True] * (n//2)
for i in range(3,int(n**0.5)+1,2):
if sieve[i//2]:
sieve[i*i//2::i] = [False] * ((n-i*i-1)//(2*i)+1)
return [2] + [2*i+1 for i in range(1,n//2) if sieve[i]]
primedin105 = primesfrom2to(10 ** 5)
def primedec(n):
ret = dict()
if n == 1:
return ret
s = ceil(sqrt(n))
for i in primedin105:
if s < i:
return {n : 1}
if not (n % i):
d = primedec(n // i)
ret = d
try:
ret[i] += 1
except Exception:
ret[i] = 1
return ret
def solution(n):
p = primedec(n)
k = sorted(p.keys())
a = k[0];
if p[a] == 1:
if len(k) == 1:
print('NO')
return
b = k[1]
elif p[a] == 2:
if len(k) == 1:
print('NO')
return
b = k[0] * k[1]
else:
b = k[0] ** 2
c = n // (a * b)
if c == 1 or c == a or c == b:
print("NO")
return
print("YES")
print(a, b, c)
t = int(input())
for i in range(t):
solution(int(input()))
| 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 i in range(int(input())) :
n = int(input())
a,b = 0,0
for i in range(2, math.floor(math.sqrt(n))+1) :
if n%i == 0 :
a = i
break
if a != 0 :
for j in range(a+1, math.floor(math.sqrt(n//a))+1) :
if (n//i)%j == 0 :
b = j
break
if b != 0 and a != b and b != n//(a*b) :
print('YES')
print(a, b, n//(a*b))
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 math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,copy,functools
# import time,random,resource
sys.setrecursionlimit(10**7)
inf = 10**20
eps = 1.0 / 10**10
mod = 10**9+7
mod2 = 998244353
dd = [(-1,0),(0,1),(1,0),(0,-1)]
ddn = [(-1,0),(-1,1),(0,1),(1,1),(1,0),(1,-1),(0,-1),(-1,-1)]
def LI(): return list(map(int, sys.stdin.readline().split()))
def LLI(): return [list(map(int, l.split())) for l in sys.stdin.readlines()]
def LI_(): return [int(x)-1 for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def S(): return input()
def pf(s): return print(s, flush=True)
def pe(s): return print(str(s), file=sys.stderr)
def JA(a, sep): return sep.join(map(str, a))
def JAA(a, s, t): return s.join(t.join(map(str, b)) for b in a)
def IF(c, t, f): return t if c else f
class Prime():
def __init__(self, n):
self.M = m = int(math.sqrt(n)) + 10
self.A = a = [True] * m
a[0] = a[1] = False
self.T = t = [2]
for j in range(4, m, 2):
a[j] = False
for i in range(3, m, 2):
if not a[i]:
continue
t.append(i)
for j in range(i*i,m,i):
a[j] = False
self.ds_memo = {}
self.ds_memo[1] = set([1])
def is_prime(self, n):
return self.A[n]
def division(self, n):
d = collections.defaultdict(int)
for c in self.T:
while n % c == 0:
d[c] += 1
n //= c
if n < 2:
break
if n > 1:
d[n] += 1
return d.items()
# memo
def divisions(self, n):
if n in self.ds_memo:
return self.ds_memo[n]
for c in self.T:
if n % c == 0:
rs = set([c])
for cc in self.divisions(n // c):
rs.add(cc)
rs.add(cc * c)
self.ds_memo[n] = rs
return rs
rs = set([1, n])
self.ds_memo[n] = rs
return rs
def sowa(self, n):
r = 1
for k,v in self.division(n):
t = 1
for i in range(1,v+1):
t += math.pow(k, i)
r *= t
return r
def main():
t = I()
pr = Prime(10**10)
rr = []
for _ in range(t):
n = I()
ds = sorted(pr.division(n))
r = []
if len(ds) == 1:
k,v = ds[0]
if v < 6:
rr.append("NO")
continue
r = [k, k**2, k**(v-3)]
else:
t = n // ds[0][0]
t //= ds[1][0]
if t < 2 or t == ds[0][0] or t == ds[1][0]:
rr.append("NO")
continue
r = [ds[0][0], ds[1][0], t]
rr.append("YES")
rr.append(JA(r, " "))
return JA(rr, "\n")
print(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 | __author__ = 'Devesh Bajpai'
'''
https://codeforces.com/problemset/problem/1294/C
Solution: We first find the factor for n using a standard sqrt time complexity stub. If we could
find one, that becomes a. And then we find factor for n/a which is not equal to a. If we could
find oen, that becomes b. By then, the value of n is made (n/a)/b or n/ab. Hence if that value of
n is not equal to a,b or 1 then it is c. We add it to the distinct factor's set and return the result.
At any step if we fail, the answer returned is NO.
