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| description
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| source
int64 1
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| difficulty
int64 0
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| solution
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1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | def smallestDivisor(n, a):
if not n & 1 and a != 2:
return 2
i = 3
while i*i <= n:
if n % i == 0 and a != i:
return i
i += 1
return n
t = int(input())
for _ in range(t):
n = int(input())
a = smallestDivisor(n, 1)
b = smallestDivisor(n//a, a)
c = n//(a*b)
if a < b and b < c:
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;
const long long N = 1e5 + 10;
vector<long long> ans;
pair<long long, long long> a[N];
long long n, T;
void f(long long n) {
ans.clear();
long long x = n;
for (long long i = 2; i * i <= x; i++) {
while (x % i == 0) {
ans.push_back(i);
x = x / i;
}
}
ans.push_back(x);
if (x == n) {
cout << "NO" << endl;
} else {
if (ans.size() >= 3) {
long long k = 2;
long long x1 = ans[0];
long long x2 = ans[1];
if (x2 == x1) x2 = ans[1] * ans[2], k = 3;
long long x3 = 1;
for (long long i = k; i < ans.size(); i++) {
x3 *= ans[i];
}
if (x2 != x3 && x1 != x3 && x3 != 1) {
cout << "YES" << endl;
cout << x1 << " " << x2 << " " << x3 << endl;
} else {
cout << "NO" << endl;
}
} else {
cout << "NO" << endl;
}
}
}
int main() {
cin >> T;
while (T--) {
cin >> n;
f(n);
}
return 0;
}
| CPP |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import java.io.*;
import java.util.*;
public class Solution{
static FastReader in = new FastReader();
//static BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
public static void main(String[] args) throws Exception{
int t = in.nextInt();
while(t-- > 0){
int n = in.nextInt();
int i = 2;
int a = -1;
int b = -1;
while(i*i<=n){
if(n%i==0){
a = i;
b = getB(n/i,a);
if(b!=-1)
break;
}
i++;
}
if(a!=-1 && b!=-1)
System.out.println("YES\n"+a+" "+b+" "+(n/(a*b)));
else
System.out.println("NO");
}
in.close();
}
private static int getB(int n,int a){
int i = 2;
while(i*i<n){
if(i==a){i++;continue;}
if(n%i==0 && (n/i)!=a)return i;
i++;
}
return -1;
}
static class FastReader {
final private int BUFFER_SIZE = 1 << 16;
private DataInputStream din;
private byte[] buffer;
private int bufferPointer, bytesRead;
public FastReader()
{
din = new DataInputStream(System.in);
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public FastReader(String file_name) throws IOException
{
din = new DataInputStream(new FileInputStream(file_name));
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public String readLine() throws IOException
{
byte[] buf = new byte[64]; // line length
int cnt = 0, c;
while ((c = read()) != -1)
{
if (c == '\n')
break;
buf[cnt++] = (byte) c;
}
return new String(buf, 0, cnt);
}
public int nextInt() throws IOException
{
int ret = 0;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do
{
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public long nextLong() throws IOException
{
long ret = 0;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
}
while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public double nextDouble() throws IOException
{
double ret = 0, div = 1;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
}
while ((c = read()) >= '0' && c <= '9');
if (c == '.')
{
while ((c = read()) >= '0' && c <= '9')
{
ret += (c - '0') / (div *= 10);
}
}
if (neg)
return -ret;
return ret;
}
private void fillBuffer() throws IOException
{
bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
if (bytesRead == -1)
buffer[0] = -1;
}
private byte read() throws IOException
{
if (bufferPointer == bytesRead)
fillBuffer();
return buffer[bufferPointer++];
}
public void close() throws IOException
{
if (din == null)
return;
din.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 | t = int(input())
for _ in range(t):
n = int(input())
l1 = []
i = 2
while len(l1)<2 and i*i <n:
if n%i == 0:
l1.append(i)
n = n//i
i+=1
if len(l1)==2:
print('YES')
for i in l1:
print (i,end = ' ')
print(n)
else:
print("NO")
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | def find_div(x, j):
for i in range(2, int(x**.5)+1):
if x % i == 0 and i != j and x//i !=j and i != x//i:
return True, i, x//i
return False, -1, -1
t = int(input())
for _ in range(t):
n = int(input())
for i in range(2, int(n**.5)+1):
if n % i == 0:
ok, a, b = find_div(n//i, i)
if ok:
print('YES')
print(i, a, b)
break
else:
print('NO')
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | t=int(input())
while(t):
t-=1
n=int(input())
f=set()
for i in range(2,int(pow(n,0.5))+1):
if(n%i==0):
f.add(i)
f.add(n//i)
flag=0
for i in f:
x=n//i
for j in f:
if(x%j==0):
z=[i,x//j,j]
p=i*j*(x//j)
if(len(set(z))==3 and p==n and i!=1 and x!=1 and x//j!=1):
flag=1
break
if(flag==1):
