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| description
stringlengths 29
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| source
int64 1
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| difficulty
int64 0
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| solution
stringlengths 7
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| language
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1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | for i in range(int(input())):
n = int(input())
j = 2
a = []
while j*j < n:
if n % j == 0:
n //= j
a.append(j)
if len(a) == 2:
a.append(n)
break
j += 1
if len(a) < 3:
print('NO')
elif a[1] == a[2]:
print('NO')
else:
print('YES')
print(a[0],a[1],a[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 sys
import time
import math
start_time = time.time()
input = sys.stdin.readline
t = int(input())
for _ in range( t):
n = int(input())
sqrtn = math.ceil(n**0.5)
a,b,c=1,1,1
mod = 2
while(True):
if n == 1 or mod > sqrtn:
break
if n % mod == 0 :
if a == 1 :
a*= mod
else:
b*= mod
n//= mod
if(a>1 and b > 1 and a!= b):
c = n
break
else:
if mod == 2 :
mod = 3
else:
mod += 2
# print(a,b,c)
if(a==1 or b==1 or c==1 or a==b or b==c or a==c):
print("NO")
else:
print("YES")
print(a,b,c)
# print("--- %s seconds ---" % (time.time() - start_time))
| 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 | kl = int(input())
for l in range(kl):
pr = 0
n = int(input())
t = int(n ** 0.5)
a = []
for i in range(2, t + 1):
if n % i == 0:
a += [i]
n = n // i
if len(a) == 2:
pr = 1
break
if pr and a[1] < n:
print('YES')
print(a[0], a[1], n)
else:
print('NO') | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 |
import java.util.ArrayList;
import java.util.Scanner;
public class Main {
public static void main(String[] args){
int t;
Scanner sc = new Scanner(System.in);
t = sc.nextInt();
for(int i=0;i<t;i++)
{
int n = sc.nextInt();
ArrayList<Integer> ans = new ArrayList<>();
int count=0;
int cuberoot = (int)Math.pow(n, 0.33)+1;
int j=2;
int temp = n;
while(j*j<n)
{
if(temp%j==0)
{
temp = temp/j;
ans.add(j);
count++;
if(count==2)
break;
}
j++;
}
if(count!=2)
{
System.out.println("NO");
continue;
}
else
{
if(temp!=ans.get(1) && temp!=ans.get(0) && temp>2)
{
System.out.println("YES");
System.out.println(ans.get(0)+" "+ans.get(1)+" "+n/(ans.get(0)*ans.get(1)));
}
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 | t = int(input())
for _ in range(t):
n = int(input())
on = n
a = []
for i in range(2, 1+int(n**0.5)):
if not n%i:
a.append(i)
n//=i
if len(a) == 2:
break
if len(a) < 2 or on//(a[0]*a[1]) == 1 or on//(a[0]*a[1]) in a:
print("NO")
else:
print("YES")
a.append(on//(a[0]*a[1]))
print(*a) | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
inline int mulmd(long long a, long long b) {
long long ret = (a * b) % 1000000007;
return (int)ret;
}
inline int power(long long x, long long y, int m) {
long long res = 1;
x = x % m;
while (y > 0) {
if (y & 1) {
res = mulmd(res, x);
}
y = y >> 1;
x = mulmd(x, x);
}
return (int)res;
}
inline int submd(long long a, long long b) {
long long ret = (a - b);
if (ret < 0) ret += 1000000007;
return (int)ret;
}
inline int addmd(long long a, long long b) {
long long ret = (a + b) % 1000000007;
return (int)ret;
}
inline int invPow(long long a) { return power(a, 1000000007 - 2, 1000000007); }
inline int divmd(long long a, long long b) {
long long ret = mulmd(a, invPow(b));
return (int)ret;
}
const int N = 1e5 + 5;
int arr[N];
void solve() {
int n;
cin >> n;
vector<int> ans;
int r = n;
for (int i = 2; i <= 100000; i++) {
if (n % i == 0) {
n /= i;
ans.push_back(i);
break;
}
}
if (ans.size() < 1) {
cout << "NO\n";
return;
}
for (int i = 2; i <= 100000; i++) {
if (n % i == 0 && i != ans[0]) {
n /= i;
ans.push_back(i);
break;
}
}
if (ans.size() < 2) {
cout << "NO\n";
return;
}
int prod = ans[0] * ans[1];
int fin = r / prod;
if (fin != ans[0] && fin != ans[1] && fin >= 2) {
cout << "YES\n";
cout << ans[0] << " " << ans[1] << " " << fin << "\n";
} else {
cout << "NO\n";
}
}
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 |
import java.util.*;
public class C {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int t = in.nextInt();
while (t-- > 0) {
int n = in.nextInt();
int a = 1, b = 1, c = 1;
for (int i = 2; i <= Math.sqrt(n); i++) {
if (n % i == 0) {
a = i;
break;
}
}
int x = n / a;
for (int i = 2; i <= Math.sqrt(x); i++) {
if (x % i == 0 && i != a) {
b = i;
break;
}
}
c = x / b;
if (c != a && c != b && a >= 2 && b >= 2 && c >= 2) {
System.out.println("YES");
if (a > b) {
int temp = a;
a = b;
b = temp;
}
if (b > c) {
int temp = b;
b = c;
c = temp;
}
System.out.println(a + " " + b + " " + c);
}
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 | for _ in range(int(input())):
n = int(input())
f = 0
i = 2
li = []
while i*i<=n:
if n%i == 0:
f+=1
li.append(i)
n = n//i
i+=1
if f == 2:
break
if f==2 and n!=1 and n not in li:
print("YES")
print(*li, 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 sys import stdin , stdout
from os import path
if path.exists("input.txt"):
stdin = open("input.txt",'r')
for _ in range(int(stdin.readline())):
x = int(stdin.readline())
i = 2
q = []
while len(q)< 2 and i*i < x:
if x%i == 0 :
q.append(i)
x //= i
i+=1
if len(q) == 2 and x not in q :
print("YES")
print(*q,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 math
def is_prime(n):
isPrime=1
for i in range(2,math.floor(math.sqrt(n))+1):
if n%i==0:
isPrime=1
break
if isPrime==1:
return False
return True
for _ in range(int(input())):
n=int(input())
isPrime = is_prime(n)
if isPrime:
print("NO")
else:
a,b,c = 0,0,0
l=[]
k=n
for i in range(2,math.floor(math.sqrt(n))+1):
if n%i==0:
if (n//i==i):
l.append(i)
else:
l.append(i)
l.append(n//i)
l.append(k)
a=min(l)
if (a!=0):
l=[]
n=k//a
for i in range(2,math.floor(math.sqrt(n))+1):
if n%i==0:
if (n//i==i):
if i>a:
l.append(i)
else:
if i>a:
l.append(i)
if n//i>a:
l.append(n//i)
l.append(k//a)
b=min(l)
if (b!=0):
l=[]
n=k//(a*b)
for i in range(2,math.floor(math.sqrt(n)+1)):
if n%i==0:
if (n//i==i):
if i>b:
l.append(i)
else:
if i>b:
l.append(i)
if n//i>b:
l.append(n//i)
l.append(k//(a*b))
c=max(l)
if (len(set([a,b,c]))<3) or a<2 or b<2 or c<2:
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 java.util.ArrayList;
import java.util.Scanner;
public class ewtry {
public static void main(String[] args) {
Scanner sc=new Scanner (System.in);
int t=sc.nextInt();
while(t-->0)
{
int n=sc.nextInt();
ArrayList <Integer> l1=new ArrayList<Integer>();
for(int i=2;i<=Math.sqrt(n);i++)
{
if(n%i==0)
{
n=n/i;
l1.add(i);
if(l1.size()==2&&n>1&&n!=i)
{
l1.add(n);
break;
}
}
}
if(l1.size()==3)
{
System.out.println("YES");
String s="";
for(int i=0;i<3;i++)
{
s=s+l1.get(i)+" ";
}
System.out.println(s);
}
else{
System.out.println("NO");
}
}
}
}
| JAVA |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | from math import ceil
def factorize(n):
ans = []
sq = ceil(n ** 0.5)
while n > 1 and len(ans) < 2:
prime = True
for i in range(2, sq + 1):
if n % i == 0:
if i not in ans:
ans.append(i)
n //= i
prime = False
break
if prime:
if n != 1 and n not in ans:
ans.append(n)
n = 1
else:
break
if n != 1 and n not in ans:
ans.append(n)
return ans
#print(factorize(2 * 2 * 2 * 2 * 2 * 69 * 228 * 1337))
t = int(input())
for i in range(t):
n = int(input())
#print(factorize(n))
if len(factorize(n)) == 3:
print('YES')
print(*factorize(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 | #include <bits/stdc++.h>
using namespace std;
int main() {
int t;
cin >> t;
while (t--) {
int n, a, b, c, co = 0, te = 0;
cin >> n;
int x[3];
for (int j = 2; j <= sqrt(n); j++) {
if (n % j == 0) {
x[co++] = j;
n /= j;
te++;
if (te == 2) break;
}
}
if (te == 2 && n != 1 && n != x[0] && n != x[1]) {
cout << "YES" << endl;
cout << x[0] << " " << x[1] << " " << n << endl;
} else
cout << "NO" << endl;
}
}
| CPP |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
int t;
cin >> t;
while (t--) {
long long int n, x, i, y;
cin >> n;
x = n;
vector<pair<long long int, int>> v;
for (i = 2; i * i <= x; i++) {
y = 0;
if (x % i == 0) {
while (x % i == 0) {
x /= i;
y++;
}
v.push_back({i, y});
}
}
if (x > 1) v.push_back({x, 1});
if (v.size() >= 3) {
x = v[0].first;
y = v[1].first;
cout << "YES\n" << x << " " << y << " " << n / (x * y) << "\n";
} else if (v.size() == 2) {
x = v[0].first;
y = v[1].first;
long long int c0 = v[0].second, c1 = v[1].second;
if (c0 >= 3) {
cout << "YES\n" << x << " " << x * x << " " << n / (x * x * x) << "\n";
} else if (c1 >= 3) {
x = y;
cout << "YES\n" << x << " " << x * x << " " << n / (x * x * x) << "\n";
} else if (c0 >= 2 && c1 >= 2) {
cout << "YES\n" << x << " " << y << " " << x * y << "\n";
} else
cout << "NO\n";
} else if (v.size() == 1) {
x = v[0].first;
y = v[0].second;
if (y >= 6) {
cout << "YES\n" << x << " " << x * x << " " << n / (x * x * x) << "\n";
} else
cout << "NO\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 | import math
for _ in range(int(input())):
n=int(input())
r=math.ceil(math.sqrt(n))+1
ou=[]
for i in range(2,r):
if len(ou)==2:
break
else:
if n%i==0:
n=n//i
ou.append(i)
if n>1 and len(ou)==2 and n not in ou:
print('YES')
print(n,*ou)
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)
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<=m:
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 | #include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cout.tie(NULL);
cout.tie(NULL);
;
long long tt;
cin >> tt;
while (tt--) {
long long n;
cin >> n;
long long a = -1;
for (long long i = 2; i * i <= n; i++) {
if (n % i == 0) {
a = i;
n /= a;
break;
}
}
if (a != -1) {
long long l = 0;
for (long long i = 2; i * i <= n; i++) {
if (n % i == 0 && i != a && a != n / i && i != n / i) {
cout << "YES\n";
cout << a << " " << n / i << " " << i << "\n";
l = 1;
break;
}
}