'''
def solve(n):
distinct_factors = set()
done, a = find_factors(n, distinct_factors)
if not done:
return "NO"
else:
n /= a
done, b = find_factors(n, distinct_factors)
if not done:
return "NO"
else:
n /= b
if n in distinct_factors or n == 1:
return "NO"
else:
distinct_factors.add(n)
return "YES" + "\n" + " ".join(str(_) for _ in distinct_factors)
'''
Helper to find distinct factor for the current value of n
Time Complexity: O(sqrt(n))
'''
def find_factors(n, distinct_factors):
factor = 2
f = 2
done = False
while f * f <= n and not done:
if n % f == 0 and f not in distinct_factors:
distinct_factors.add(f)
factor = f
done = True
f += 1
return done, factor
if __name__ == "__main__":
t = int(raw_input())
results = list()
for _ in xrange(0, t):
n = int(raw_input())
results.append(solve(n))
for result in results:
print result
| 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 | t = int(input())
for i in range(0,t):
n = int(input())
tmp = n
lst = []
j = 2
while(j*j<=n) :
if(tmp%j==0):
tmp = int(tmp/j)
lst.append(j)
if(len(lst)>=2):
if(tmp>j): lst.append(tmp)
break
j = j+1
if(len(lst)==3):
print("YES")
print(lst[0]," ",lst[1]," ",lst[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 | #include <bits/stdc++.h>
using namespace std;
int main() {
int t;
cin >> t;
while (t--) {
long long n;
cin >> n;
int cnt = 0;
vector<long long> ans;
for (long long i = 2; i * i <= n; i++) {
if (n % i == 0) {
if (cnt == 0) {
ans.push_back(i);
n = n / i;
cnt++;
} else {
ans.push_back(i);
ans.push_back(n / i);
break;
}
}
}
if ((ans.size() == 3) && (ans[0] != ans[1]) && (ans[1] != ans[2]) &&
(ans[0] != ans[2])) {
cout << "YES" << endl;
cout << ans[0] << " " << ans[1] << " " << ans[2] << 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 | for x in range(int(input())):
n = int(input())
a = 1
for i in range(2, int(n ** 0.5) + 2):
if n % i == 0:
a = i
n /= i
break
if a != 1:
b = 1
for i in range(a + 1, int(n ** 0.5) + 2):
if n % i == 0:
if n // i != a and n // i != i:
b = i
n /= i
break
if b != 1:
if n != 1:
print('YES')
print(a, b, int(n))
else:
print('NO')
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 collections import defaultdict
def solve(n):
table = set()
i = 2
while i*i <= n:
if n % i == 0:
table.add(i)
n /= i
break
i += 1
i = 2
while i*i <= n:
if n % i == 0 and i not in table:
table.add(i)
n /= i
break
i += 1
if len(table) < 2 or n == 1 or n in table:
print("NO")
else:
print("YES")
table.add(int(n))
print(" ".join([str(t) for t in table]))
T = int(input())
for t in range(T):
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 | import java.util.*;
public class Main
{
public static void main(String[] args)
{
Scanner sc=new Scanner(System.in);
int t=sc.nextInt();
while (t-- > 0)
{
long n=sc.nextLong();
long a=0L,b=0L,c=0L;
boolean found=false;
for (long i=2L;i<=Math.sqrt((long)n);i++)
{
if (n % i == 0)
{
n=n/i;
a=i;
found=true;
break;
}
}
if (found)
{
found=false;
for (long i=2;i<=Math.sqrt((long)n);i++)
{
if (n%i==0 && i!=a)
{
b=i;
c=n/i;
if (c!=a && c!=b)
{
found=true;
break;
}
}
}
if (found)
{
System.out.println("YES");
System.out.println(a + " " + b + " " + c);
}
else
System.out.println("NO");
}
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 | cases = int(input())
def prime(n):
for i in range(2,int((n+1)**0.5)):
if n%i == 0:
return False
return True
for i in range(cases):
num = int(input())
if prime(num):
print("NO")
continue
if num < 24:
print("NO")
continue
n = num
d = {}
for j in range(2, int((n+1)**0.5)):
while num%j == 0:
if j not in d:
d[j] = 1
else:
d[j] += 1
num /= j
if num > 1:
if num not in d:
d[num] = 1
else:
d[num] += 1
ans = []
if len(d) >= 3:
two = 2
print("YES")
for key in d:
if two == 0:
break
ans.append(key)
d[key] -= 1
two -= 1
c = 1
for ele in d:
val = d[ele]
c *= (ele**val)
ans.append(int(c))
print(int(ans[0]),int(ans[1]),int(ans[2]))
elif len(d) == 2:
for key in d:
ans.append(key)
d[key] -= 1
c = 1
for key in d:
val = d[key]
c *= key**val
if c not in ans and c != 1:
ans.append(int(c))
print("YES")
print(int(ans[0]),int(ans[1]),int(ans[2]))
elif c in ans or c == 1:
print("NO")
elif len(d) == 1:
val = list(d.values())
if val[0] < 6:
print("NO")
else:
a = int(list(d.keys())[0])
b = int(list(d.keys())[0]**2)
c = int(list(d.keys())[0]**(val[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 | t=int(input())
for _ in range(t):
n=int(input())
a=[]
i=2
while len(a)<2 and i*i<n:
if n%i==0:
a.append(i)
n//=i
i+=1
if len(a)==2 and n not in a:
print('YES')
print(*a,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 *
for _ in range(int(input())):
n = int(input())
nn = n
l = []
now = 2
m = []
sq = int(sqrt(n))
while n > 1:
if n % now == 0:
while n % now == 0:
l.append(now)
n //= now
else:
now += 1
if now > sq:
break
if len(l) < 2:
print("NO")
continue
m.append(l[0])
if l[0] != l[1]:
m.append(l[1])
elif len(l) == 2:
print("NO")
continue
else:
m.append(l[1] * l[2])
t = nn // (m[0] * m[1])
if (t == 1) or (t in m):
print("NO")
continue
else:
print("YES")
print(m[0], m[1], t) | 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())
a=n
i=2
while i**2<n:
if n%i==0:
a=i
n//=i
break
i+=1
if a==n:
print('NO')
continue
b=n
i=a+1
while i**2<n:
if n%i==0:
b=i
n//=i
break
i+=1
if b==n:
print('NO')
continue
print('YES')