break
if(flag==0):
print("NO")
else:
print("YES")
print(*z)
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 |
import java.util.Scanner;
public class ProductofThreeNumbers {
public static int findFirstDivisor(int n, int t) {
for (int i = 2; i <= Math.sqrt(n); i++) {
if (n % i == 0 && n / i != t && i != t && n != i && n / i != i) {
return i;
}
}
return 0;
}
public static void main(String args[]) {
Scanner scan = new Scanner(System.in);
int testCase = scan.nextInt();
StringBuffer sb = new StringBuffer();
while (testCase-- > 0) {
int n = scan.nextInt();
int firstSmallestDivisor = 0;
int firstBigestDivisor = 0;
boolean ch = false;
for (int i = 2; i <= Math.sqrt(n); i++) {
if (n % i == 0) {
firstSmallestDivisor = i;
firstBigestDivisor = n / i;
break;
}
}
if (firstSmallestDivisor == 0) {
sb.append("NO\n");
continue;
} else {
int f = findFirstDivisor(firstSmallestDivisor, firstBigestDivisor);
int l = findFirstDivisor(firstBigestDivisor, firstSmallestDivisor);
if (f == 0 && l == 0) {
sb.append("NO\n");
continue;
} else {
if (f > 0) {
if (f != firstBigestDivisor && firstSmallestDivisor / f > 1 && firstSmallestDivisor / f != firstBigestDivisor) {
sb.append("YES\n");
sb.append(f + " " + firstSmallestDivisor / f + " " + firstBigestDivisor + "\n");
continue;
}
}
if (l > 0 && !ch) {
if (l != firstSmallestDivisor && firstBigestDivisor / l > 1 && firstBigestDivisor / l != firstSmallestDivisor) {
sb.append("YES\n");
sb.append(l + " " + firstSmallestDivisor + " " + firstBigestDivisor / l + "\n");
continue;
}
}
sb.append("NO\n");
}
}
}
System.out.println(sb);
}
}
| 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
from collections import Counter
def primeFactors(n):
# Print the number of two's that divide n
l = []
while n % 2 == 0:
l.append(2)
n = n / 2
# n must be odd at this point
# so a skip of 2 ( i = i + 2) can be used
for i in range(3, int(math.sqrt(n)) + 1, 2):
# while i divides n , print i ad divide n
while n % i == 0:
l.append(int(i))
n = n / i
# Condition if n is a prime
# number greater than 2
if n > 2:
l.append(int(n))
return l, Counter(l)
if __name__ == '__main__':
for _ in range(int(input().strip())):
n = int(input().strip())
ll, pf = primeFactors(n)
ll.sort()
if len(pf) >= 3:
print("YES")
a = ll[0]
b = ll[-1]
c = 1
for i in range(1, len(ll) - 1):
c *= ll[i]
print(int(a), int(b), int(c))
elif len(pf) == 2:
if len(ll) >= 4:
print("YES")
a = ll[0]
b = ll[-1]
c = 1
for i in range(1, len(ll) - 1):
c *= ll[i]
print(int(a), int(b), int(c))
else:
print("NO")
else:
if len(ll) >= 6:
print("YES")
tot = ll[0] ** len(ll)
print(ll[0], ll[0] ** 2, (tot // (ll[0] * (ll[0] * ll[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 | for _ in range(int(input())):
n = int(input())
ok = False
for i in range(2, int(n**0.5)+1):
if n % i ==0:
p = n//i
if p == i:
continue
for j in range(i+1, int(p**0.5)+1):
if p%j == 0:
if j*j!=p and p//j != i:
print('YES')
print(i,j,p//j)
ok = True
break
if ok:
break
if not ok:
print('NO') | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import math
for _ in range(int(input())):
n=int(input())
l=[]
for i in range(2,int(math.sqrt(n))+1):
if(n%i==0):
t=n
l.append(i)
t=t//i
for j in range(2,int(math.sqrt(t)+1)):
if(t%j==0):
l.append(j)
l.append(t//j)
if(len(set(l))==3):
break
if(len(set(l))==3):
break
else:
l=[]
if(len(l)==3):
print("YES")
print(l[0],l[1],l[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 whatever //do not write package name here */
import java.util.*;
public class GFG {
public static void main (String[] args) {
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
while(t-- > 0){
long num = sc.nextLong();
long n = num;
int count = 0;
long[] arr = new long[3];
for(long i = 2;i * i <= num;i++){
if(n % i == 0){
n /= i;
arr[count] = i;
count++;
}
if(count == 2){
arr[count] = n;
break;
}
}
if(count < 2 || arr[0] == arr[2] || arr[1] == arr[2]){
System.out.println("NO");
}else{
System.out.println("YES");
System.out.println(arr[0] + " " + arr[1] + " " + arr[2]);
}
}
}
} | JAVA |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | from sys import stdin, stdout
cin = stdin.readline
cout = stdout.write
mp = lambda:list(map(int, cin().split()))
t, = mp()
for _ in range(t):
n, = mp()
if n < 24:
cout('NO\n')
continue
y = int(n**.5)
l = []
for i in range(2, y+1):
if not n%i:
l += [i, n//i]
if len(l)<4:
cout('NO\n')
continue
l.sort()
#print(l)
for i in range(len(l)):
if not n%(l[i]*l[i+1]) and l[i]*l[i+1] < n and (n//(l[i]*l[i+1]) != l[i] and n//(l[i]*l[i+1]) != l[i+1]):
cout('YES\n')
cout(str(l[i]) + ' ' + str(l[i+1]) + ' ' + str(n//(l[i]*l[i+1])) + '\n')
break
elif l[i]*l[i+1] >= n:
cout('NO\n')
break
#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 | import java.util.*;
import java.io.*;
public class Solution{
public static void main(String[] args) throws Exception{