if (l == 0) cout << "NO\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 | t = int(input())
for i in range(0,t):
n = int(input())
a = 1
j = 2
while((j*j) <= n):
if(n % j == 0):
a = j
break
j = j + 1
if(a == 1):
print("NO")
else:
b = 1
n = n / a
j = a + 1
while((j*j) <= n):
if(n % j == 0):
b = j
break
j = j + 1
n = int(n / b)
if(b == 1):
print("NO")
elif n <= b:
print("NO")
else:
print("YES")
print(str(a) + " " + str(b) + " " + str(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
def getFactors(n):
factors = []
while n%2==0:
factors.append(2)
n = n//2
k=3
while k <= math.sqrt(n):
if n%k==0:
n= n//k
factors.append(k)
else:
k+=2
if n>1:
factors.append(n)
return factors
t = int(input())
numbers = [ int(input()) for _ in range(t) ]
for n in numbers:
factors = getFactors(n)
distinct = set()
i=0
num = 1
while i < len(factors):
num *= factors[i]
if len(distinct)==2:
pass
elif num not in distinct:
distinct.add(num)
num=1
else:
pass
i+=1
if num!=1: distinct.add(num)
if len(distinct)==3:
print('YES')
print(" ".join(map(str,sorted(list(distinct)))))
else:
print('NO') | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | t = int(input())
for _ in range(t) :
n = int(input())
i = 2
k = 1
x = 1
y = 1
z = 1
flag = 0
while i*i<=n :
if n%i==0 :
if k==1 :
x = i
n = n//i
k = 0
elif n//i != i and x!=i and x!= n//i:
y = i
z = n//i
break
else :
flag = 1
break
i += 1
if flag == 0 and x>1 and y>1 and z>1:
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 | def solve(n):
a, b, c = None, None, None
for i in range(2, int(n**0.5) + 1):
if n%i == 0:
a = i
n = n/i
break
else:
return None
for i in range(a+1, int(n**0.5) + 1):
if n%i == 0:
b = i
c = int(n/i)
break
else:
return None
if b != c and a != c:
return (a, b, c)
else:
return None
t = int(input())
for _ in range(t):
n = int(input())
res = solve(n)
if res:
print('YES')
print (*res)
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())
f = ""
for _ in range(t):
n = int(input())
m = n
r = 0
x = []
for i in range(2, int(sqrt(n))+1):
if i>m or r==2:
break
else:
if m%i==0:
x.append(i)
m = m//i
r += 1
if r==2 and (n//(x[0]*x[1])>x[1]):
f += "YES\n"
f += str(x[0]) + " " +str(x[1]) + " " + str(n//(x[0]*x[1]))+"\n"
else:
f += "NO\n"
print(f)
| 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 d[1000001] = {0};
int sum[1000001] = {0};
int main() {
int t;
cin >> t;
while (t--) {
int n;
cin >> n;
int a = -1, b = -1, c;
for (int i = 2; i <= sqrt(n); i++) {
if (n % i == 0 && a == -1) {
a = i;
} else if (a != -1 && (n / a) % i == 0) {
b = i;
break;
}
}
c = n / (a * b);
if (c == 1 || b == -1 || c == b || c == a) a = -1;
if (a == -1)
cout << "NO\n";
else
cout << "YES"
<< "\n"
<< a << ' ' << b << ' ' << c << 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 | from functools import reduce
import io
import os
import sys
from atexit import register
import random
import math
import itertools
##################################### Flags #####################################
# DEBUG = True
DEBUG = False
STRESSTEST = False
# STRESSTEST = True
##################################### IO #####################################
if not DEBUG:
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
sys.stdout = io.BytesIO()
register(lambda: os.write(1, sys.stdout.getvalue()))
tokens = []
tokens_next = 0
def nextStr():
global tokens, tokens_next
while tokens_next >= len(tokens):
tokens = input().split()
tokens_next = 0
tokens_next += 1
if type(tokens[tokens_next - 1]) == str:
return tokens[tokens_next - 1]
return tokens[tokens_next - 1].decode()
def nextInt():
return int(nextStr())
def nextIntArr(n):
return [nextInt() for i in range(n)]
def print(*argv, end='\n'):
for arg in argv:
sys.stdout.write((str(arg) + ' ').encode())
sys.stdout.write(end.encode())
##################################### Helper Methods #####################################
def genTestCase():
return random.randint(1, 100)
def bruteforce(n):
for c in itertools.combinations(range(2, n), 3):
if reduce(lambda x, y: x * y, c) == n:
return list(c)
return None
def check(mySoln, bruteforceSoln):
if bruteforceSoln is None or mySoln == None:
return mySoln == bruteforceSoln
n = reduce(lambda x, y: x * y, bruteforceSoln)
return mySoln[0] * mySoln[1] * mySoln[2] == n and len(set(mySoln)) == 3
def doStressTest():
while True:
curTest = genTestCase()
mySoln = solve(curTest)
bruteforceSoln = bruteforce(curTest)
if not check(mySoln, bruteforceSoln):
print('### Found case ###')
print(curTest)
print(f'{mySoln} should have been: {bruteforceSoln}')
return
def genPrimes(limit):
limit = int(limit)
isPrime = [0, 0] + [1] * (limit + 10)
for i in range(len(isPrime)):
if isPrime[i]:
yield i
for j in range(i * i, len(isPrime), i):
isPrime[j] = 0
primes = [i for i in genPrimes(10**5)]
def factorize(n):
res = []
ps = iter(primes)
while n > 1:
p = next(ps, -1)
if p == -1:
res.append(n)
break
while n % p == 0:
n = n // p
res.append(p)
return res
def solve(n):
factors = factorize(n)
if len(factors) > 2:
p1 = factors[0]
p2 = next((i for i in factors if i != p1), -1)
if p2 == -1:
p2 = factors[1] * factors[2]
res = [p1, p2, n // p1 // p2]
if reduce(lambda x, y: x * y, res) == n and len(
set(res)) == 3 and 1 not in res:
return res
return None
##################################### Driver #####################################
if __name__ == "__main__":
if not DEBUG and STRESSTEST:
raise Exception('Wrong flags!')
if STRESSTEST:
doStressTest()
else:
### Read input here
t = nextInt()
for _ in range(t):
n = nextInt()
res = solve(n)
if res is None:
print('NO')
else:
print('YES')
print(' '.join(map(str, res)))
sys.stdout.flush() | 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.math.*;
import java.util.*;
// @author : Dinosparton [ SHIVAM BHUVA ]
public class test {
static class Pair{
int x;
int y;
Pair(int x,int y){
this.x = x;
this.y = y;
}
}
static class Compare {
void compare(Pair arr[], int n)
{
// Comparator to sort the pair according to second element
Arrays.sort(arr, new Comparator<Pair>() {
@Override public int compare(Pair p1, Pair p2)
{
return p1.x - p2.x;
}
});
// for (int i = 0; i < n; i++) {
// System.out.print(arr[i].x + " " + arr[i].y + " ");
// }
// System.out.println();
}
}
public static void main(String args[]) {
Scanner sc = new Scanner(System.in);
StringBuffer res = new StringBuffer();
int tc= sc.nextInt();
while(tc-->0) {
long n = sc.nextLong();
int a = 0;
int b = 0;
for(int i = 2;i*i<=n;i++) {
if(n%i==0) {
a = i;
n = n/i;
break;
}
}
for(int i=a+1;i*i<=n;i++) {
if(n%i==0) {
b = i;
n = n/i;
break;
}
}
if(a!=n && b!=n && n!=1 && a!=0 && b!=1 && b!=0)
res.append("YES").append("\n").append(a+" "+b+" "+n).append("\n");
else
res.append("NO").append("\n");
}
System.out.println(res);
}
}
| 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 | # ------------------- fast io --------------------
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")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# ------------------- fast io --------------------
import math
import bisect
primes=[2]
for j in range(3,10**5):
indy=min(bisect.bisect_left(primes,math.ceil(math.sqrt(j))),len(primes)-1)
broke=False
for s in range(indy+1):
if j%primes[s]==0:
broke=True
break
if broke==False:
primes.append(j)
minny=len(primes)
testcases=int(input())
for j in range(testcases):
#ok we will find its prime divisors
n=int(input())
indy=min(bisect.bisect_left(primes,math.ceil(math.sqrt(n))),minny-1)
facs=[]
for s in range(indy+1):
if n%primes[s]==0:
facs.append(primes[s])
exi=True
if len(facs)>=3:
a,b,c=facs[0],facs[1],n//(facs[0]*facs[1])
elif len(facs)==2:
a,b,c=facs[0],facs[1],n//(facs[0]*facs[1])
if c==0 or c==1 or c==a or c==b:
exi=False
elif len(facs)==1:
a,b,c=facs[0],facs[0]**2,n//(facs[0]**3)
if c==0 or c==1 or c==a or c==b:
exi=False
elif len(facs)==0:
a,b,c=0,0,0
exi=False
if a*b*c==n and exi==True:
print("YES")
print(str(a)+" "+str(b)+" "+str(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 | import sys
I=sys.stdin.readline
ans=""
for _ in range(int(I())):
n=int(input())
fac=[]
for i in range(2,int(n**.5)):
if n%i==0:
fac.append((i,n//i))
break
if len(fac)!=0:
x=fac[0][1]
for i in range(2,int(x**.5)+1):
if x%i==0 and i!=fac[0][0]:
if i!=x//i:
ans+="YES\n"
ans+="{} {} {}\n".format(fac[0][0],i,x//i)
break
else:
ans+="NO\n"
else:
ans+="NO\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 | #include <bits/stdc++.h>
using namespace std;
const long long MOD = 1e9 + 7;
int main() {
ios_base::sync_with_stdio(false), cin.tie(0);
;
int t;
cin >> t;
while (t--) {
int n;
cin >> n;
int curr = n;
vector<int> ans;
bool ok = true;
while (ans.size() < 2 && ok) {
int currMax = (int)sqrt(curr) + 1;
int i = 2;
if (ans.size()) {
i = ans[ans.size() - 1] + 1;
}
while (i < currMax) {
if (curr % i == 0) {
ans.push_back(i);
curr /= i;
break;
}
i++;
}
if (i >= currMax) {
ok = false;
}
}
if (ans.size() == 2 && ans[1] != n / ans[0] / ans[1]) {
cout << "YES\n"
<< ans[0] << " " << ans[1] << " " << n / ans[0] / ans[1] << "\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 | for i in range(int(input())):
n=int(input());c=0;x=[];y=n
for i in range(2,int(n**0.5)+1):
if y%i==0:
c+=1
y=y//i
x.append(i)
if c>=2:
break
print('YES' if c>=2 and y not in x else 'NO')
if c>=2 and y not in x:
print(*x[:2],y ) | 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=input()
for p in xrange(t):
n=input()
if n<20:
print "NO"
continue
i=2
k=n
fac=[]
d=0
while i*i<=n and d==0:
while k%i==0 and k!=0:
# print k
k=k/i
fac.append(i)
if len(fac)==3:
d=1
break
i+=1
# print k
if k!=0 and k!=1:
fac.append(k)
# print fac
if len(fac)<3:
print 'NO'
continue
m=1
if fac[0]!=fac[1]:
for i in xrange(2, len(fac)):
m*=fac[i]
if m>1 and m!=fac[1] and m!=fac[0]:
print 'YES'
print fac[0], fac[1], m
else:
print "NO"
else:
for i in xrange(3, len(fac)):
m*=fac[i]
if fac[1]*fac[2]!=m and m!=fac[0] and m!=1:
print 'YES'
print fac[0], fac[1]*fac[2], m
else:
print "NO" | 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 java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashSet;
public class ProductofThreeNumbers {
static boolean[] isPrime;
static ArrayList<Integer> primes;
static void sieve(int N) {
isPrime = new boolean[N + 1];
Arrays.fill(isPrime, true);
isPrime[0] = false;
isPrime[1] = false;
for (int i = 2; i * i <= N; i++) {
if (isPrime[i]) {
for (int j = i * i; j <= N; j += i) {
isPrime[j] = false;
}
}
}
for (int i = 2; i < isPrime.length; i++) {
if (isPrime[i])
primes.add(i);
}
}
static ArrayList<Integer> primeFactors(int N){ //O(sqrt(N)/log(sqrt(N))
int idx = 0;
int p = primes.get(idx++);
ArrayList<Integer> factors = new ArrayList<Integer>();
while(p*p <= N) {
while(N % p == 0) {
factors.add(p);
N /= p;
}
p = primes.get(idx++);
}
if(N != 1)
factors.add(N);
return factors;
}
public static void main(String[] args) throws NumberFormatException, IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
PrintWriter out = new PrintWriter(System.out);
int t = Integer.parseInt(br.readLine());
isPrime=new boolean[100000];
primes=new ArrayList<Integer>();
sieve(100000);
//int counter=;
while(t-->0) {
int n=Integer.parseInt(br.readLine());
//out.println(n);
ArrayList<Integer> a=primeFactors(n);
HashSet<Long> h=new HashSet();
//System.out.println(a);
if(a.size()>3) {long number=1;
h.add(a.get(0)*1l);
if(a.get(1)==a.get(0)) {
h.add(a.get(1)*a.get(2)*1l);
for(int i=3;i<a.size();i++)
number*=a.get(i);
}
else {
h.add(a.get(1)*1l);
for(int i=2;i<a.size();i++)
number*=a.get(i);
}
h.add(number);
if(h.size()==3) {
out.println("YES");
for(long l:h) {
out.print(l+" ");
}
out.println();
}
else {
out.println("NO");
}
}
else if(a.size()==3) {
h.add(a.get(0)*1l);h.add(a.get(1)*1l);h.add(a.get(2)*1l);
if(h.size()==3) {
out.println("YES");
for(long l:h) {
out.print(l+" ");
}
out.println();
}
else {
out.println("NO");
}
}
else {
out.println("NO");
}
}
out.flush();
}
}
| 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):
used = []
n = int(input())
for i in range(2,round(n/2)):
if i*i>=n: break
if n%i==0 and (i not in used):
used.append(i)
n=n/i
break
for i in range(2,round(n/2)):
if i*i>=n: break
if n%i==0 and (i not in used):
used.append(i)
n=n/i
break
if len(used) < 2 or (n in used) or n==1:
print("NO")
else:
print("YES")
used.append(int(n))
for item in used:
print(item, end=" ")
print() | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | #include <bits/stdc++.h>
#pragma GCC optimize("O2")
using namespace std;
const int INF = int(2e9) + 99;
void test_case() {
long long n;
cin >> n;
set<long long> st;
for (long long i = 2; i * i <= n; i++) {
if (n % i == 0 && st.find(i) == st.end()) {
st.insert(i);
n /= i;
break;
}
}
for (long long i = 2; i * i <= n; i++) {
if (n % i == 0 && st.find(i) == st.end()) {
st.insert(i);
n /= i;
break;
}
}
if (st.size() < 2 || n == 1 || st.find(n) != st.end()) {
cout << "NO\n";
return;
}
cout << "YES\n";
st.insert(n);
for (auto x : st) cout << x << ' ';
cout << '\n';
return;
}
int main() {
cin.tie(0);
ios_base::sync_with_stdio(false);
int t;
cin >> t;
while (t--) test_case();
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 | # maa chudaaye duniya
import math
def pf(n):
cat = []
while n%2 == 0:
cat.append(2)
n //= 2
for i in range(3, int(math.sqrt(n)) + 1, 2):
while n%i == 0:
cat.append(i)
n //= i
if n > 2:
cat.append(n)
return cat
for _ in range(int(input())):
n = int(input())
factors = pf(n)
factors.sort()
ss = list(set(factors))
# print(factors)
if len(ss) >= 3 and len(factors) >= 3:
a = ss[0]
b = ss[1]
d = a*b
c = n // d
print('YES')
print(a, b, c)
elif len(ss) == 1 and len(factors) >= 6:
a = ss[0]
b = pow(a, 2)
c = n // pow(a, 3)
print('YES')
print(a, b, c)
elif len(ss) == 2 and len(factors) >= 4:
a = ss[0]
b = ss[1]
c = n // (a*b)
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())
while t!=0:
n = int(input())
p = int(pow(n,.5))
a,b,c = -1,-1,-1
for i in range(2,p+1):
if n%i==0:
a= i
n = n//i
break
q = int(pow(n,.5))
for i in range(2,q+1):
if i!=a and n%i==0 and i!=n//i:
b=i
c=n//i
break
if a!=-1 and b!=-1 and c!=-1:
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 | for _ in range(int(input())):
n=int(input())
s=set()
for i in range(2,int(pow(n,0.5))+1):
if(n%i==0):
s.add(i)
s.add(n//i)
l1=list(s)
n1=len(l1)
if(n1<3):
print("NO")
else:
flag=0
for i in range(n1-2):
if(flag==1):
break
for j in range(i+1,n1-1):
if(flag==1):
break
for k in range(j+1,n1):
v=l1[i]*l1[j]*l1[k]
if(n==v):
flag=1
print("YES")
print(l1[i],l1[j],l1[k])
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 | //package CodeForces.RoadMap.D2_B;
import java.util.Arrays;
import java.util.Scanner;
/**
* @author Syed Ali.
* @createdAt 16/04/2021, Friday, 01:40
*/
public class ProductOfThreeNumsSolution {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int t=sc.nextInt();
int a[]=new int[3];
while(t-->0) {
Arrays.fill(a,0);
int cnt=0;
int n=sc.nextInt();
for(int i=2;i<=Math.sqrt(n);i++) {
if(n%i==0) {
a[cnt++]=i;
n/=i;
if(cnt==2) {a[cnt]=n;break;}
}
}
if(cnt==2&&a[2]>a[1]) {
System.out.println("YES\n"+a[0]+" "+a[1]+" "+a[2]);
}
else System.out.println("NO");
}
sc.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 | #include <bits/stdc++.h>
using namespace std;
int t, n;
int main() {
cin >> t;
while (t--) {
cin >> n;
int a[3] = {}, j = 0;
for (int i = 2; i * i < n && j < 2; i++) {
if (n % i == 0) a[j++] = i, n /= i;
}
if (j != 2)
cout << "NO\n";
else {
cout << "YES\n";
cout << a[0] << " " << a[1] << " " << n << endl;
}
}
}
| CPP |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | t = int(input())
for i in range(t):
n = int(input())
a=0
b=0
c=0
s=0
ss=0
nn=int(n**0.5)
for i in range(2,nn):
if n%i==0:
a=i
ss=50
break
if ss==0:
print('NO')
continue
ss=0
for i in range(a+1,nn):
if (n/a)%i==0:
b=i
ss=50
break
if ss==0:
print('NO')
continue
if n/(a*b)>b:
print('YES')
print(a,b,int(n/(a*b)))
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 solveOne(n):
ret = []
d = 2
while d * d < n and len(ret) < 3:
if n % d == 0:
ret.append(d)
n //= d
dd = 2
while dd * dd < n:
if n % dd == 0 and dd not in ret and n // dd not in ret:
ret.extend((dd, n // dd))
break
dd += 1
d += 1
return 'YES\n%s' % ' '.join(map(str, ret)) if len(ret) == 3 else 'NO'
for _ in range(int(input())):
print(solveOne(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 | from sys import stdin, stdout
from collections import defaultdict
import math
rl = lambda: stdin.readline()
rll = lambda: stdin.readline().split()
def main():
cases = rll()
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")
stdout.close()
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 | t = int(input())
for _ in range(t):
n = int(input())
a, b, c = 0, 0, 0
i = 1
while i < n and i < 10 ** 4:
i += 1
if n % i == 0:
n = n // i
if a == 0:
a = i
elif b == 0:
b = i
c = n
break
if c > 2 and b != c and c != a:
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;
long long gcd(long long a, long long b);
long long nCr(long long n, long long r);
long long pow(long long b, long long n);
void read(long long a[], long long n);
void solve() {
long long n;
cin >> n;
vector<long long> v;
for (long long i = 2; i * i <= n; i++) {
if (n % i == 0) {
v.push_back(i);
n /= i;
if (v.size() == 2) {
v.push_back(n);
break;
}
}
}
if (v.size() != 3) {
cout << "NO" << endl;
return;
}
if (v[0] != v[1] && v[0] != v[1] && v[1] != v[2]) {
cout << "YES" << endl;
cout << v[0] << " " << v[1] << " " << v[2] << endl;
} else {
cout << "NO" << endl;
}
}
signed main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
long long t;
cin >> t;
while (t--) {
solve();
}
}
long long gcd(long long a, long long b) { return (b == 0) ? a : gcd(b, a % b); }
long long nCr(long long n, long long r) {
if (r > n) return 0;
if (r * 2 > n) r = n - r;
if (r == 0) return 1;
long long ans = n;
for (long long i = 2; i <= r; i++) {
ans *= (n - i + 1);
ans /= i;
}
return ans;
}
long long pow(long long b, long long n) {
long long p = b, ans = 1;
while (n) {
if (n & 1) {
ans *= p;
}
p *= p;
n = n >> 1;
}
return ans;
}
void read(long long a[], long long n) {
for (long long i = 0; i < n; i++) {
cin >> a[i];
}
}
| CPP |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import java.io.*;
import java.util.*;
//import javafx.util.*;
import java.math.*;
//import java.lang.*;
public class Main
{
//static int n;
// static ArrayList<Integer> adj[];
// static boolean vis[];
// static long ans[];
static int arr[];
static long mod=1000000007;
static final long oo=(long)1e18;
// static int n;
public static void main(String[] args) throws IOException {
// Scanner sc=new Scanner(System.in);
PrintWriter pw = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
br = new BufferedReader(new InputStreamReader(System.in));
int test=nextInt();
// int test=1;
outer: while(test--!=0){
long n=nextLong();
long ans[]=new long[3];
// int j=0;
for(long i=2;i<=Math.sqrt(n);i++){
if(n%i==0){
long b=n/i;
// if(isPrime(b))
// pw.println("NO");
// else{
for(long j=i+1;j<=Math.sqrt(b);j++){
if(b%j==0&&j!=b/j){
pw.println("YES");
pw.println(i+" "+j+" "+b/j);
continue outer;
}
}
pw.println("NO");
continue outer;
//}
}
}
pw.println("NO");
// pw.println(ans[0]+" "+ans[1]+" "+ans[2]);
// if(j==3){
// pw.println("YES");
// }
// else{
// }
}
pw.close();
}
static long ncr(long n,long r){
if(r==0)
return 1;
long val=ncr(n-1,r-1);
val=(n*val)%mod;
val=(val*modInverse(r,mod))%mod;
return val;
}
static int find(ArrayList<Integer> a,int i){
if(a.size()==0||i<0)return 0;
ArrayList<Integer> l=new ArrayList<Integer>();
ArrayList<Integer> r=new ArrayList<Integer>();
for(int v:a){
if((v&(1<<i))!=0)l.add(v);
else r.add(v);
}
if(l.size()==0)return find(r,i-1);
if(r.size()==0)return find(l,i-1);