print(a,b,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 | from functools import reduce
def fac(n):
return sorted(set(reduce(list.__add__,
([i, n//i] for i in range(1, int(n**0.5) + 1) if n % i == 0))))
t=int(input())
for i in range(t):
n=int(input())
x=fac(n)
x=x[1:-1]
flag=1
if len(x)<3:
flag=0
else:
a,b=x[0],0
for j in range(1,len(x)):
if n%(a*x[j])==0:
b=x[j]
break
if b:
te=a*b
if n%te==0:
c=n//te
if a!=b and a!=c and b!=c and a>=2 and b>=2 and c>=2:
pass
else:
flag=0
else:
flag=0
else:
flag=0
if flag:
pass
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;
using namespace std;
string chk[10] = {"1110111", "0010010", "1011101", "1011011", "0111010",
"1101011", "1101111", "1010010", "1111111", "1111011"};
void solve() {
long long n;
cin >> n;
for (int a = 2; a * (a + 1) * (a + 2) <= n; a++)
if (n % a == 0) {
int num = n / a;
for (int b = a + 1; b * (b + 1) <= num; b++)
if (num % b == 0) {
int c = num / b;
printf("YES\n");
printf("%d %d %d\n", a, b, c);
return;
}
}
printf("NO\n");
}
int main() {
int t;
cin >> t;
while (t--) {
solve();
}
}
| 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 | from functools import reduce
def factors(n):
return set(reduce(list.__add__, ([i, n//i] for i in range(1, int(n**0.5) + 1) if n % i == 0)))
t = int(input())
#for each test cases
for i in range(t):
n=int(input())
def x(n):
x=factors(n)
counter=1
if (len(x)<5):
return True
else:
x=x-set([1])-set([n])
x=list(x)
for i in range(len(x)-2):
for j in range (i+1,len(x)-1):
for k in range (j+1,len(x)):
if(x[i]*x[j]*x[k]==n):
print("YES")
print(x[i],x[j],x[k])
return False
return True
if x(n):
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
from _collections import deque
mod = 10**9 + 7
import sys
sys.setrecursionlimit(10**5)
from queue import PriorityQueue
from bisect import bisect_right
from bisect import bisect_left
from _collections import defaultdict
from math import sqrt,factorial,gcd,log2,inf,ceil
import heapq
input = lambda : sys.stdin.readline().rstrip()
from sys import stdin, stdout
from heapq import heapify,heappush,heappop
# min = 0
# n,m = map(int,input().split())
# maxi = n
# z = m
# while z>0:
#
# yo = (1+sqrt(1+8*z))//2
# maxi-=yo
# print(yo)
# z-=((yo)*(yo+1))//2
#
#
# maxi = max(int(maxi),0)
#
#
#
# if m>=n-1:
# min = 0
# else:
# min = (n)-2*(m)
# #
# if m == 0:
# print(n,n)
# exit()
# if n == 1 and m == 1:
# min = 1
# print(min,maxi)
# t = int(input())
#
# for _ in range(t):
#
# a,b,c,n = map(int,input().split())
# if (a+b+c+n)%3 == 0:
# print('YES')
# else:
# print('NO')
#
# t = int(input())
# for _ in range(t):
#
# n = int(input())
# la = []
# for i in range(n):
#
# a,b = map(int,input().split())
# la.append([a,b])
#
# la.sort()
# flag = 0
# for i in range(1,n):
# if la[i][1]<la[i-1][1]:
# flag = 1
# break
#
# if flag:
# print('NO')
# else:
# x1,y1 = 0,0
# ans = []
# while la!=[]:
# x,y = la.pop(0)
# z1 = x-x1
# z2 = y-y1
# for i in range(z1):
# ans.append('R')
# for i in range(z2):
# ans.append('U')
#
# x1,y1 = x,y
# print('YES')
# print(''.join(ans))
def solve(n):
la = []
for i in range(2,int(sqrt(n))+1):
if n%i == 0:
if n//i!=i:
la.append((i,n//i))
return la
t = int(input())
for _ in range(t):
n = int(input())
l = solve(n)
flag = 0
x,y,z = -1,-1,-1
for a,b in l:
la = solve(a)
flag = 0
for c,d in la:
if b!=c and b!=d:
x = b
y = c
z = d
flag = 1
break
if flag:
break
la = solve(b)
for c,d in la:
if a!=c and a!=d:
x = a
y = c
z = d
flag = 1
break
if flag:
break
if x*y*x>0:
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 | for _ in range(int(input())):
n=int(input())
l=[]
if(n%2==0):
l.append(2)
for i in range(3,int(pow(n,0.5))+1):
if(n%i==0):
if(n//i==i):
l.append(i)
else:
l.append(i)
l.append(n//i)
f=0
for i in range(0,len(l)):
for j in range(i+1,len(l)):
for k in range(j+1,len(l)):
if(l[i]*l[j]*l[k]==n):
f=1
break
if(f==1):
break
if(f==1):
break
if(f==1):
print("YES")
print(l[i],l[j],l[k])
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 fin(n):
for a in range(2,int(n**(1/3))+1):
if n%a==0:
x=n/a
for b in range(2,int(x**(1/2))+1):
if x%b==0:
y=int(x/b)
if a!=b and b!=y and y!=a:
return a,b,y
return 0,0,0
for _ in range(int(input())):
n=int(input())
p,q,r=0,0,0
p,q,r=fin(n)
if p!=0:
print("YES")
print(p,q,r)
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 | n=int(input())
for i in range (n):
a=int(input())
k=a
p=[]
j=2
while (j*j<k):
if (k%j==0 and j not in p):
p.append(j)
k=k//j
if (len(p)==2):
g=a//(p[0]*p[1])
if (g!=p[0] and g!=p[1]):
break
j+=1
if len(p)<2:
print ("NO")
elif len (p)==2 and p[0]*p[1]==a:
print ("NO")
elif p[0]!=a//(p[0]*p[1]) and p[1]!=a//(p[0]*p[1]):
print ("YES")
print(p[0],p[1],a//(p[0]*p[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 | from math import gcd,sqrt
t = int(input())
for _ in range(t):
n = int(input())
flag = 0
for i in range(2,int(sqrt(n))+1):
if flag:
break
if n%i==0:
divided = n//i
for j in range(2,int(sqrt(n))+1):
if divided%j==0 and divided//j > 1 and j!=i and j!= divided//j and i!=divided//j:
print("YES")
print(j,i,divided//j)
flag = 1
break
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 | #include <bits/stdc++.h>
using namespace std;
int main() {
int T, t;
cin >> T;
for (t = 0; t < T; t++) {
int n, i;
cin >> n;
int numPrime = 0;
int num = n;
int f = 1;
vector<int> v;
map<int, int> mp;
for (i = 2; i <= sqrt(n); i++) {