BufferedReader buffer = new BufferedReader(new InputStreamReader(System.in));
int t = Integer.parseInt(buffer.readLine());
while(t-- > 0){
long num = Long.parseLong(buffer.readLine());
int a=-1,b= -1, c=-1;
boolean check = false;
for(int i=2; i*i<=num; i++){
if(num % i == 0){
a = i;
int n = (int)num/a;
for(int j=2; j*j<=n && j!= a && n/j != a; j++){
if(n % j == 0){
if(j != n/j){
b = j;
c = n/j;
check = true;
break;
}
}
}
if(check)
break;
}
}
if(check){
System.out.println("YES");
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())
a=[]
f=n
i=2
while len(a)<2 and i*i<=n:
if n%i==0:
a.append(i)
n=n//i
i+=1
if len(a)<2 or n in a:
print('NO')
else:
print('YES')
print(*a,n)
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import java.io.BufferedReader;
import java.io.FileNotFoundException;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.math.BigInteger;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Comparator;
import java.util.HashSet;
import java.util.Scanner;
import java.util.StringTokenizer;
public class Main {
static int MOD = (int) (1e9 +7);
public static class Point implements Comparable<Point>
{
public static final double EPS = 1e-9;
public double x, y;
public Point(double a, double b) { x = a; y = b; }
public int compareTo(Point p)
{
if(Math.abs(x - p.x) > EPS) return x > p.x ? 1 : -1;
if(Math.abs(y - p.y) > EPS) return y > p.y ? 1 : -1;
return 0;
}
}
public static void main(String[] args) throws NumberFormatException, IOException
{
//BufferedReader br = new BufferedReader(new FileReader("input.txt"));
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st ;
Scanner sc = new Scanner(System.in);
//PrintWriter out = new PrintWriter("output.txt");
PrintWriter out = new PrintWriter(System.out);
int test = sc.nextInt();
sieve(1000000);
while(test-->0)
{
int n = sc.nextInt();
ArrayList<Integer> fact = new ArrayList<>();
fact = primeFactors(n);
if(fact.size() >= 3)
{
int a = fact.get(0);
int b = fact.get(1);
int i = 2;
while(b == a && i<fact.size())
{
b*=fact.get(i);
i++;
}
int c = n/(b*a);
if(a == b || a == c || c == b || c == 1)
out.println("NO");
else
{
out.println("YES");
out.println(a+" " +b+" "+c);
}
}
else
out.println("NO");
}
out.flush();
out.close();
}
static ArrayList<Integer> primes;
static int[] isComposite;
/*
* 1. Generating a list of prime factors of N
*/
static ArrayList<Integer> primeFactors(int N) // O(sqrt(N) / ln sqrt(N))
{
ArrayList<Integer> factors = new ArrayList<Integer>(); //take abs(N) in case of -ve integers
int idx = 0, p = primes.get(idx);
while(p * p <= N)
{
while(N % p == 0) { factors.add(p); N /= p; }
p = primes.get(++idx);
}
if(N != 1) // last prime factor may be > sqrt(N)
factors.add(N); // for integers whose largest prime factor has a power of 1
return factors;
}
static void sieve(int N) // O(N log log N)
{
isComposite = new int[N+1];
isComposite[0] = isComposite[1] = 1; // 0 indicates a prime number
primes = new ArrayList<Integer>();
for (int i = 2; i <= N; ++i) //can loop till i*i <= N if primes array is not needed O(N log log sqrt(N))
if (isComposite[i] == 0) //can loop in 2 and odd integers for slightly better performance
{
primes.add(i);
if(1l * i * i <= N)
for (int j = i * i; j <= N; j += i) // j = i * 2 will not affect performance too much, may alter in modified sieve
isComposite[j] = 1;
}
}
/*
* 2. Primality Test
*
* Preprocessing: call sieve with sqrt(N), O(sqrt(N) log log sqrt(N))
* Query: best case O(1), worst case O(sqrt(N) / log sqrt(N))
*/
static boolean isPrime(int N)
{
if(N < isComposite.length)
return isComposite[N] == 0;
for(int p: primes) //may stop if p * p > N
if(N%p==0)
return false;
return true;
}
/*
* 3. Sieve of Eratostheses in linear time
*/
static void sieveLinear(int N)
{
ArrayList<Integer> primes = new ArrayList<Integer>();
int[] lp = new int[N + 1]; //lp[i] = least prime divisor of i
for(int i = 2; i <= N; ++i)
{
if(lp[i] == 0)
{
primes.add(i);
lp[i] = i;
}
int curLP = lp[i];
for(int p: primes)
if(p > curLP || p * i > N)
break;
else
lp[p * i] = i;
}
}
} | JAVA |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | /* package codechef; // 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 Codechef
{
public static ArrayList<Long> primeFactors(long n)
{
ArrayList<Long> arr=new ArrayList<>();
// Print the number of 2s that divide n
while (n%2==0)
{
arr.add(2l);
n /= 2;
}
// n must be odd at this point. So we can
// skip one element (Note i = i +2)
for (long i = 3; i <= Math.sqrt(n); i+= 2)
{
// While i divides n, print i and divide n
while (n%i == 0)
{
arr.add(i);
n /= i;
}
}
// This condition is to handle the case whien
// n is a prime number greater than 2
if (n > 2)
arr.add(n);
return arr;
}
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc=new Scanner(System.in);
int t=sc.nextInt();
for(int y=0;y<t;++y)
{
long n=sc.nextLong();
ArrayList<Long> arr = primeFactors(n);