return Math.min(find(l,i-1),find(r,i-1))+(1<<i);
}
public static BufferedReader br;
public static StringTokenizer st;
public static String next() {
while (st == null || !st.hasMoreTokens()) {
try {
st = new StringTokenizer(br.readLine());
} catch (Exception e) {
throw new RuntimeException(e);
}
}
return st.nextToken();
}
public static Integer nextInt() {
return Integer.parseInt(next());
}
public static Long nextLong() {
return Long.parseLong(next());
}
public static Double nextDouble() {
return Double.parseDouble(next());
}
// static class Pair{
// int x;int y;
// Pair(int x,int y,int z){
// this.x=x;
// this.y=y;
// // this.z=z;
// // this.z=z;
// // this.i=i;
// }
// }
// static class sorting implements Comparator<Pair>{
// public int compare(Pair a,Pair b){
// //return (a.y)-(b.y);
// if(a.y==b.y){
// return -1*(a.z-b.z);
// }
// return (a.y-b.y);
// }
// }
public static int[] na(int n)throws IOException{
int[] a = new int[n];
for(int i = 0;i < n;i++)a[i] = nextInt();
return a;
}
static class query implements Comparable<query>{
int l,r,idx,block;
static int len;
query(int l,int r,int i){
this.l=l;
this.r=r;
this.idx=i;
this.block=l/len;
}
public int compareTo(query a){
return block==a.block?r-a.r:block-a.block;
}
}
static class Pair implements Comparable<Pair>{
int x;int y;
Pair(int x,int y){
this.x=x;
this.y=y;
//this.z=z;
}
public int compareTo(Pair p){
return (x-p.x);
//return (x-a.x)>0?1:-1;
}
}
// static class sorting implements Comparator<Pair>{
// public int compare(Pair a1,Pair a2){
// if(o1.a==o2.a)
// return (o1.b>o2.b)?1:-1;
// else if(o1.a>o2.a)
// return 1;
// else
// return -1;
// }
// }
static boolean isPrime(int n) {
if (n <= 1)
return false;
if (n <= 3)
return true;
if (n % 2 == 0 ||
n % 3 == 0)
return false;
for (int i = 5;
i * i <= n; i = i + 6)
if (n % i == 0 ||
n % (i + 2) == 0)
return false;
return true;
}
static long gcd(long a, long b) {
if (b == 0)
return a;
return gcd(b, a % b);
}
// To compute x^y under modulo m
static long power(long x, long y, long m){
if (y == 0)
return 1;
long p = power(x, y / 2, m) % m;
p = (p * p) % m;
if (y % 2 == 0)
return p;
else
return (x * p) % m;
}
static long fast_pow(long base,long n,long M){
if(n==0)
return 1;
if(n==1)
return base;
long halfn=fast_pow(base,n/2,M);
if(n%2==0)
return ( halfn * halfn ) % M;
else
return ( ( ( halfn * halfn ) % M ) * base ) % M;
}
static long modInverse(long n,long M){
return fast_pow(n,M-2,M);
}
} | 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 | # cook your dish here
import math
t=int(input())
for _ in range(t):
count=0
a=[]
n=int(input())
for i in range(2,int(math.sqrt(n))):
if (n%i==0 and count<2):
n=int(n/i)
a.append(i)
count+=1
if n not in a and n!=1:
a.append(n)
count+=1
if count == 3:
print('YES')
for i in a:
print(i,end=" ")
print("")
else:
print("NO") | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | def prime_decomposition(n):
i = 2
table = []
while i * i <= n:
while n % i == 0:
n /= i
table.append(i)
i += 1
if n > 1:
table.append(n)
return table
t = int(input())
for _ in range(t):
n = int(input())
ansl = []
pl = prime_decomposition(n)
k = 1
for p in pl:
k *= p
if k not in ansl and len(ansl) < 2:
ansl.append(int(k))
k = 1
else:
if k in ansl or k == 1:
print("NO")
continue
ansl.append(int(k))
if len(ansl) < 3:
print("NO")
else:
print("YES")
print(str(ansl[0])+ " " + str(ansl[1]) + " " + str(ansl[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 math
t=int(input())
for i in range(t):
n=int(input())
s=set()
for j in range(2,int(math.sqrt(n))+1):
if n%j==0 and not list(s).count(j):
s.add(j)
n/=j
break
for k in range(2,int(math.sqrt(n))+1):
if n%k==0 and not list(s).count(k):
s.add(k)
n/=k
break
if len(s)<2 or list(s).count(n) or n==1:
print("NO")
else:
print("YES")
s.add(int(n))
print(*list(s),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 | for i in range(0,int(input())):
n=int(input())
f1=0
f2=0
root=int(n**0.5)
for i in range(2,root+1):
if n%i==0:
k=i
f1=1
break
if f1==1:
n=int(n/k)
root=int(n**0.5)
k1=k
for i in range(2,root+1):
if n%i==0 and i*i!=n and i!=k1:
k=i
f2=1
break
if f2==1:
print("YES")
print(k1,k,int(n/k))
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 | def smallestDivisor(start,n):
# if divisible by 2
if (n % 2 == 0 and start == 2):
return 2;
# iterate from 3 to sqrt(n)
i = start;
while(i * i <= n):
if (n % i == 0):
return i;
i += 1;
return n;
t = int(input())
for i in range(t):
n = int(input())
a = smallestDivisor(2, n)
b = smallestDivisor(a+1, n/a)
c = n//(a*b)
if c!=1 and c!=a and c!=b:
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 | import math
def factors(n):
l = []
for i in range(2, int(math.sqrt(n))+1):
if n%i == 0:
l.append(i)
l.append(n//i)
return l
for _ in range(int(input())):
n = int(input())
l = factors(n)
nn = len(l)
flag = 0
for i in range(nn):
for j in range(nn):
for k in range(nn):
if l[i] * l[j] * l[k] == n:
if not (i == j or j == k or k == i):
print("YES")
print(l[i], l[j], l[k])
flag = 1
break
if flag == 1:
break
if flag == 1:
break
if flag == 0:
print("NO")
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import sys
import math
t = int(sys.stdin.readline())
for _ in range(t):
n = int(sys.stdin.readline())
a = None
for i in range(2, int(math.sqrt(n))):
if n % i == 0:
a = i
break
if a == None:
print("NO")
continue
b = None
for j in range(2, int(math.sqrt(n))):
if (n//i) % j == 0 and j != a:
b = j
break
if b == None:
print("NO")
continue
if n // (a*b) != a and n // (a*b) != b and n // (a*b) != 1:
c = n // (a*b)
else:
print("NO")
continue
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
def printDivisors(n) :
i = 1
l=[]
while i <= math.sqrt(n):
if (n % i == 0) :
if (n / i == i) :
l.append(i)
else :
l.append(i)
l.append(n//i)
i = i + 1
return l
t=int(input())
for _ in range(t):
n=int(input())
a=printDivisors(n)
a.sort()
#print(a)
z=-1
d={}
d[a[0]]=1
l=[]
c=0
for i in range(1,len(a)):
for j in range(i+1,len(a)):
if n%(a[i]*a[j])==0:
if (n//(a[i]*a[j])) in d and (a[i]>=2 and a[j]>=2 and (n//(a[i]*a[j]))>=2):
print("YES")
z=0
l.append(a[i])
l.append(a[j])
l.append((n//(a[i]*a[j])))
l.sort()
print(*l)
if z==0:
c=1
break
if c==1:
break
d[a[i]]=1
if z==-1:
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
queries = int(sys.stdin.readline())
def numberToFactors(number,fromTop=False):
factor1 = number
factor2 = 2
if fromTop:
while factor2<=factor1**0.5:
factor2+=1
while factor2>2:
if factor1%factor2==0:
if factor1//factor2!=factor2:
return [factor1//factor2,factor2]
factor2-=1
else:
factor2 = 2
while factor2<=factor1**0.5:
if factor1%factor2==0:
if factor1//factor2!=factor2:
return [factor1//factor2,factor2]
factor2+=1
return [factor1]
for query in range(queries):
number = int(sys.stdin.readline())
factors = numberToFactors(number)
if len(factors)==2:
factorsLeft = numberToFactors(factors[0],True)
factorsRight = numberToFactors(factors[1],True)
next = True
if len(factorsLeft)==2:
if factorsLeft[0]!=factors[1] and factorsLeft[1]!=factors[1]:
del factors[0]
factors = factorsLeft+factors
next = False
if len(factorsRight)==2 and next:
if factorsRight[0]!=factors[0] and factorsRight[1]!=factors[0]:
del factors[1]
factors = factors+factorsRight
if len(factors) == 3:
print('YES')
print(" ".join(map(str,factors)))
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())
while t > 0:
res = []
n = int(input())
root = int(n**.5)
while root > 1:
if n % root == 0:
rem = n/root
rem_root = int(rem**.5)
if rem == rem_root**2:
root -= 1
continue
else:
while rem_root > 1:
if rem % rem_root == 0:
if rem_root == root or (rem/rem_root) == root:
rem_root -= 1
continue
else:
res = [root, int(rem_root), int(rem/rem_root)]
print(f'YES\n{root} {int(rem_root)} {int(rem/rem_root)}')
break
rem_root -= 1
if len(res) == 3:
break
root -= 1
t -= 1
if len(res) != 3:
print('NO') | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
vector<pair<long long, long long>> arr;
vector<pair<long long, long long>> arr2;
vector<long long> dels;
signed main() {
ios_base::sync_with_stdio(false);
cin.tie(nullptr);
cout.tie(nullptr);
long long q;
cin >> q;
while (q--) {
long long n;
cin >> n;
long long v = -1;
for (long long i = 2; i * i < n; i++) {
if (n % i == 0) {
for (long long j = 2; j * j <= i; j++) {
if (i % j == 0 && i / j != 1 && i / j != n / i && n / i != j &&
j != i / j) {
cout << "YES\n" << i / j << ' ' << n / i << ' ' << j << endl;
goto r;
}
}
for (long long j = 2; j * j <= n / i; j++) {
if ((n / i) % j == 0 && (n / i) / j != 1 && (n / i) / j != i &&
i != j && j != (n / i) / j) {
cout << "YES\n" << (n / i) / j << ' ' << i << ' ' << j << endl;
goto r;
}
}
}
}
cout << "NO\n";
r:;
}
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 _ in range(int(input())):
num=int(input())
c=2
p=[]
while len(p)<2 and c*c<=num:
if num%c == 0:
num=num//c
p.append(c)
c+=1
if len(p)==2 and num not in p:
print("YES")
print(*p,num)
else:
print("NO") | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 |
import java.awt.Desktop;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.net.URI;
import java.net.URISyntaxException;
import java.sql.Array;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Comparator;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Iterator;
import java.util.LinkedHashMap;
import java.util.LinkedHashSet;
import java.util.LinkedList;
import java.util.List;
import java.util.Map;
import java.util.PriorityQueue;
import java.util.Queue;
import java.util.Scanner;
import java.util.Set;
import java.util.Stack;
import java.util.StringTokenizer;
import java.util.TreeSet;
import java.util.Vector;
public class codechef3 {
static class comp implements Comparator<Integer>
{
@Override
public int compare(Integer o1, Integer o2) {
if(Math.abs(o1)>Math.abs(o2))
return -1;
else return 1;
}
}
//=======================================================