f = 1;
if (num % i == 0) {
f = f * i;
while (num % i == 0) {
if (f != 1 && mp[f] == 0) {
v.push_back(f);
mp[f] = 1;
f = 1;
} else {
f = f * i;
}
num = num / i;
}
}
}
if (mp[num] == 0 && num != 1) {
v.push_back(num);
mp[num] = 1;
}
if (v.size() == 2) {
int r = n / (v[0] * v[1]);
if (mp[r] == 0 && r != 1) v.push_back(r);
}
sort(v.begin(), v.end());
if (v.size() > 2) {
cout << "YES" << endl;
cout << v[0] << " " << v[1] << " " << n / (v[0] * v[1]) << 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 | from sys import stdin
t = int(stdin.readline())
for _ in range(t):
n = int(stdin.readline())
if n < 24:
print("NO")
continue
a = []
for i in range(2, int(n**0.5)+1):
if n%i == 0:
n//=i
a.append(i)
break
if len(a) == 1:
for i in range(a[0]+1, int(n**0.5)+1):
if n%i == 0:
n//=i
a.append(i)
break
a.append(n)
if len(a) == 3 and a[1] < a[2]:
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 |
t = int(input())
for i in range(t):
f = 0
n = int(input())
if n % 2 == 0:
a = 2
f = 1
else:
for j in range(3 ,int((n**(1/3)) + 2), 2):
if (n % j == 0):
a = j
f = 1
break
if (f == 0):
print("NO")
else:
n = n // a
f = 0
for j in range(a ,int((n**(1/2)) + 2)):
c = n // j
if (n % j == 0) and (j != a) and (c) != 1:
b = j
f = 1
break
if (f == 0) or (b == c) or (a == c):
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 | import math
for _ in range(int(input())):
n = int(input())
def prime(n):
primes = {2:0}
while n % 2 == 0:
primes[2] += 1
n = n // 2
for i in range(3, int(math.sqrt(n)+1), 2):
while n % i == 0:
if i in primes:
primes[i] += 1
else:
primes[i] = 1
n = n // i
if n > 2:
if n in primes:
primes[n] += 1
else:
primes[n] = 1
return primes
def prod(l):
p = 1
for x in l:
p *= x
return p
primes = prime(n)
# print(primes)
ar = []
keys = sorted(primes.keys())
for key in keys:
ar.extend([key]*primes[key])
# print(ar)
if len(ar) < 3:
print('NO')
else:
a = ar[0]
i = 1
b = ar[1]
if ar[0] == ar[1]:
b *= ar[2]
i += 1
c = prod(ar[i+1:])
if i == len(ar) - 1 or b == c or len(set([a, b, c])) < 3:
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 | import sys
ip = lambda : sys.stdin.readline().rstrip()
for _ in range(int(ip())):
n=int(ip())
ans=[]
for i in range(2,int(n**0.5)+1):
if n%i==0:
v=n//i
for j in range(2,int(v**0.5)+1):
if v%j==0:
st=set([i,j,v//j])
if len(st)==3:
ans=sorted([i,j,v//j])
break
if ans!=[]:
break
if ans!=[]:
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 | t = int(input())
for i in range(t):
n = int(input())
f = 2
li = []
c = 0
while c < 2 and n >= f * f:
if n % f == 0:
n = n // f
c += 1
li.append(f)
if c == 2:
if n not in li and n >= 2:
li.append(n)
f += 1
if len(li) == 3:
print("YES")
print(li[0], li[1], li[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 collections,math
for _ in range(int(input())):
n=int(input())
factors=[]
flag=False
d=collections.defaultdict(int)
for i in range(2,int(math.sqrt(n))+1):
if n%i==0:
d[i]=1
factors.append(i)
if (n//i) != i:
d[n//i]=1
factors.append(n//i)
factors.sort()
for i in range(len(factors)-1):
for j in range(i+1,len(factors)):
k=n//(factors[i]*factors[j])
if d[k]==1 and factors[i]*factors[j]*k==n and factors[i]!=k and factors[j]!=k:
flag=True
print('YES')
print(factors[i],factors[j],k)
break
if flag:
break
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 java.util.*;
import java.io.*;
public class Main {
static class pair implements Comparable<pair>{
int a;
int b;
public pair(int a, int b){
this.a=a;
this.b=b;
}
public int compareTo(pair p){
return (a-p.a==0)?b-p.b:a-p.a;
}
}
public static void main(String[] args) throws IOException,InterruptedException{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
PrintWriter out = new PrintWriter(System.out);
// String s = br.readLine();
// char[] arr=s.toCharArray();
// ArrayList<Integer> arrl = new ArrayList<Integer>();
// TreeSet<Integer> ts1 = new TreeSet<Integer>();
// HashSet<Integer> h = new HashSet<Integer>();
// HashMap<Integer, Integer> map= new HashMap<>();
// PriorityQueue<String> pQueue = new PriorityQueue<String>();
// LinkedList<String> object = new LinkedList<String>();
// StringBuilder str = new StringBuilder();
//*******************************************************//
// StringTokenizer st = new StringTokenizer(br.readLine());
// int t = Integer.parseInt(st.nextToken());
// while(t-->0){
// st = new StringTokenizer(br.readLine());
// int n = Integer.parseInt(st.nextToken());
// int[] arr = new int[n];
// st = new StringTokenizer(br.readLine());
// for(int i=0; i<n; i++){
// arr[i] =Integer.parseInt(st.nextToken());
// }
// int ans =0;
// out.println(ans);
// }
StringTokenizer st = new StringTokenizer(br.readLine());
int t = Integer.parseInt(st.nextToken());
loop: while(t-->0){
st = new StringTokenizer(br.readLine());
int n = Integer.parseInt(st.nextToken());
Factors(n);
}
out.flush();
}
static void Factors(int n)
{ int k =n;
ArrayList<Integer> P = new ArrayList<Integer>();
while (n % 2 == 0)
{
P.add(2);
n /= 2;
}
for (int i = 3; i * i <= n; i = i + 2)
{
while (n % i == 0)
{
n = n / i;
P.add(i);
}
}
if (n > 2)
P.add(n);
if (P.size() < 3)
{
System.out.println("NO");
return;
}
HashSet<Integer> h = new HashSet<Integer>();
int pro = 1;
int i =0;
for (; i<P.size()&&h.size()<2; i++) {
pro*=(int)P.get(i);
if(!h.contains(pro)){h.add(pro); pro =1;}
}