if(arr.size()<3)
{
System.out.println("NO");
continue;
}
if(arr.size()==3)
{ long a=arr.get(0);
long b=arr.get(1);
long c=arr.get(2);
if(a==b || b==c || a==c)
{
System.out.println("NO");
continue;
}
else{
System.out.println("YES");
System.out.println(a+" "+b+" "+c);
continue;
}
}
long a=arr.get(0);
long b=0;
if(arr.get(0)==arr.get(1))
{
b=arr.get(1)*arr.get(2);
}
else{
b=arr.get(1);
}
long c=n/(a*b);
//System.out.println(a+" akshay "+b+" "+c);
if(a==b || b==c || a==c)
{
System.out.println("NO");
continue;
}
else{
System.out.println("YES");
System.out.println(a+" "+b+" "+c);
continue;
}
}
}
}
| JAVA |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | from math import sqrt
def two_dist(n, k):
ans = []
for i in range(2, int(sqrt(n))+5):
if n % i == 0:
if i != n//i and i != k and n//i != k and n//i >= 2:
return [i, n//i]
return False
for _ in range(int(input())):
n = int(input())
ans = []
for p in range(2, int(sqrt(n))+4):
if n % p == 0:
find = two_dist(n//p, p)
if find:
ans = [p]+find
break
if ans:
print("YES")
print(" ".join(str(k) for k in ans))
else:
print("NO")
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 |
def foo(n):
x=pow(n,.5)
x=int(x)+1
l=[]
for i in range(2,x):
if n%i==0:
z=n//i
l.append(i)
if z!=i:
l.append(z)
l.sort()
if len(l)>=3:
for i in range(0,len(l)-1):
k=l[i]*l[i+1]
z=n//k
if (z in l) and z!=l[i] and z!=l[i+1]:
print("YES")
k=[l[i],l[i+1],z]
print(*k)
return
print("NO")
else:
return print("NO")
t=int(input())
for t1 in range(0,t):
n=int(input())
foo(n)
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.PriorityQueue;
import java.util.StringTokenizer;
public class productOfThree {
static BufferedReader b = new BufferedReader(new InputStreamReader(System.in));
static PrintWriter p = new PrintWriter(System.out);
public static void main(String arg[]) throws IOException {
int t = Integer.parseInt(b.readLine());
while(t > 0){
t--;
int prod = Integer.parseInt(b.readLine());
int k = prod;
int a = 0;
int b = 0;
int c = 0;
for(int x = 2; x < Math.sqrt(k); x++){
if(k % x == 0){
if(a == 0){
a = x;
} else {
b = x;
break;
}
k /= x;
}
}
if(a == 0 || b == 0){
p.println("NO");
} else {
c = prod / a / b;
if(c == a || c == b || c == 1){
p.println("NO");
} else {
p.println("YES");
p.println(a + " " + b + " " + c);
}
}
}
p.close();
}
}
| JAVA |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 |
import java.util.ArrayList;
import java.util.Collections;
import java.util.Scanner;
import java.util.TreeSet;
public class C {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int t = in.nextInt();
for (int test = 0; test < t; test++) {
n = in.nextLong();
long original = n;
//TreeSet<Long> factors = factors(n);
//TreeSet<Long> res = new TreeSet();
TreeSet<Long> res = factors2(n);
for (long x: res){
original/=x;
}
if (original == 1 && res.size()==3){
System.out.println("YES");
for (long x: res){
System.out.print(x+" ");
}
System.out.println();
}
else System.out.println("NO");
// if (factors.size()>2) {
// long smallestDiv = factors.pollFirst();
// res.add(smallestDiv);
// n /= smallestDiv;
//
// for (Long x : factors) {
// if (!res.contains(x)) {
// if (factors.contains(n / x)) {
// res.add(x);
// res.add(n / x);
// break;
// }
// }
// }
//
// if (res.size()==3){
// System.out.println("YES");
// for (long val: res){
// System.out.print(val+" ");
// }
// System.out.println();
// }
// else System.out.println("NO");
// }else System.out.println("NO");
}
}
static long n;
public static TreeSet<Long> factors2(long n) {
long original = n;
TreeSet<Long> list = new TreeSet<>();
for (int i = 2; i <= Math.sqrt(original); i++) {
if (n % i == 0) {
if (!list.contains((long)i)){
list.add((long)i);
n/=i;
}
if (list.size()==2){
if (original % n == 0){
list.add(n);
break;
}
}
}
}
//list.add(n);
return list;
}
public static TreeSet<Long> factors(long n) {
TreeSet<Long> list = new TreeSet<>();
for (int i = 2; i <= Math.sqrt(n); i++) {
if (n % i == 0) {
if (n / i == i) {
list.add((long)i);
} else {
list.add((long)i);
list.add(n / i);
}
}
}
return list;
}
static long countTwo = 0;
public static TreeSet<Long> primeFactors(Long n) {
ArrayList<Long> list = new ArrayList<>();
TreeSet<Long> tree = new TreeSet();
for (int i = 2; i * i <= n; i++) {
while (n % i == 0) {
if (i == 2)countTwo++;
tree.add((long)i);
n /= i;
}
}
if (n > 1) {
if (n == 2) countTwo++;
tree.add(n);
}
return tree;
}
}
| 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.math.*;
import java.util.*;
import java.lang.*;
public class Main implements Runnable {
static class InputReader {
private InputStream stream;
private byte[] buf = new byte[1024];
private int curChar;
private int numChars;
private SpaceCharFilter filter;
private BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
public InputReader(InputStream stream) {
this.stream = stream;
}
public int read() {
if (numChars == -1)