//sorting Pair
static class comp1 implements Comparator<Pair<Integer,Integer>>
{
@Override
public int compare(Pair<Integer, Integer> o1, Pair<Integer, Integer> o2) {
if(o1.k>o2.k)
return 1;
else return -1;
}
}
//=======================================================
//Creating Pair class
//----------------------------------------------------------------------
static class Pair<Integer,Intetger>
{
int k=0;
int v=0;
public Pair(int a,int b)
{
k=a;
v=b;
}
}
//--------------------------------------------------------------------------
static class FastReader
{BufferedReader br;
StringTokenizer st;
public FastReader()
{ 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());
}
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;
}
}
//gcd of two number
public static int gcd(int a, int b)
{
if (b == 0)
return a;
return gcd(b, a % b);
}
//--------------------------------------------------------------------------------------------
//lcm of two number
static int x;
public static int lcm(int a, int b)
{
return (a / gcd(a, b)) * b;
}
//-------------------------------------------------------------------------------------------
public static void main(String[] args)
{
FastReader s=new FastReader();
int t=s.nextInt();
while(t-->0)
{
int n=s.nextInt();
int x=1;
int y=1;
for(int i=2;i*i<=n;i++)
{
if(n%i==0)
{
x=i;
break;
}
}
for(int i=3;i*i<=n;i++)
{
if(n%(x*i)==0&&x!=i&&x!=1)
{y=i;
break;}
}
int flag=0;
//System.out.println(x+" "+y);
if(n/(x*y)!=x&&n/(x*y)!=y&&x!=1&&y!=1)
{
System.out.println("YES");
System.out.println(x+" "+y+" "+n/(x*y));
}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 | def primfacs(n):
i = 2
primfac = []
while i * i <= n:
while n % i == 0:
primfac.append(i)
n = n / i
i = i + 1
if n > 1:
primfac.append(n)
return primfac
def mul(arr):
p = 1
if not len(arr):
return 0
for i in arr:
p = p * i
return p
t = int(input())
for _ in range(t):
n = int(input())
dels = primfacs(n)
result = [dels[0]]
tmp = []
for i in range(1, len(dels)):
tmp.append(dels[i])
p = mul(tmp)
if not p or p in result:
continue
else:
if len(result) < 2:
result.append(p)
tmp = []
else:
continue
p = mul(tmp)
if not p or p in result:
print("NO")
else:
result.append(p)
print("YES")
print(' '.join(str(int(i)) for i in result))
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import math
from collections import Counter
t = int(input())
for _ in range(t):
n = int(input())
divisors = Counter()
for p in range(2, int(math.sqrt(n))+1):
while n % p == 0:
divisors[p] += 1
n //= p
if n > 1:
divisors[n] += 1
ans = [1, 1, 1, 1]
i = 1
for k, v in divisors.items():
for j in range(v):
ans[i] *= k
if all(ans[i] != ans[j] for j in range(i)) and i < 3:
i += 1
if len(set(ans)) == 4:
print('YES')
print(*ans[1:])
else:
print('NO')
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 |
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.StringTokenizer;
public class pre372
{
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
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());
}
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;
}
}
public static void main(String args[])
{
FastReader obj = new FastReader();
int tc = obj.nextInt();
while(tc--!=0)
{
int n = obj.nextInt();
boolean flag = true;
ArrayList<Integer> arr = new ArrayList<>();
for(int i=2;i<=Math.sqrt(n);i++)
{
if(n%i==0 && !arr.contains(i))
{
n /=i;
arr.add(i);
break;
}
}
for(int i=2;i<=Math.sqrt(n);i++)
{
if(n%i==0 && !arr.contains(i))
{
n /=i;
arr.add(i);
break;
}
}
if(arr.size()<2 || arr.contains(n))System.out.println("NO");
else
{
System.out.println("YES");
for(int i=0;i<2;i++)
System.out.print(arr.get(i)+" ");
System.out.print(n);
System.out.println();
}
}
}
}
| JAVA |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import math
def ld(a, last):
found = False
for div in range(last,int(math.sqrt(a))+1):
if a % div == 0:
found = True
return div
if not found:
return False
def do(int):
x = ld(int,2)
if x:
y = ld(int//x, x+1)
if y:
z = int//(x*y)
if x != y and x != z and y != z:
print("Yes")
print(x,y,z)
else:
print("No")
else: print("No")
else:
print("No")
t = int(input())
a = []
for inp in range(0,t):
inp = int(input())
a.append(inp)
for element in a:
do(element)
| 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 = []
i = 2
while i < math.sqrt(n) and len(l) < 2:
if (n % i == 0):
l.append(i)
n = n // i
i = i + 1
if len(l) == 2 and n > l[1]:
print("YES")
print(*l, 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 sys import stdin, stdout
fin = lambda: stdin.readline()
fout = lambda *args: stdout.write(' '.join(str(i) for i in args) + '\n')
def f(n, s=2):
a, b = -1, -1
for i in range(s, 22361):
if n % i == 0:
if a == -1:
a = i
n //= i
elif i < n // i:
return a, i, n // i
return -1
for _ in range(int(fin())):
n = int(fin())
a = f(n)
if a == -1:
fout('NO')
else:
fout('YES')
fout(a[0], a[1], a[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 | for _ in range(int(input())):
n = int(input())
fac = []
rem = 1
if n % 2 == 0:
n /= 2
fac.append(2)
tot = 1
tot *= 2
added = 2
x = 1
while n > 1 and n % 2 == 0:
x *= 2
n /= 2
tot *= 2
if fac[-1] < x:
fac.append(x)
# app = True
added *= x
x = 1
if len(fac) >= 3:
rem *= n
n = 1
break
rem *= tot / added
i = 3
while n > 1:
if n % i == 0:
n /= i
fac.append(i)
tot = 1
tot *= i
added = i
x = 1
while n > 1 and n % i == 0:
x *= i
n /= i
tot *= i
if fac[-1] < x:
fac.append(x)
# app = True
added *= x
x = 1
if len(fac) >= 3:
break
rem *= tot / added
if len(fac) >= 3:
rem *= n
n = 1
break
if i * i >= n:
if n > 1 and n not in fac:
fac.append(n)
else:
rem *= n
n = 1
break
i += 2
# print(fac, rem)
if len(fac) >= 3:
a = fac[0]
b = fac[1]
c = fac[2]
for x in fac[3:]:
c *= x
c *= rem
print("YES")
print(int(a), int(b), int(c))
elif len(fac) >= 2 and rem > 1 and rem not in fac:
# print()
a = fac[0]
b = fac[1]
c = rem
print("YES")
print(int(a), int(b), int(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())
for q in range(t):
n=int(input())
lists=[]
i=2
while len(lists)<2 and i*i<n:
if n%i==0:
lists.append(i)
n//=i
i+=1
if len(lists)==2 and n not in lists:
print('YES')
print(*lists,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 | #include <bits/stdc++.h>
using namespace std;
template <typename T>
std::ostream &operator<<(std::ostream &out, vector<T> &v) {
for (typename vector<T>::size_type i = 0; i < v.size(); ++i)
out << v[i] << " ";
out << "\n";
return out;
}
template <typename T>
std::ostream &operator<<(std::ostream &out, vector<pair<T, T> > &v) {
for (size_t i = 0; i < v.size(); ++i)
out << "(" << v[i].first << ", " << v[i].second << ") ";
out << "\n";
return out;
}
template <typename T>
std::ostream &operator<<(std::ostream &out, vector<vector<T> > &v) {
for (size_t i = 0; i < v.size(); ++i) {
for (size_t j = 0; j < v[i].size(); ++j) {
out << v[i][j] << " ";
}
out << "\n";
}
return out;
}
int main() {
std::ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
int t;
cin >> t;
while (t--) {
int n;
cin >> n;
int a = 1;
for (int i = 2; i * i <= n; ++i) {
if (n % i == 0) {
a = i;
break;
}
}
if (a == 1) {
cout << "NO\n";
continue;
}
int b = 1;
for (int i = a + 1; i * i <= n / a; ++i) {
if ((n / a) % i == 0) {
b = i;
break;
}
}
if (b == 1) {
cout << "NO\n";
continue;
}
int c = n / (a * b);
if (c && c != 1 && c != a && c != b) {
cout << "YES\n";
cout << a << " " << b << " " << c << "\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 | t = int(input())
for _ in range(t):
n = int(input())
arr = [1] * 3
i = 0
p = 2
while n > 1:
if n % p != 0:
p += 1
if p >= int(n**0.5) + 1:
break
continue
arr[i] *= p
n //= p
if i > 0 and arr[1] == arr[0]:
continue
i += 1
if i == 2:
break
if i < 2 or n == 1 or n == arr[0] or n == arr[1]:
print('NO')
else:
print('YES')
print(f'{arr[0]} {arr[1]} {n}')
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import math,sys
from collections import Counter, defaultdict, deque
from sys import stdin, stdout
input = stdin.readline
lili=lambda:list(map(int,sys.stdin.readlines()))
li = lambda:list(map(int,input().split()))
#for deque append(),pop(),appendleft(),popleft(),count()
I=lambda:int(input())
S=lambda:input().strip()
mod = 1000000007
# A Better (than Naive) Solution to find all divisiors
import math
# method to print the divisors
def printDivisors(n) :
a=[]
i = 1
while i <= math.sqrt(n):
if (n % i == 0) :
# If divisors are equal, print only one
if (n / i == i) :
a.append(i)
else :
# Otherwise print both
a.append(i)
a.append(n//i)
i = i + 1
return (a)
for i in range(I()):
n=I()
a=printDivisors(n)
d=Counter(a)
for i in a:
f=0
if(i==1 or i==n):
continue
p=n//i
#print(p)
j=1
while(j<=math.sqrt(p)):
if(p%j==0 and j!=1 and j in d and (p//j) in d and j!=i and i!=(p//j) and j!=(p//j) and j!=p):
print("YES")
print(i,j,p//j)
f=1
break
j+=1
if(f==1):
break
if(f==0):
print("NO")
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | from functools import reduce
def factors(n):
l = []
for i in range(2,int(n**0.5)+1):
if n%i==0:
l.append(i)
l.append(n//i)
return l
for _ in range(int(input())):
n = int(input())
l = factors(n)
flag = False
flag1 = False
flag2 = False
for i in range(len(l)):
a = l[i]
b = n//a
if a!=b:
flag = True
break
if flag:
# print(a,b,'****')
p = factors(b)
l1 = p
for i in range(len(l1)):
c = l1[i]
d = b//c
if c!=d and d!=a and c!=a:
flag1 = True
# print(a,c,d,'%%%%%%%%%%%5')
break
if flag1:
print('YES')
print(a,c,d)
else:
p = factors(a)
# p.remove(1)
# s.remove(a)
l1 = list(p)
for i in range(len(l1)):
c = l1[i]
d = a // c
if d!=c and b!=d and c!=b:
flag2 = True
break
if flag2:
print('YES')
print(c,d,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 java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.*;
public class PrimeFactorization {
private static List<Integer> primes = new ArrayList();
// Prepoulates the primes array with all prime numbers uptil N.