int product = 1;
for (; i < P.size(); i++)
product = product * (int)P.get(i);
if(product!=1)h.add(product);
if(h.size()==3){
Iterator value = h.iterator();
StringBuilder str = new StringBuilder();
int check =1;
while (value.hasNext()) {
int x =(int)value.next();
check*=x;
str.append(x+" ");
}
if(check==k){System.out.println("YES");System.out.println(str);}
else System.out.println("NO");
}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 | import java.util.*;
import java.io.*;
public class Main {
public static void solve(InputReader in) {
int n = in.readInt();
HashSet<Integer> set = new HashSet<>();
for(int i = 2; i*i <= n; i++) {
if(n % i == 0) {
set.add(i);
set.add(n/i);
}
}
for(Integer i : set) {
int a = i;
for(Integer b : set) {
if((b != a) && n / (a*b) != 0 && n /(a*b) != a && n/(a*b) != b && n/(a*b) != 1 && n%(a*b) == 0) {
System.out.println("YES");
System.out.println(a + " " + b + " " + n/(a*b));
return;
}
}
}
System.out.println("NO");
}
public static void main(String ...strings) {
InputReader in = new InputReader(System.in);
int t = in.readInt();
while(t-- > 0) {
solve(in);
}
}
}
class InputReader{
private InputStream stream;
private byte[] buf = new byte[1024];
private int curChar;
private int numChars;
private SpaceCharFilter filter;
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 int readInt() {
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 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 long readLong() {
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 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);
}
}
| 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 math import sqrt
def main():
def solve():
n = int(input())
a = 0
if n%2 == 0:
a = 2
else:
for i in range(3, int(n ** (1./3))+1, 2):
if n%i == 0:
a = i
break
if a == 0:
print("NO")
return
b = 0
for i in range(a+1, int(sqrt(n//a))+1):
if (n//a)%i == 0:
b = i
break
if b == 0:
print("NO")
return
c = n//(a*b)
if b == c:
print("NO")
return
print("YES")
print(a, b, c)
q = int(input())
for _ in range(q):
solve()
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 | import java.util.*;
public class ProductOfThreeNumbers {
public static void main(String[] args) {
Scanner sc=new Scanner(System.in);
int t=sc.nextInt();
while(t-->0) {
int n=sc.nextInt();
int k=n;
int ans=0;
int count=0;
outer:
for(int i=2;i<Math.sqrt(n)+1;i++) {
n=k;
if(n%i==0 && i!=n) {
ans=i;
n=n/i;
for(int j=2;j<Math.sqrt(n)+1;j++) {
//System.out.println(i+" "+j+" "+n+" "+(n/j));
if(j!=i && n%j==0 && (n/j)!=j && (n/j)!=i) {
count++;
System.out.println("YES");
System.out.println(ans+" "+j+" "+(n/j));
break outer;
}
}
}
}
if(count==0) {
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 | import math
for _ in range(int(input())):
n=int(input())
a=[]
s=round(math.sqrt(n))
for i in range(2,s+1):
if len(a)==2:
if n not in a:
a.append(n)
break
else:
if n%i==0:
a.append(i)
n=n//i
if len(a)==3:
print("Yes")
for j in range(0,3):
print(a[j],end=" ")
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 | num = int(input())
answer = []
for i in range(num):
target = int(input())
lis = []
k = 2
while len(lis)<2 and k**2<target:
if target % k == 0:
target = target //k
lis.append(k)
k += 1
if len(lis) == 2 and target not in lis:
answer.append([lis[0],lis[1],target])
else:
answer.append('NO')
for i in answer:
if i == "NO":
print(i)
else:
print('YES')
print(i[0],i[1],i[2])
| 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.*;
public class Main
{
final static int mod = 1000000007;
static FastReader sc;
static PrintWriter out;
static boolean test_case_input = true;
public static void solution() throws IOException
{
int n = sc.nextInt();
int a = -1, b = -1, c = -1;
int temp = n;
for(int i=2; (long) i*i <= temp; i++) {
if(n == 1) break;
if(n % i == 0) {
if(a == -1) {
a = i;
n /= i;
}
while(n % i == 0) {
if(b == -1) {
b = i;
n /=i;
}
else if(b == a) {
b *= i;
n /= i;
}
else {
if(c == -1) {
c = i;
n /=i;
}
else {
c *= i;
n /= i;
}
}
}
}
}
if(n != 1) {
if(c == -1) c = n;
else c *= n;
}
if(a != b && b != c && a != c && a != -1 && b != -1 && c != -1) {
out.println("YES\n" + a + " " + b + " " + c);
}
else out.println("NO");
}
// GCD
public static int __gcd(int a, int b)
{
BigInteger n1 = BigInteger.valueOf(a);
BigInteger n2 = BigInteger.valueOf(b);
BigInteger gcd = n1.gcd(n2);
return gcd.intValue();
}
public static long __gcd(long a, long b)
{
BigInteger n1 = BigInteger.valueOf(a);
BigInteger n2 = BigInteger.valueOf(b);
BigInteger gcd = n1.gcd(n2);
return gcd.longValue();
}
public static void main(String args[]) throws IOException
{
long start = 0, end = 0;
try
{
File output = new File("output.txt");
sc = new FastReader();
if (output.exists())
{
out = new PrintWriter(new FileOutputStream("output.txt"));
start = System.nanoTime();
} else
{
out = new PrintWriter(System.out);
}
int test_cases = 1;
if (test_case_input) test_cases = sc.nextInt();
while (test_cases-- > 0)
{
solution();
}
if (output.exists())
{
end = System.nanoTime();
out.println("Execution time: " + (end - start) / 1000000 + " ms");
}
out.flush();
out.close();
} catch (Exception e)
{
out.println("Exception: " + e);
out.flush();
out.close();
return;
}
}
// Fast IO
static class FastReader
{
BufferedReader br;
StringTokenizer st;
public FastReader() throws FileNotFoundException