throw new InputMismatchException();
if (curChar >= numChars) {
curChar = 0;
try {
numChars = stream.read(buf);
} catch (IOException e) {
throw new InputMismatchException();
}
if (numChars <= 0)
return -1;
}
return buf[curChar++];
}
public String nextLine() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
public int nextInt() {
int c = read();
while (isSpaceChar(c))
c = read();
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
int res = 0;
do {
if (c < '0' || c > '9')
throw new InputMismatchException();
res *= 10;
res += c - '0';
c = read();
}
while (!isSpaceChar(c));
return res * sgn;
}
public long nextLong() {
int c = read();
while (isSpaceChar(c))
c = read();
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
long res = 0;
do {
if (c < '0' || c > '9')
throw new InputMismatchException();
res *= 10;
res += c - '0';
c = read();
}
while (!isSpaceChar(c));
return res * sgn;
}
public double nextDouble() {
int c = read();
while (isSpaceChar(c))
c = read();
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
double res = 0;
while (!isSpaceChar(c) && c != '.') {
if (c == 'e' || c == 'E')
return res * Math.pow(10, nextInt());
if (c < '0' || c > '9')
throw new InputMismatchException();
res *= 10;
res += c - '0';
c = read();
}
if (c == '.') {
c = read();
double m = 1;
while (!isSpaceChar(c)) {
if (c == 'e' || c == 'E')
return res * Math.pow(10, nextInt());
if (c < '0' || c > '9')
throw new InputMismatchException();
m /= 10;
res += (c - '0') * m;
c = read();
}
}
return res * sgn;
}
public String readString() {
int c = read();
while (isSpaceChar(c))
c = read();
StringBuilder res = new StringBuilder();
do {
res.appendCodePoint(c);
c = read();
}
while (!isSpaceChar(c));
return res.toString();
}
public boolean isSpaceChar(int c) {
if (filter != null)
return filter.isSpaceChar(c);
return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
}
public String next() {
return readString();
}
public interface SpaceCharFilter {
public boolean isSpaceChar(int ch);
}
}
public static void main(String args[]) throws Exception {
new Thread(null, new Main(), "Main", 1 << 26).start();
}
public void run() {
InputReader sc = new InputReader(System.in);
// Scanner sc=new Scanner(System.in);
// Random sc=new Random();
PrintWriter out = new PrintWriter(System.out);
int t=sc.nextInt();
while (t-->0){
int n=sc.nextInt();
int m=n;
int i=2;
ArrayList<Integer> l=new ArrayList<>();
while(i*i<=n){
if(n%i==0){
while (n%i==0){
l.add(i);
n/=i;
}
}
if(i==2)i=3;
else i+=2;
}
if(n!=1){
l.add(n);
}
if(l.size()==1){
out.println("NO");
}
else{
int mul=1;
TreeSet<Integer> h=new TreeSet<>();
for (int j = 0; j <l.size() ; j++) {
mul*=l.get(j);
if(!h.contains(mul)){
h.add(mul);
mul=1;
}
}
// out.println("h.size "+h.size()+" "+l);
if(h.size()<3){
out.println("NO");
}
else{
out.println("YES");
int f=h.pollFirst();
int s=h.pollFirst();
out.println(f+" "+s+" "+(m/(f*s)));
}
}
}
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 |
import java.util.*;
public class S{
static ArrayList<Integer> al=new ArrayList<>();
public static ArrayList<Integer> primeFactors(int n)
{
// Print the number of 2s that divide n
while (n%2==0)
{
al.add(2);
//System.out.print(2 + " ");
n /= 2;
}
// n must be odd at this point. So we can
// skip one element (Note i = i +2)
for (int i = 3; i <= Math.sqrt(n); i+= 2)
{
// While i divides n, print i and divide n
while (n%i == 0)
{
//System.out.print(i + " ");
al.add(i);
n /= i;
}
}
// This condition is to handle the case when
// n is a prime number greater than 2
if (n > 2)
al.add(n);
//System.out.print(n);
return al;
}
public static void main(String[] args){
Scanner sc=new Scanner(System.in);
int t=sc.nextInt();
for(int i=0;i<t;i++) {
int n = sc.nextInt();
ArrayList<Integer> list = primeFactors(n);
Set<Integer> s = new HashSet<>(list);
//System.out.println(list);
//System.out.println(s);
int l = list.size();
if (l < 3) {
System.out.println("NO");
} else if (l == 3) {
if (s.size() == l) {
System.out.println("YES");
//System.out.println(list);
for(int j=0;j<list.size();j++)
{
System.out.print(list.get(j)+" ");
}
System.out.println("");
} else {
System.out.println("NO");
}
} else {
ArrayList<Integer> ll=new ArrayList<>();
int str=0;
ll.add(list.get(0));
if(list.get(1)!=list.get(0))
{
ll.add(list.get(1));
str=n/(ll.get(0)*ll.get(1));
}
else
{
ll.add(list.get(1)*list.get(2));
str=n/(ll.get(0)*ll.get(1));
}
// ll.add(list.get(1)*list.get(2));
// int str=n/(ll.get(0)*ll.get(1));
if(str!=ll.get(0) && str!=ll.get(1) && str>2)
{
System.out.println("YES");
ll.add(str);
for(int j=0;j<ll.size();j++)
{
System.out.print(ll.get(j)+" ");
}
System.out.println("");
}
else
{
System.out.println("NO");
}
}
list.clear();
s.clear();
}
}
}
| JAVA |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import java.util.Scanner;
/**
* Created by abhay.kumar on 26/01/20.