private static void prepopulatePrimes(int N) {
BitSet isPrime = new BitSet();
isPrime.set( 0, false);
isPrime.set( 1, false );
isPrime.set( 2, N, true );
for (int i=2; i*i<N; i++ ) {
if (isPrime.get( i )) {
for( int j=i*i; j<N; j+=i )
{
isPrime.clear( j );
}
}
}
for (int i=0; i<N; i++) {
if (isPrime.get(i)) {
// System.out.println(i);
primes.add(i);
}
}
// System.out.println(count);
}
// Uses the prepopulated primes array to factorize N.
public static List<Long> primeFactors(long N) {
List<Long> factors = new ArrayList<>();
if (N==0) {
factors.add(0L);
return factors;
}
if (N==1) {
factors.add(1L);
return factors;
}
for (long prime : primes) {
while( N%prime == 0 ) {
factors.add(prime);
N = N/prime;
}
}
if (N!=1) {
factors.add(N);
}
for (long x : factors) {
// System.out.print(x + " ");
}
// System.out.println();
return factors;
}
public static void main( String[] args ) throws Exception {
int MAX = 100000 + 1;
prepopulatePrimes(MAX);
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int T = Integer.parseInt(br.readLine());
for (int t=0; t<T; t++) {
long X = Long.parseLong(br.readLine());
List<Long> factors = primeFactors(X);
if (factors.size() <= 2) {
System.out.println("NO");
} else {
long a = factors.get(0);
long b_part = 1;
int index = 1;
boolean fail = false;
while ((b_part == 1) || (b_part == a)) {
if (index >= factors.size()) {
fail = true;
break;
}
b_part = b_part * factors.get(index);
index++;
}
long c_part = 1;
while (true) {
if (index >= factors.size()) {
break;
}
c_part = c_part * factors.get(index);
index++;
}
if (fail || (c_part == a) || (c_part == b_part) || (c_part == 1)) {
System.out.println("NO");
} else {
System.out.println("YES");
System.out.println(a + " " + b_part + " " + c_part);
}
}
}
}
} | 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;
using ll = long long int;
void solve() {
int n;
cin >> n;
if (n < 24) {
cout << "NO\n";
return;
}
set<int> div;
for (int i = 2; i * i <= n; ++i) {
if (n % i == 0 and !div.count(i)) {
div.insert(i);
n /= i;
break;
}
}
for (int i = 2; i * i <= n; ++i) {
if (n % i == 0 and !div.count(i)) {
div.insert(i);
n /= i;
break;
}
}
if (int(div.size()) < 2 || div.count(n)) {
cout << "NO" << endl;
} else {
cout << "YES" << endl;
div.insert(n);
for (auto it : div) cout << it << " ";
cout << endl;
}
}
int main() {
ios_base::sync_with_stdio(false);
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 | #include <bits/stdc++.h>
using namespace std;
const long long inf = 1e18;
const long long mod = 1e9 + 7;
const long long N = 2e5;
signed main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
long long t;
cin >> t;
;
while (t--) {
long long n;
cin >> n;
;
long long cpy = n;
map<long long, long long> mp;
long long sqr = sqrt(n + 1);
for (long long i = (2); i <= (sqr); i++) {
while (n % i == 0) {
n /= i;
mp[i]++;
}
}
if (n > 1) mp[n]++;
if (mp.size() >= 3) {
cout << "YES\n";
long long x = (*mp.begin()).first, y = (*(++mp.begin())).first;
cout << x << " " << y << " " << cpy / (x * y) << '\n';
} else if (mp.size() == 1) {
long long x = (*mp.begin()).first, y = (*mp.begin()).second;
if (y < 6)
cout << "NO\n";
else {
cout << "YES\n";
cout << x << " " << x * x << " " << cpy / (x * x * x) << '\n';
}
} else {
long long x = (*mp.begin()).first, y = (*(++mp.begin())).first,
z = cpy / (x * y);
if (z == x || z == y || z == 1)
cout << "NO\n";
else {
cout << "YES\n";
cout << x << " " << y << " " << z << '\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 |
t = int(input())
def prime_decomposition(n):
table = []
for i in range(2, int(n ** 0.5) + 1):
while n % i == 0:
table += [i]
n //= i
if n == 1:
break
if n != 1:
table += [n]
return table
for i in range(t):
n = int(input())
s = prime_decomposition(n)
s.sort()
flg = 0
if len(s) < 3:
print ("NO")
continue
#print (s)
#if len(set(s)) == 3:
# print ("YES")
# print (" ".join(map(str,list(s))))
# continue
for j in range(1,len(s)):
for k in range(j+1,len(s)):
z = s[0]
x = 1
y = 1
for a in s[j:k]:
x *= a
for b in s[k:]:
y *= b
if x != y and y != z and x != z:
print ("YES")
print (z,x,y)
flg = 1
break
if flg == 1:
break
if flg == 0:
print ("NO")
#print (s) | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import math
t = int(input())
for _ in range(t):
n = int(input())
factors = []
for factor in range (2, int(math.sqrt(n))):
if n <= factor:
break
if n % factor == 0:
if n // factor <= factor:
break
factors.append(factor)
n = n // factor
if len(factors) == 2:
break
if len(factors) == 2:
factors.append(n)
print('YES')
print(' '.join(map(str, factors)))
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())
while(t):
l=set()
tr=0
n=int(input())
ce=n
if n>=12:
i=2
while(n>0 and tr<3 ):
if tr<2 and i*i>ce:
break
if tr==2:
l.add(int(n))
tr=tr+1
break
elif n%i==0:
l.add(i)
tr=tr+1
n=n/i
elif tr==2:
l.add(int(n))
tr=tr+1
break
i=i+1
if tr==3 and len(l)==3:
print("YES")
l=list(l)
l.sort()
for i in range (3):
print(l[i],end=" ")
print()
else:
print("NO")
t=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.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Scanner;
/**
* Built using CHelper plug-in
* Actual solution is at the top
*/
public class Main {
public static void main(String[] args) {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
Scanner in = new Scanner(inputStream);
PrintWriter out = new PrintWriter(outputStream);
TaskC solver = new TaskC();
solver.solve(1, in, out);
out.close();
}
static class TaskC {
public void solve(int testNumber, Scanner in, PrintWriter out) {
int q = in.nextInt();
while (q > 0) {
int n = in.nextInt();
int a = -1;
int b = -1;
int c = -1;
boolean flag = false;
for (int i = 2; i <= Math.sqrt(n); i++) {
int temp = n;
if (temp % i == 0) {
a = i;
temp /= i;
boolean f = false;
for (int j = i + 1; j <= Math.sqrt(temp); j++) {
if (temp % j == 0 && temp / j != j) {
b = j;
c = (n / b) / a;
flag = true;
f = true;
break;
}
}
if (f) break;
}
}
if (flag) {
out.println("YES");
out.println(a + " " + b + " " + c);
} else {
out.println("NO");
}
q--;
}
}
}
}
| 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 | for _ in range(int(input())):
n=int(input())
l=[]
i=2
while i**2<=n and len(l)<2:
if n%i==0:
l.append(i)
n//=i
i+=1
if len(l)<2 or n==1 or l.count(n):
print("NO")
else:
print("YES")
l.append(n)
print(*l)
| 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 s(a,b):
i = b+1
c = int(a**0.5)
while 1:
if a%i==0:
return i
else:
if i<c:
i+=1
else:
return 1
for _ in range(int(input())):
n = int(input())
a = s(n,1)
#print(a)
if a==1 or a==n:
print('NO')
else:
a1 = n//a
a2 = s(a1,a)
if a2==1 or a2==a1:
print('NO')
else:
a3 = a1//a2
b = sorted([a,a2,a3])
if len(set(b))==3:
print('YES')
print(b[0],b[1],b[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 | // package Div3_615;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.HashMap;
public class ProblemC {
public static void main(String[] args)throws IOException {
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
int test=Integer.parseInt(br.readLine());
int seive[]=new int[100001];
ArrayList<Integer> primes=new ArrayList<>();
for(int i=2;i<=100000;i++){
if(seive[i]==0){
primes.add(i);
for(int j=2*i;j<=100000;j+=i){
seive[j]=1;
}
}
}
StringBuilder print=new StringBuilder();
// System.out.println(primes.size());
while(test--!=0){
int n=Integer.parseInt(br.readLine());
HashMap<Integer,Integer> power=new HashMap<>();
int temp=n;
for(int i=0;i<primes.size();i++){
int prime=primes.get(i);
int count=0;
while(temp%prime==0){
count++;
temp/=prime;
}
if(count!=0){
power.put(prime,count);
}
if(temp==1){
break;
}
}
if(temp!=1){
power.put(temp,1);
}
// System.out.println(power.toString());
if(power.size()==1){
for(int k:power.keySet()){
if(power.get(k)<6){
print.append("NO\n");
}
else{
print.append("YES\n");
int a=(int)Math.pow(k,1);
int b=(int)Math.pow(k,2);
int c=(int)Math.pow(k,power.get(k)-3);
print.append(a+" "+b+" "+c+"\n");
}
}
}
else if(power.size()==2){
int pow1=0,pow2=0;
int prime1=0,prime2=0;
for(int k:power.keySet()){
if(pow1==0){
pow1=power.get(k);
prime1=k;
continue;
}
else{
pow2=power.get(k);
prime2=k;
}
}
if(pow1>=3||pow2>=3){
print.append("YES\n");
if(pow1>=3){
int a=(int)Math.pow(prime1,1);
int b=(int)Math.pow(prime1,pow1-1);
int c=(int)Math.pow(prime2,pow2);
print.append(a+" "+b+" "+c+"\n");
}
else{
int a=(int)Math.pow(prime2,1);
int b=(int)Math.pow(prime2,pow2-1);
int c=(int)Math.pow(prime1,pow1);
print.append(a+" "+b+" "+c+"\n");
}
}
else if(pow1==2&&pow2==2){
print.append("YES\n");
int a=(int)Math.pow(prime1,1);
int b=(int)Math.pow(prime2,1);
int c=a*b;
print.append(a+" "+b+" "+c+"\n");
}
else{
print.append("NO\n");
}
}
else{
int a=0,b=0,c=1;
for(int k:power.keySet()){
if(a==0){
a=(int)Math.pow(k,power.get(k));
continue;
}
if(b==0){
b=(int)Math.pow(k,power.get(k));
continue;
}
int t=(int)Math.pow(k,power.get(k));
c*=t;
}
print.append("YES\n");
print.append(a+" "+b+" "+c+"\n");
}
}