{
File in = new File("input.txt");
if (in.exists())
{
br = new BufferedReader(new InputStreamReader(new FileInputStream("input.txt")));
} else
{
br = new BufferedReader(new InputStreamReader(System.in));
}
}
String next()
{
while (st == null || !st.hasMoreElements())
{
try
{
st = new StringTokenizer(br.readLine());
} catch (IOException e)
{
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt()
{
return Integer.parseInt(next());
}
float nextFloat()
{
return Float.parseFloat(next());
}
long nextLong()
{
return Long.parseLong(next());
}
double nextDouble()
{
return Double.parseDouble(next());
}
String nextLine()
{
String str = "";
try
{
str = br.readLine();
} catch (IOException e)
{
e.printStackTrace();
}
return 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 | import sys
import math
import heapq
import collections
def inputnum():
return(int(input()))
def inputnums():
return(map(int,input().split()))
def inputlist():
return(list(map(int,input().split())))
def inputstring():
return([x for x in input()])
def inputstringnum():
return([ord(x)-ord('a') for x in input()])
def inputmatrixchar(rows):
arr2d = [[j for j in input().strip()] for i in range(rows)]
return arr2d
def inputmatrixint(rows):
arr2d = []
for _ in range(rows):
arr2d.append([int(i) for i in input().split()])
return arr2d
t = int(input())
for q in range(t):
n = inputnum()
i = 2
a = 0
b = 0
while i*i <= n and (a == 0 or b == 0):
if n%i == 0 and a == 0:
n //= i
a = i
i += 1
continue
elif n%i == 0 and b == 0:
n //= i
b = i
i += 1
continue
i += 1
if (a == 0 or b == 0) or (n == a or n == b):
print("NO")
else:
print("YES")
print(a, b, 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 math
t = int(input())
for ii in range(0,t):
n = int(input())
a = 0
b = 0
c = 0
if n<24:
print("NO")
continue
ifFind = 0
for i in range(2,int(math.sqrt(n))+1):
if ifFind == 1:
break
if n%i!=0:
continue
n2 = n // i
for j in range(i+1,int(math.sqrt(n2))+1):
if n2%j!=0:
continue
n3 = n2//j
if j==n3:
break
ifFind = 1
a=i
b=j
c=n3
break
if ifFind == 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 | t=int(input())
from math import sqrt
primes=[True for i in range(10**5)]
for i in range(2,len(primes)):
jumper=i
ind=2
while i*ind<len(primes):
primes[i*ind]=False
ind+=1
primes=set([i for i in range(len(primes)) if primes[i]])
def divider(x):
if x%2==0:
return x//2,2
for i in range(3,int(sqrt(x)+1),2):
#print("Inside loop:",i)
if x%i==0:
return x//i,i
return -1,-1
for _ in range(t):
n=int(input())
flag=False
ans1,ans2,ans3=0,0,0
for i in range(2,int(sqrt(n)+1)):
#print(i)
reali=n//i
if n%i==0 and n//i not in primes:
ans2,ans3=divider(n//i)
if ans2==-1 and ans3==-1:
continue
ans1=i
if ans1!=ans2 and ans2!=ans3 and ans3!=ans1:
flag=True
break
if flag:
print("YES")
print(ans1,ans2,ans3)
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 R(): return map(int, input().split())
def I(): return int(input())
def S(): return str(input())
def L(): return list(R())
from collections import Counter
import math
import sys
for _ in range(I()):
n=I()
div=[]
n2=n
for i in range(2,int(math.sqrt(n))+1):
if n%i ==0:
div.append(i)
n/=i
if n>1:
div.append(n)
if len(div)<3 or n2//(div[0]*div[1])== div[0] or n2//(div[0]*div[1])==div[1]:
print('NO')
else:
print('YES')
print(div[0],div[1],n2//(div[0]*div[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 | for i in range(int(input())):
n=int(input())
i=2
arr=[]
while n>0 and i<=(int(n**0.5)):
if n%i==0:
n/=i
arr.append(i)
i+=1
if len(arr)==2:
break
if len(arr)==2:
if n not in arr:
print('YES')
print(arr[0],arr[1],int(n))
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 as S
def div(x):
d=[]
if x==1:
return [1]
if x==2:
return [1,2]
cnt=[1,n]
for i in range(2,int(S(x))+1):
if x %i==0:
cnt.append(i)
return cnt
def check(arr):
if 1 in arr or 0 in arr:
return 0
return len(set(arr))==3
import sys
input=sys.stdin.readline
t=int(input())
ans=[]
for i in range(t):
n=int(input())
f=0
cnt=div(n)
for x in cnt :
y=n//x
if f: break
for z in div(y):
if check([x,z,y//z]):
ans=[x,z,y//z]
f=1
break
if f:
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 | def main():
t = int(input())
for z in range (t):
n = int(input())
arr = []
i=2;
while i*i <= n:
while n%i==0:
arr.append(i)
n //= i
i += 1
if n>1:arr.append(n)
a,b,c = 1,1,1
for x in arr:
if a==1: a *= x
elif b==1 or a==b: b *= x
else: c *= x
arr = set([a,b,c])
if a!=1 and b!=1 and c!=1 and len(arr)==3:
print("YES")
print(a,b,c)
else:
print("NO")
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 | import math
import os
def get_array():
return list(map(int, input().split(' ')))
def get_divisions(a):
i = 2
while i <= math.sqrt(a):
if a % i == 0:
yield i
i += 1
yield -1
if __name__ == '__main__':
t = int(input())
for i in range(t):
n = int(input())
divs = get_divisions(n)
while True:
a = next(divs)
f = False
if a != -1:
divs_a = get_divisions(n / a)
while True:
b = next(divs_a)
if b != -1:
c = int(n / a / b)
if b != a and c != b:
print('YES')
print(a, ' ', b, ' ', c)
f = True
break
else:
f = False
break
if not f:
print('NO')
break
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 | /*
If you want to aim high, aim high
Don't let that studying and grades consume you
Just live life young
******************************
If I'm the sun, you're the moon