* 1294 C
*/
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int[] query = new int[n];
for (int i = 0; i < n; i++) {
query[i] = sc.nextInt();
}
productOfThree(query);
}
private static void productOfThree(int[] query) {
for (int val : query) {
int a = 0;
for (int i = 2; i * i <= val; i++) {
if (val % i == 0) {
a = i;
val = val / a;
break;
}
}
if (a < 2) {
System.out.println("NO");
continue;
}
boolean found = false;
for (int i = 2; i * i <= val; i++) {
if (val % i == 0 && i != a && val / i != i) {
System.out.println("YES");
System.out.println(a + " " + i + " " + val / i);
found = true;
break;
}
}
if (!found) {
System.out.println("NO");
}
}
}
} | JAVA |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import java.util.HashSet;
import java.util.Iterator;
import java.util.Scanner;
public class Solution {
private static HashSet<Integer> set = new HashSet<>();
private static void process(int num, int init){
for(int i = init; i * i <= num; i++){
if(num % i == 0){
if(set.size() == 1){
if(i != num){
set.add(i);
set.add(num/ i);
break;
}
}
set.add(i);
process(num / i, i + 1);
break;
}
}
}
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
for(int i = 0; i < n; i++){ //test cases
int num = sc.nextInt();
process(num, 2);
if(set.size() >= 3){
System.out.println("YES");
Iterator<Integer> it = set.iterator();
while(it.hasNext()){
System.out.print(it.next() + " ");
}
} else {
System.out.println("NO");
}
set.clear();
}
}
}
| JAVA |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | #include <bits/stdc++.h>
const int INF = 0x3f3f3f3f;
using namespace std;
const int MAXN = 1000;
int prime[MAXN + 1];
void getPrime() {
memset(prime, 0, sizeof(prime));
for (int i = 2; i <= MAXN; i++) {
if (!prime[i]) prime[++prime[0]] = i;
for (int j = 1; j <= prime[0] && prime[j] <= MAXN / i; j++) {
prime[prime[j] * i] = 1;
if (i % prime[j] == 0) break;
}
}
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int t;
cin >> t;
getPrime();
for (int te = 0; te < t; te++) {
int n;
cin >> n;
int res[3];
int flag = 0;
for (int i = 2; i <= 10000; i++) {
if (n % i != 0) {
continue;
} else {
int temp = n / i;
for (int j = i + 1; j <= 10000; j++) {
if (temp % j != 0) {
continue;
} else if (temp / j <= j) {
break;
} else {
res[0] = i;
res[1] = j;
res[2] = temp / res[1];
flag = 1;
break;
}
}
if (flag == 1) {
break;
}
}
}
if (flag == 1) {
cout << "YES"
<< "\n";
cout << res[0] << " " << res[1] << " " << res[2] << "\n";
} else {
cout << "NO"
<< "\n";
}
}
return 0;
}
| CPP |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import java.util.*;
public class MyClass {
public static void main(String args[]) {
Scanner sc=new Scanner(System.in);
int t=sc.nextInt();
for(int i=0;i<t;i++)
{
int n=sc.nextInt();
int b=0;
int arr[]=new int[3];
for(int j=2;j*j<n;j++)
{
if(n%j==0)
{
arr[0]=j;
int h=n/j;
for(int k=j+1;k*k<h;k++)
{
if(h%k==0)
{
arr[1]=k;
arr[2]=h/k;
break;
}
}
break;
}
}
int g=0;
Arrays.sort(arr);
for(int j=0;j<2;j++)
{
if(arr[j]==arr[j+1]||arr[j]==0||arr[j+1]==0)
g=1;
}
if(g==0)
{
System.out.println("YES");
System.out.println(arr[0]+" "+arr[1]+" "+arr[2]);
}
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 | #include <bits/stdc++.h>
using namespace std;
template <class T>
void read(T &u) {
u = 0;
char c = getchar();
long long flag = 0;
while (c < '0' || c > '9') flag |= (c == '-'), c = getchar();
while (c >= '0' && c <= '9')
u = (u << 3) + (u << 1) + (c ^ 48), c = getchar();
if (flag) u = -u;
}
template <class T>
void write(T u) {
if (u < 0) printf("-"), u = -u;
if (u < 10)
putchar(48 + u);
else
write(u / 10), putchar(48 + u % 10);
}
void write(char u) { putchar(u); }
template <class T, class... Arg>
void read(T &u, Arg &...v) {
read(u);
read(v...);
}
template <class T, class... Arg>
void write(T u, Arg... v) {
write(u);
write(v...);
}
template <class T>
bool checkmin(T &a, T b) {
return (b < a ? a = b, 1 : 0);