System.out.println(print.toString());
}
}
| 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 SieveOfEratosthenes(n):
data=[]
prime = [True for i in range(n + 1)]
p = 2
while (p * p <= n):
if (prime[p] == True):
for i in range(p * 2, n + 1, p):
prime[i] = False
p += 1
prime[0]= False
prime[1]= False
for p in range(n + 1):
if prime[p]:
data.append(p)
return data
# print(len(SieveOfEratosthenes(10**6)))
if __name__=="__main__":
l = SieveOfEratosthenes(10**6)
for t in range(int(input())):
data=[]
n = int(input())
d=n
for i in l :
if i*i>n:
break
if n%i==0:
while n%i==0:
data.append(i)
n = n/i
if n>1:
data.append(n)
a,b,c=data[0],0,0
if len(data)<=2:
print("NO")
else:
b = data[1]
if a==data[1]:
b = data[1]*data[2]
c = d//(a*b)
if c==1 or c==b or c==a:
print("NO")
else:
print("YES")
print(a,b,c)
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | /* package whatever; // don't place package name! */
import java.util.*;
import java.lang.*;
import java.io.*;
/* Name of the class has to be "Main" only if the class is public. */
public class Ideone
{
public static void main (String[] args)
{
// your code goes here
Scanner s=new Scanner(System.in);
int t=s.nextInt();
while(t-->0)
{
int n=s.nextInt();
int jo=0;
for(int i=2;i<=Math.sqrt(n);i++)
{
if(n%i==0)
{
int r=n/i;
for(int j=2;j<=Math.sqrt(r);j++)
{
if(r%j==0&&j!=i&&r/j!=i&&j!=r/j)
{
System.out.println("YES");
System.out.println(i+" "+j+" "+r/j);
jo=1;
break;
}
}
if(jo==1)
break;
}
}
if(jo==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 java.io.*;
import java.util.*;
public class Main
{
public static void main(String[] args) throws Exception
{
Scanner sc = new Scanner(System.in) ;
PrintWriter out = new PrintWriter(System.out) ;
int t = sc.nextInt() ;
while(t-->0)
{
int [] factors = new int [32] ;
int id = 0 ;
int n = sc.nextInt() ;
for(int i = 2 ; 1l * i * i <= n ; i++)
while(n % i == 0)
{
n /= i ;
factors[id++] = i ;
}
if(n > 1)
factors[id++] = n;
id = 0 ;
int a = factors[id++] ;
int b = factors[id++] ;
if(a == b)
b *= factors[id++] ;
int c = factors[id++] ;
if(c == 0 || b == 0 || a == 0)
out.println("NO");
else
{
while(id < 32 && factors[id] != 0) c *= factors[id++];
if(a != b && a != c && b != c)
out.printf("YES \n%d %d %d\n" , a , b , c);
else
out.println("NO");
}
}
out.flush();
out.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 | from collections import Counter
def primes(n):
primfac = []
d = 2
while d*d <= n:
while (n % d) == 0:
primfac.append(d)
n //= d
d += 1
if n > 1:
primfac.append(n)
return primfac
for _ in range(int(input())):
n = int(input())
a = primes(n)
if(len(a) < 3):
print("NO")
else:
c = Counter(a)
y = list(c.keys())
if(len(c) >= 3):
print("YES")
x = 1
for i in a[2:]:
x *= i
print(y[0], y[1], n//(y[0]*y[1]))
else:
# print(c, list(c.keys())[0])
if(len(c) == 1):
if(c[y[0]] > 5):
print("YES")
print(a[0], a[0]*a[1], n//(a[0]*a[0]*a[0]))
else:
print("NO")
else:
if(c[y[0]] + c[y[1]] >= 4):
print("YES")
print(y[0], y[1], n//(y[0]* y[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 sys
input = sys.stdin.readline
from collections import *
def factorize(n):
fct = []
for i in range(2, int(n**0.5)+1):
c = 0
while n%i==0:
n //= i
c += 1
if c>0:
fct.append((i, c))
if n>1:
fct.append((n, 1))
return fct
t = int(input())
for _ in range(t):
n = int(input())
factors = factorize(n)
if len(factors)==1:
p, c = factors[0]
if c>=6:
print('YES')
print(p, p**2, p**(c-3))
else:
print('NO')
elif len(factors)==2:
p1, c1 = factors[0]
p2, c2 = factors[1]
if c1+c2>=4:
print('YES')
print(p1, p2, n//p1//p2)
else:
print('NO')
else:
p1, _ = factors[0]
p2, _ = factors[1]
print('YES')
print(p1, p2, n//p1//p2) | 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
# method to print the divisors
def printDivisors(n) :
# Note that this loop runs till square root
i = 2
l=[]
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
def find3(A, arr_size, s):
A.sort()
p=0
for i in range(0, arr_size-2):
l = i + 1
r = arr_size-1
while (l < r):
if( A[i] * A[l] * A[r] == s):
print("YES")
print(A[i],A[l],A[r])
p=1
break
elif (A[i] * A[l] * A[r] < s):
l += 1
else:
r -= 1
if p==1:
break
if p==0:
print("NO")
t=int(input())
while t:
t-=1
n=int(input())
l=printDivisors(n)
find3(l,len(l),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 | for tc in range(int(input())):
n=int(input())
a=[]
for i in range(2,int(n**0.5)+1):
if n%i==0:
a.append(i)
if n//i!=i:
a.append(n//i)
a.sort()
if len(a)<3:
print("NO")
else:
flag=0
for i in range(len(a)):
for j in range(i+1,len(a)):
k=n//(a[i]*a[j])
if k in a and k!=a[i] and k!=a[j]:
print("YES")
print(k,a[i],a[j])
flag=1
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 | def solution():
temp = [True]*31623
for i in range(2, 31623):
if temp[i]:
for j in range(i+i, 31623, i):
temp[j] = False
er = [i for i, v in enumerate(temp) if v][2:]
for _ in range(int(input())):
n = int(input())
ans = set()
er_iter = er.__iter__()
current = er_iter.__next__()
l = 1
while current < n:
if n%current == 0:
if current in ans:
if l == 1:
l = current
else:
ans.add(current*l)
l = 1
else:
ans.add(current)
n //= current
if len(ans) == 2 and n not in ans: break
else:
try:
current = er_iter.__next__()
except StopIteration:
break
if l:
n *= l
if len(ans) == 2 and n not in ans:
print("YES")
ans.add(n)
print(*ans)
else:
print("NO")
solution()
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | from sys import stdin
import math
t=int(stdin.readline().rstrip())
while t>0:
n=int(stdin.readline().rstrip())
p=int(math.floor(math.sqrt(n)))
l=[]
a,b=-1,-1
for i in range(2,p+1):
if n%i==0:
a=i
n//=a
break
for i in range(2,p+1):
if n%i==0 and i!=a:
b=i
n//=b
break
f=0
if n==1 or n==a or n==b:
f=1
if a==-1 or b==-1:
f=1
if f==0:
print("YES")
print(str(a)+" "+str(b)+" "+str(n))
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 | #include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
srand(time(NULL));
;
int q;
cin >> q;
while (q--) {
long long n;
cin >> n;
vector<long long> ans;
for (long long i = 2; i <= sqrt(n); i++) {
if (n % i == 0) {
ans.push_back(i);
n = n / i;
}
if (n == 1) break;
if (ans.size() == 2) break;
}
if (ans.size() < 2 || n == ans[0] || n == 1 || n == ans[1])
cout << "NO" << endl;
else {
cout << "YES" << endl << ans[0] << " " << ans[1] << " " << n << endl;
}
}
}
| CPP |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | # -*- coding: utf-8 -*-
"""
Created on Sun Feb 2 11:27:04 2020
@author: pc
"""
'''
times = int(input())
for _ in range(times):
testified_number = int(input())
able = 0
for factor in range(2,testified_number):
if testified_number % factor == 0:
testified_number_new = testified_number // factor
for factor_new in range(factor + 1,testified_number_new):
if testified_number_new % factor_new == 0 and testified_number_new // factor_new > factor_new:
able = 1
b = factor_new
c = testified_number_new // factor_new
break
if able == 1:
a = factor
break
if able:
print("YES")
print(a,b,c)
else:
print('NO')
'''
'''
times = int(input())
testified_numbers = [0] * times
for i in range(times):
testified_numbers[i] = int(input())
M = max(testified_numbers)
judge = [False] * 2 + [True] * (M-1)
for i in range(M+1):
if judge[i]:
for j in range(2*i,M+1,i):
judge[j]=False
for i in range(times):
able = 0
number = testified_numbers[i]
factors = {}
if not judge[number]:
for factor in range(2,int(M**(1/3))+1):
if judge[factor] and number % factor == 0:
while number % factor == 0:
number = number // factor
if factor in factors:
factors[factor] += 1
else:
factors[factor] = 1
if len(factors) == 3:
able = 1
a = list(factors.keys())[0]
b = list(factors.keys())[1]
c = testified_numbers[i] // (a*b)
break
elif len(factors) == 2 and list(factors.values()) != [1,1] and list(factors.values()) != [1,2] and list(factors.values()) != [2,1]:
able = 1
a = list(factors.keys())[0]
b = list(factors.keys())[1]
c = testified_numbers[i] // (a*b)
break
elif len(factors) == 1 and list(factors.values())[0] >= 6:
able = 1
a = list(factors.keys())[0]
b = list(factors.keys())[0]**2
c = testified_numbers[i] // (a*b)
break
if able:
print("YES")
print(a,b,c)
else:
print('NO')
'''
for _ in range(int(input())):
n = int(input())
a = []
for i in range(2, int(n**0.5)+1):
if n % i == 0:
a.append(i)
n = n//i
if len(a) == 2:
break
if len(a) == 2 and n > a[1]:
print('YES')
print(a[0],a[1],n)
else:
print('NO')
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | from bisect import *
from collections import *
from itertools import *
import functools
import sys
import math
from decimal import *
from copy import *
from heapq import *
from fractions import *
getcontext().prec = 30
MAX = sys.maxsize
MAXN = 10**5+10
MOD = 10**9+7
spf = [i for i in range(MAXN)]
spf[0]=spf[1] = -1
def sieve():
for i in range(2,MAXN,2):
spf[i] = 2
for i in range(3,int(MAXN**0.5)+1):
if spf[i]==i:
for j in range(i*i,MAXN,i):
if spf[j]==j:
spf[j]=i
def fib(n,m):
if n == 0:
return [0, 1]
else:
a, b = fib(n // 2)
c = ((a%m) * ((b%m) * 2 - (a%m)))%m
d = ((a%m) * (a%m))%m + ((b)%m * (b)%m)%m
if n % 2 == 0:
return [c, d]
else:
return [d, c + d]
def charIN(x= ' '):
return(sys.stdin.readline().strip().split(x))
def arrIN(x = ' '):
return list(map(int,sys.stdin.readline().strip().split(x)))
def ncr(n,r):
num=den=1
for i in range(r):
num = (num*(n-i))%MOD
den = (den*(i+1))%MOD
return (num*(pow(den,MOD-2,MOD)))%MOD
def flush():
return sys.stdout.flush()
'''*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*'''
for _ in range(int(input())):
n = int(input())
ans = set()
i = 2
while i*i<=n:
if len(ans)==2 and n not in ans:
ans.add(n)
break
if n%i==0:
if i not in ans:
ans.add(i)
n//=i
i = 2
else:
i+=1
else:
i+=1
if len(ans)==3:
print('YES')
print(*list(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 | # cook your dish here
from sys import stdin, stdout
import math
from itertools import permutations, combinations
from collections import defaultdict
from bisect import bisect_left
from bisect import bisect_right
def L():
return list(map(int, stdin.readline().split()))
def In():
return map(int, stdin.readline().split())
def I():
return int(stdin.readline())
P = 1000000007
def main():
for t in range(I()):
n = I()
lis = []
for i in range(2, int(math.sqrt(n))):
if n%i == 0:
lis.append(i)
n //= i
if len(lis) == 2:
lis.append(n)
break
lis = list(set(lis))
lis.sort()
if len(lis) == 3:
print('YES')
print(*lis)
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 |
# Author : raj1307 - Raj Singh
# Date : 13.02.2020
from __future__ import division, print_function
import os,sys
from io import BytesIO, IOBase
if sys.version_info[0] < 3:
from __builtin__ import xrange as range
from future_builtins import ascii, filter, hex, map, oct, zip
def ii(): return int(input())
def si(): return input()
def mi(): return map(int,input().strip().split(" "))
def msi(): return map(str,input().strip().split(" "))
def li(): return list(mi())
def dmain():
sys.setrecursionlimit(100000000)
threading.stack_size(40960000)
thread = threading.Thread(target=main)
thread.start()
from collections import deque, Counter, OrderedDict,defaultdict
#from heapq import nsmallest, nlargest, heapify,heappop ,heappush, heapreplace
from math import ceil,floor,log,sqrt,factorial,pi
#from bisect import bisect,bisect_left,bisect_right,insort,insort_left,insort_right
#from decimal import *,threading
#from itertools import permutations
#Copy 2D list m = [x[:] for x in mark] .. Avoid Using Deepcopy
abc='abcdefghijklmnopqrstuvwxyz'
abd={'a': 0, 'b': 1, 'c': 2, 'd': 3, 'e': 4, 'f': 5, 'g': 6, 'h': 7, 'i': 8, 'j': 9, 'k': 10, 'l': 11, 'm': 12, 'n': 13, 'o': 14, 'p': 15, 'q': 16, 'r': 17, 's': 18, 't': 19, 'u': 20, 'v': 21, 'w': 22, 'x': 23, 'y': 24, 'z': 25}
mod=1000000007
#mod=998244353
inf = float("inf")
vow=['a','e','i','o','u']
dx,dy=[-1,1,0,0],[0,0,1,-1]
def getKey(item): return item[0]
def sort2(l):return sorted(l, key=getKey)
def d2(n,m,num):return [[num for x in range(m)] for y in range(n)]
def isPowerOfTwo (x): return (x and (not(x & (x - 1))) )
def decimalToBinary(n): return bin(n).replace("0b","")
def ntl(n):return [int(i) for i in str(n)]
def powerMod(x,y,p):
res = 1
x %= p
while y > 0:
if y&1:
res = (res*x)%p
y = y>>1
x = (x*x)%p
return res
def gcd(x, y):
while y:
x, y = y, x % y
return x
def isPrime(n) : # Check Prime Number or not
if (n <= 1) : return False
if (n <= 3) : return True
if (n % 2 == 0 or n % 3 == 0) : return False
i = 5
while(i * i <= n) :
if (n % i == 0 or n % (i + 2) == 0) :
return False
i = i + 6
return True
def read():
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
def main():
for _ in range(ii()):
n=ii()
f=0
for i in range(2,int(sqrt(n))+1):
if n%i==0:
x=n//i
for j in range(2,int(sqrt(x))+1):
if x%j==0 and j!=i:
if x//j>=2 and x//j!=j and x//j!=i:
print('YES')
print(i,j,x//j)
f=1
break
if f==1:
break
if f==0:
print('NO')
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
def print(*args, **kwargs):
"""Prints the values to a stream, or to sys.stdout by default."""
sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
at_start = True
for x in args:
if not at_start:
file.write(sep)
file.write(str(x))
at_start = False
file.write(kwargs.pop("end", "\n"))
if kwargs.pop("flush", False):
file.flush()
if sys.version_info[0] < 3:
sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# endregion
if __name__ == "__main__":
#read()
main()
#dmain()
# Comment Read()
| 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())
#if n < 24:
# print('NO')
a = 0
b = 0
c = 0
sqrt = int(n ** 0.5)
for i in range(2, sqrt + 1):
if n % i == 0:
a = i
n /= a
break
if a:
for i in range(a +1, int(n ** 0.5) + 1):
if n % i == 0:
b = i
c = int(n // i)
break
if a and b and c:
if a != b != c:
print('YES')
print(a, b, c)
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
def pd(n,not_include) :
li = []
for i in range(2, int(math.sqrt(n) + 1)) :
if (n % i == 0) :
if (n / i == i) :
if n!=not_include:
lis=[]
lis.append(i)
lis.append(i)
#li.append(lis)
else :
lis=[]
if i!=not_include and n//i!=not_include:
lis.append(i)
lis.append(n//i)
li.append(lis)
return li
t=(int)(input())
for wirfb in range(t):
n=(int)(input())
li=pd(n,1)
f=0
if len(li)>=1:
for x in li:
lis=pd(x[1],x[0])
if len(lis)>=1:
f=1
break;
else:
f=0
else:
f=0
if f==0:
print("NO")
else:
print("YES")
print(x[0],lis[0][0],lis[0][1]) | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | """
// Author : snape_here - Susanta Mukherjee
"""
from __future__ import division, print_function
import os,sys
from io import BytesIO, IOBase
if sys.version_info[0] < 3:
from __builtin__ import xrange as range
from future_builtins import ascii, filter, hex, map, oct, zip
def ii(): return int(input())
def si(): return input()
def mi(): return map(int,input().split())
def li(): return list(mi())
def gcd(x, y):
while y:
x, y = y, x % y
return x
def read():
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
import math
# method to print the divisors
def printDivisors(n) :
l=[]
# 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
#This code is contributed by Nikita Tiwari.
def main():
for _ in range(ii()):
n=ii()
b=e=f=0
a=printDivisors(n)
f=0
for i in range(1,len(a)):
for j in range(i+1,len(a)):
x,y=a[i],a[j]
z=n/(x*y)
if math.ceil(z)==math.floor(z) and z!=1 and z!=x and z!=y:
f=1
print("YES")
print(x,y,int(z))
break
if f:
break
if f==0:
print("NO")
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
def print(*args, **kwargs):
"""Prints the values to a stream, or to sys.stdout by default."""
sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
at_start = True
for x in args:
if not at_start:
file.write(sep)
file.write(str(x))
at_start = False
file.write(kwargs.pop("end", "\n"))
if kwargs.pop("flush", False):
file.flush()
if sys.version_info[0] < 3:
sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# endregion
if __name__ == "__main__":
#read()
main()
#Comment read() | 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 | def Fenjie(n):
k = {}
if (n == 1):
return {}
a = 2
while (n >= 2):
b = n%a
if (a*a > n):
if (n in k):
k[n] += 1
else:
k[n] = 1
break
if (b == 0):
if (a in k):
k[a] += 1
else:
k[a] = 1
n = n//a
else:
a += 1
return k
n = int(input())
for _ in range(n):
m = int(input())
k = Fenjie(m)
s = 0
le = len(k)
l = [i for i in k]
if (le >= 3):
print("YES")
flag = 0
print(l[0], l[1], m//l[0]//l[1])
elif (le==2):
if (k[l[0]]+k[l[1]] >= 4):
print("YES")
print(l[0], l[1], m//l[0]//l[1])
else:
print("NO")
else:
if (k[l[0]] >= 6):
print("YES")
print(l[0], l[0]*l[0], m//l[0]//l[0]//l[0])
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 fact(n):
i=2
for i in range(2,int(n**0.5)+1):
if n%i==0 and i not in l:
return i
return 0
for _ in range(int(input())):
l=[]
n=int(input())
a=fact(n)
l.append(a)
if a!=0:
b=fact(n/a)
else:
print("NO")
continue
if b!=0:
z=(n//(a*b))
else:
print("NO")
continue
if z!=a and z!=b and z!=1:
print("YES")
print(a,b,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())
flag=0
for i in range(2,1001):
if(n%i==0):
a=i
n//=i
flag=1
break
#print(n)
if(flag==0):
print("NO")
else:
i=a+1
flag1=0
while(i*i<=n):
if(n%i==0):
if(i!=n//i):
print("YES")
print(a,i,n//i)
flag1=1
break
i+=1
if(flag1==0):
print("NO") | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import math
t = int(input())
while t > 0:
n = int(input())
fact = []
for i in range(2, int(math.sqrt(n))+1):
if i*int(n/i) == n:
if i != int(n/i):
fact.append(i)
fact.append(int(n/i))
else:
fact.append(i)
fact.sort()
flag = 0
for i in range(len(fact)-1):
for j in range(i+1, len(fact)):
if fact[i]*fact[j] < n and (n % (fact[i]*fact[j]) == 0):
num = int(n/(fact[i]*fact[j]))
if num != fact[i] and num != fact[j]:
flag = 1
break
if flag == 1:
break
if flag == 0:
print('NO')
else:
print('YES')
print(fact[i], fact[j], num)
t -= 1
| PYTHON3 |
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