Because when I go up, you go down
*******************************
I'm working for the day I will surpass you
****************************************
*/
import java.util.*;
import java.awt.Point;
import java.lang.Math;
import java.util.Arrays;
import java.util.Arrays;
import java.util.Scanner;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.io.OutputStream;
import java.util.Comparator;
import java.util.stream.IntStream;
public class Main {
static int oo = (int)1e9;
public static void main(String[] args) {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
Scanner in = new Scanner(inputStream);
PrintWriter out = new PrintWriter(outputStream);
MuskA solver = new MuskA();
solver.solve(1, in, out);
out.close();
}
static class MuskA {
static int inf = (int) 1e9 + 7;
public int curr;
public int[] visited;
public void solve(int testNumber, Scanner in, PrintWriter out) {
int T= in.nextInt();
main: for (int t=0; t<T; t++) {
long N = in.nextLong();
for (long a=2; a*a*a<N; a++) {
if (N%a == 0) {
for (long b=a+1; b*b<N/a; b++) {
if (N%(a*b) == 0) {
out.println("YES");
out.println(a + " " + b + " " + N/(a*b));
continue main;
}
}
}
}
out.println("NO");
}
}
}
public int factorial(int n) {
int fact = 1;
int i = 1;
while(i <= n) {
fact *= i;
i++;
}
return fact;
}
public static long gcd(long x,long y)
{
if(x%y==0)
return y;
else
return gcd(y,x%y);
}
public static int gcd(int x,int y)
{
if(x%y==0)
return y;
else
return gcd(y,x%y);
}
public static int abs(int a,int b)
{
return (int)Math.abs(a-b);
}
public static long abs(long a,long b)
{
return (long)Math.abs(a-b);
}
public static int max(int a,int b)
{
if(a>b)
return a;
else
return b;
}
public static int min(int a,int b)
{
if(a>b)
return b;
else
return a;
}
public static long max(long a,long b)
{
if(a>b)
return a;
else
return b;
}
public static long min(long a,long b)
{
if(a>b)
return b;
else
return a;
}
public static long pow(long n,long p,long m)
{
long result = 1;
if(p==0)
return 1;
if (p==1)
return n;
while(p!=0)
{
if(p%2==1)
result *= n;
if(result>=m)
result%=m;
p >>=1;
n*=n;
if(n>=m)
n%=m;
}
return result;
}
public static long pow(long n,long p)
{
long result = 1;
if(p==0)
return 1;
if (p==1)
return n;
while(p!=0)
{
if(p%2==1)
result *= n;
p >>=1;
n*=n;
}
return result;
}
static long sort(int a[]){
int n=a.length;
int b[]=new int[n];
return mergeSort(a,b,0,n-1);
}
static long mergeSort(int a[],int b[],long left,long right){
long c=0;
if(left<right){
long mid=left+(right-left)/2;
c= mergeSort(a,b,left,mid);
c+=mergeSort(a,b,mid+1,right);
c+=merge(a,b,left,mid+1,right);
}
return c;
}
static long merge(int a[],int b[],long left,long mid,long right){
long c=0;int i=(int)left;int j=(int)mid; int k=(int)left;
while(i<=(int)mid-1&&j<=(int)right){
if(a[i]<=a[j]){
b[k++]=a[i++];
}
else{
b[k++]=a[j++];c+=mid-i;
}
}
while (i <= (int)mid - 1)
b[k++] = a[i++];
while (j <= (int)right)
b[k++] = a[j++];
for (i=(int)left; i <= (int)right; i++)
a[i] = b[i];
return c;
}
} | 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 math import sqrt
def solve(R:list):
n = int(input())
m = n
for i in range(2,int(sqrt(m))):
if i >= n :
return
if n % i == 0:
R.append(i)
n //= i
if len(R) == 2:
break
if len(R) == 2:
if n > R[-1]:
R.append(n)
t = int(input())
for cnt in range(t):
R = []
solve(R)
# print("CASE : " + str(cnt+1))
if len(R) == 3:
print("YES")
print(" ".join(map(str,R)))
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.math.*;
import java.io.*;
public class input2 {
public static int factoiral(int a) {
if (a == 0) return 1;
else return (a * factoiral(a - 1));
}
public static void sortbyColumn(int arr[][], int col) {
Arrays.sort(arr, new Comparator<int[]>() {
@Override
public int compare(int[] entry1, int[] entry2) {
if (entry1[col] > entry2[col])
return 1;
else
return -1;
}
});
}
public static void merge(long arr[], long brr[], int l, int m, int r) {
// Find sizes of two subarrays to be merged
int n1 = m - l + 1;
int n2 = r - m;
/* Create temp arrays */
long L[] = new long[n1];
long L1[] = new long[n1];
long R[] = new long[n2];
long R1[] = new long[n2];
/* Copy data to temp arrays */
for (int i = 0; i < n1; ++i) {
L[i] = arr[l + i];
L1[i] = brr[l + i];
}
for (int j = 0; j < n2; ++j) {
R[j] = arr[m + 1 + j];
R1[j] = brr[m + 1 + j];
}
/* Merge the temp arrays */
// Initial indexes of first and second subarrays
int i = 0, j = 0;
// Initial index of merged subarry array
int k = l;
while (i < n1 && j < n2) {
if (L[i] <= R[j]) {
arr[k] = L[i];
brr[k] = L1[i];
i++;
} else {
arr[k] = R[j];
brr[k] = R1[j];
j++;
}
k++;
}
while (i < n1) {
arr[k] = L[i];
brr[k] = L1[i];
i++;
k++;
}
/* Copy remaining elements of R[] if any */
while (j < n2) {
arr[k] = R[j];
brr[k] = R1[j];
j++;
k++;
}
}
public static void sort(long arr[], long brr[], int l, int r) {
if (l < r) {
// Find the middle point
int m = (l + r) / 2;
// Sort first and second halves
sort(arr, brr, l, m);
sort(arr, brr, m + 1, r);
// Merge the sorted halves
merge(arr, brr, l, m, r);
}
}
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int t = Integer.parseInt(br.readLine());
for (int j = 0; j < t; j++) {
int n = Integer.parseInt(br.readLine());
int g = n;
int upn = (int)Math.sqrt(n);
int a = 1;
for(int i=2;i<=upn;i++){
if(n%i == 0){
a = Math.min(i,n/i);
break;
}
}
// System.out.println(a);