}
template <class T>
bool checkmax(T &a, T b) {
return (b > a ? a = b, 1 : 0);
}
template <class T>
T _min(T a, T b) {
return (a < b ? a : b);
}
template <class T>
T _max(T a, T b) {
return (a > b ? a : b);
}
const long long N = 100005;
vector<pair<long long, long long> > v;
long long n;
void fac(long long n) {
for (long long i = 2; i * i <= n; ++i) {
if (n % i == 0) {
long long tmp = 0;
while (n % i == 0) n /= i, ++tmp;
v.push_back(make_pair(i, tmp));
}
}
if (n > 1) v.push_back(make_pair(n, 1));
}
void init() {
scanf("%lld", &n);
v.clear();
fac(n);
}
void solve() {
if ((long long)v.size() >= 3) {
long long p = n;
puts("YES");
for (long long i = 0; i < 2; ++i) {
long long tmp = 1;
for (long long j = 1; j <= v[i].second; ++j) {
tmp *= v[i].first;
}
p /= tmp;
write(tmp, ' ');
}
write(p, '\n');
return;
}
if ((long long)v.size() == 1) {
if (v[0].second < 6)
printf("NO\n");
else {
long long t = v[0].first;
printf("YES\n");
write(t, ' ', t * t, ' ', n / t / t / t, '\n');
}
} else {
long long a = v[0].first, b = v[1].first, c = n / a / b;
if (a != b && b != c && c != a && c >= 2 && a >= 2 && b >= 2) {
printf("YES\n%lld %lld %lld\n", a, b, c);
} else
printf("NO\n");
}
}
int main() {
int t;
scanf("%d", &t);
while (t--) {
init();
solve();
}
return 0;
}
| CPP |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import math
for _ in range(int(input())) :
n=int(input())
li=list()
j=2
while len(li)<2 and j*j<n :
if n%j==0 :
li.append(j)
n=n//j
j+=1
if len(li)==2 and n not in li :
print("YES")
for i in li :
print(i,end=" ")
print(n)
else :
print("NO")
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | def prime_factorize(n):
a = []
while n % 2 == 0:
a.append(2)
n //= 2
f = 3
while f * f <= n:
if n % f == 0:
a.append(f)
n //= f
else:
f += 2
if n != 1:
a.append(n)
return a
# import numpy as np
t = int(input())
ans = []
for _ in range(t):
n = int(input())
l = prime_factorize(n)
L = l[0]
R = l[-1]
if L == R and len(l) > 3:
R *= l[-2]
l.pop(-1)
if len(l) >=3 and L != R:
# prod = np.prod(l[1:-1])
prod = 0
for i in range(len(l[1:-1])):
prod = prod*(l[1+i]) if prod != 0 else l[1]
if prod != L and prod != R:
ans.append('YES')
ans.append(str(L) + ' ' + str(prod) + ' ' + str(R))
else:
ans.append('NO')
else:
ans.append('NO')
for i in range(len(ans)):
print(ans[i]) | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | for _ in range(int(input())):
n=int(input())
i=2
ans=[]
while(i*i<n):
if(n%i==0):
ans.append(i)
n=n//i
if(len(ans)==2 and n not in ans):
ans.append(n)
break
i+=1
if(len(ans)==3):
print("YES")
print(*ans)
else:
print("NO") | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | for i in range(int(input())):
n = int(input())
f = []
for j in range(2, int(n**(1/2))):
while n % j == 0:
f.append(j)
n = int(n/j)
if n > 1:
f.append(n)
if len(f) >= 3:
if len(f) == 3:
if f[0] != f[1] and f[1] != f[2] and f[2] != f[0]:
print('YES')
print(f[0],f[1],f[2])
else:
print('NO')
else:
f0 = f[0]
f1 = 1
f2 = 1
if f[1] == f[0]:
f1 = f[1] * f[2]
for j in range(3, len(f)):
f2 *= f[j]
else:
f1 = f[1]
for j in range(2, len(f)):
f2 *= f[j]
if f0 != f1 and f1 != f2 and f2 != f0:
print('YES')
print(f0, int(f1), int(f2))
else:
print('NO')
else:
print('NO') | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import java.util.*;
import java.io.*;
public class Main {
static final long M = 1000000007;
static FastReader in = new FastReader();
static PrintWriter out = new PrintWriter(System.out);
// static Scanner in = new Scanner(System.in);
// File file = new File("input.txt");
// Scanner in = new Scanner(file);
// PrintWriter out = new PrintWriter(new FileWriter("output.txt"));
public static void main(String[] args) {
int t = in.nextInt();
while(t-- > 0) {
long n = in.nextLong();
int a = -1, b = -1;
for(int i = 2; i*i<=n; i++) {
if(n%i == 0) {
a = i;
n/=i;
break;
}
}
for(int i = 2; i*i<=n; i++) {
if(n%i==0 && i!=a) {