n = n/a;
upn = (int)Math.sqrt(n);
int b = 1;
for(int i=2;i<=upn;i++){
if(n%i == 0 &&(i!=a)){
b = Math.min(i,n/i);
break;
}
}
// System.out.println(b);
int x = a*b;
int c = g/x;
if(a == 1 || b == 1 || c == 1) System.out.println("NO");
else if(a == b || a == c || b == c) System.out.println("NO");
else{
System.out.println("YES");
System.out.println(a + " " + b + " " + c);
}
}
}
}
| 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())
while t:
t-=1
n = int(input())
i = 2
val = []
while(i*i<=n):
if n%i==0:
n = n//i
val.append(i)
break
i+=1
i = 2
while(i*i<=n):
if n%i==0 and i!=val[0]:
n = n//i
val.append(i)
break
i+=1
if len(val)<2 or n==1 or val[-1]==n:
print('NO')
else:
print('YES')
val.append(n)
print(*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 | import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
public class Main
{
static String s;
public static void main(String[] args) throws NumberFormatException, IOException
{
BufferedReader br= new BufferedReader(new InputStreamReader(System.in));
BufferedWriter bw= new BufferedWriter(new OutputStreamWriter(System.out));
String str= br.readLine();
int t = Integer.parseInt(str);
for(int h=0; h<t; h++)
{
int n= Integer.parseInt(br.readLine());
String s="NO";
int nn=0;
int i=2;
int m0=0;
while(i*i<=n)
{
if(nn==0)
{
if(n%i==0)
{
n /= i;
m0=i;
nn++;
}
}
else if(nn==1)
{
if(n%i==0 && n/i != i)
{
s = "YES\n" + m0 + " " + i + " " + n/i;
break;
}
}
i++;
}
System.out.println(s);
}
}
} | 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 isPrime(int n) {
if (n <= 1) return 0;
for (int i = 2; i * i <= n; i++)
if (n % i == 0) return 0;
return 1;
}
int main() {
int t;
cin >> t;
while (t--) {
int n;
cin >> n;
if (isPrime(n))
cout << "NO\n";
else {
int c = 0;
vector<int> a;
int prev = 0;
for (int i = 2; i * i < n; i++) {
if ((a.size() == 2) && (n != prev)) break;
if (n % i == 0) {
if (prev != i) {
a.push_back(i);
prev = i;
}
n = n / i;
}
}
if (a.size() < 2)
cout << "NO\n";
else {
cout << "YES\n";
cout << a[0] << " " << a[1] << " " << n << "\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.*;
public class CF1294C extends PrintWriter {
CF1294C() { super(System.out); }
Scanner sc = new Scanner(System.in);
public static void main(String[] $) {
CF1294C o = new CF1294C(); o.main(); o.flush();
}
void main() {
int t = sc.nextInt();
while (t-- > 0) {
int n = sc.nextInt();
int u = 1, ku = 0, v = 1, kv = 0, w = n;
for (int p = 2; p <= w / p; p++)
if (w % p == 0) {
int k = 0;
while (w % p == 0) {
k++;
w /= p;
}
if (u == 1) {
u = p; ku = k;
} else {
v = p; kv = k;
break;
}
}
if (w != 1) {
if (u == 1) {
u = w; ku = 1;
w = 1;
} else if (v == 1) {
v = w; kv = 1;
w = 1;
}
}
int a = -1, b = -1, c = -1;
if (u != 1 && v != 1 && w != 1 || u != 1 && v != 1 && ku + kv >= 4) {
a = u;
b = v;
c = n / a / b;
} else if (ku >= 6) {
a = u;
b = u * u;
c = n / a / b;
}
if (c == -1)
println("NO");
else {
println("YES");
println(a + " " + b + " " + c);
}
}
}
}
| 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 prime_factorize(n):
a = []
if n % 2 == 0:
a.append(2)
n //= 2
f = 3
while f * f <= n and len(a)<2:
if n % f == 0:
a.append(f)
n //= f
f += 1
else:
f += 1
if n != 1 and n not in a:
a.append(n)
return a
YN=[]
N=int(input())
for i in range(N):
P=int(input())
p=prime_factorize(P)
nums=""
num=1
if len(p)>=3:
YN.append('YES')
for j in range(2,len(p)):
num=num*p[j]
nums+= str(p[0]) + ' ' + str(p[1]) + ' ' +str(num)
YN.append(nums)
else:
YN.append('NO')
for u in YN:
print(u)
| 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
# A function to print all prime factors of
# a given number n
def primeFactors(n,l):
# Print the number of two's that divide n
while n % 2 == 0:
l.append(2)
n = n // 2
# n must be odd at this point
# so a skip of 2 ( i = i + 2) can be used
for i in range(3,int(math.sqrt(n))+1,2):
# while i divides n , print i ad divide n
while n % i== 0:
l.append(i)
n = n // i
# Condition if n is a prime
# number greater than 2
if n > 2:
l.append(n)
return l
for _ in range(int(input())):
n=int(input())
ar=[]
l=primeFactors(n,ar)
s=[]
s=set(s)
temp=1
for i in l:
temp*=i
if temp not in s and len(s)<2:
s.add(temp)
temp=1
if temp!=1:
s.add(temp)
if len(s)>2:
print("YES")
s=list(s)
for i in s:
print(i,end=" ")
else:
print("NO",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 | //Product of three numbers
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.IOException;
import java.io.PrintWriter;
import java.util.StringTokenizer;
public class ProductOfThreeNumbers
{
public static void main(String args[]) throws IOException
{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
PrintWriter out = new PrintWriter(System.out);
int t = Integer.parseInt(br.readLine());
while(t-- > 0){
int n = Integer.parseInt(br.readLine());
boolean flag = false;
for(int i = 2; i < Math.floor(n/i); ++i){
if(n/i == (double)n/i){
int[] a = factors(n/i, i);
if(a.length == 2){
out.println("YES");
out.println(i+" "+a[0]+" "+a[1]);
flag = true;
break;
}
}
}
if(!flag)
out.println("NO");
out.flush();
}
}
static int[] factors(int n, int a){
int end = (int)Math.sqrt(n);
for(int i = a+1; i <= end; ++i){
if(n%i == 0 && i != a){
if(i != n/i){
return new int[]{i, n/i};
}
}
}
return new int[]{0};
}
} | JAVA |
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