b = i;
n/=i;
break;
}
}
if(n==a || n==b || n<2 || b==-1 || a==-1) {
System.out.println("NO");
}else {
System.out.println("YES");
System.out.println(a+" "+b+" "+n);
}
}
out.close();
}
private static int gcd(int a, int b) {
if(b==0) return a;
return gcd(b, a%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;
}
}
} | JAVA |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
int cases;
cin >> cases;
for (int tt = 0; tt < cases; tt++) {
long long n;
cin >> n;
int ok = 0;
long long ax, ay, az;
for (long long i = 2; i * i <= n && !ok; i++) {
if (n % i == 0) {
for (long long j = 2; j * j <= i && !ok; j++) {
if (i % j == 0) {
long long x = i / j;
long long y = j;
long long z = n / i;
if (x != y && y != z && x != z && x > 1 && y > 1 && z > 1) {
ok = 1;
ax = x;
ay = y;
az = z;
}
}
}
for (long long j = 2; j * j <= n / i && !ok; j++) {
if ((n / i) % j == 0) {
long long x = i;
long long y = j;
long long z = n / i / j;
if (x != y && y != z && x != z && x > 1 && y > 1 && z > 1) {
ok = 1;
ax = x;
ay = y;
az = z;
}
}
}
}
}
if (ok) {
cout << "YES" << endl;
cout << ax << " " << ay << " " << az << 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 | import java.util.*;
public class Main
{
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
long t = sc.nextLong();
while(t>0){
long n = sc.nextLong();
long a[] = new long[3];
int cnt = 0;
for(long i=2;i*i<n;i++){
if(n%i==0){
a[0] = i;
cnt++;
n /= i;
break;
}
}
for(long i=a[0]+1;i*i<n;i++){
if(n%i==0){
a[1] = i;
cnt++;
n /= i;
break;
}
}
if((n<a[1] && a[1]>0)|| (n<a[0] && a[0]>0) || cnt<2){
System.out.println("NO");
}
else{
System.out.println("YES\n"+a[0]+" "+a[1]+" "+n);
}
t--;
}
}
}
| 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 i in range(int(input())):
a=int(input())
r=0
b=1
while(1):
p=0
for j in range(b+1,a+1):
if a//j<=j:
p=1
break
else:
if a%j==0:
if r==1:
c=j
d=a//j
p=1
r+=2
break
else:
b=j
a=a//b
r+=1
if p==1:
break
if r==3:
print('yes'.upper())
print(b,c,d)
else:
print('no'.upper()) | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | def product(n):
for i in range(2,int(n**(1/3))+1):
if n%i==0:
for j in range(i+1,int((n)**(1/2))):
if (n//i)%j==0 and j!=i and (n//i)/j!=i and j!=(n//i)**(1/2):
print("YES")
print(i,j,(n//i)//j)
return
print("NO")
return
t=int(input())
a=[]
for i in range(t):
a.append(int(input()))
for x in a:
product(x) | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | # -*- coding: utf-8 -*-
"""
Created on Fri Apr 3 10:55:17 2020
@author: Krishna
"""
import math
t=int(input())
for _ in range(t):
n=int(input())
m=n
d={}
while n%2==0:
if 2 not in d:
d[2]=1
else:
d[2] += 1
n=n/2
for i in range(3,int(math.sqrt(n))+1,2):
while n %i == 0:
if i not in d:
d[i]=1
else:
d[i]+=1
n=n/i
if n>2:
d[int(n)]=1
l=len(d)
f=0
if l==1:
for key in d.keys():
if d[key]>=6:
f=1
a=key
b=key**2
c=int(m/(a*b))
elif l==2:
i=0
k=0
for key in d.keys():
k=k+d[key]
if i==0:
a=key
else:
b=key
i=i+1
if k>=4:
f=1
c=int(m/(a*b))
else:
f=1
i=0
for key in d.keys():
if i==0:
a=key
elif i==1:
b=key
else:
break;
i=i+1
c=int(m/(a*b))
if f==0:
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 | t=int(input())
for _ in range(t):
n=int(input())
ad=False
for i in range(2,10001):
if (n%i==0):
ln1=n//i
n1=i
#print("n1 = ",n1)
for j in range(i+1,10001):
#print("ln1(rem)j = ",ln1%j)
if (ln1%j==0 and j*j!=ln1 and ln1//j!=i and ln1//j!=1):
ln2=ln1//j
n2=j
n3=ln2
print("YES")
print(n1,n2,n3)
ad=True
break
if (ad==True):
break
if (ad==True):
#print("NO")
break
if (ad==True):
break
if (ad==False):
print("NO") | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import math
if __name__ == "__main__":
t = int(input())
for _ in range(t):
n, a, b = int(input()), 0, 0
i = 1
for i in range(2, int(math.sqrt(n)) + 1):
if n % i == 0 and i != n // i:
n, a = n // i, i
break
for j in range(i, int(math.sqrt(n)) + 1):
if n % j == 0:
if a != j and a != n // j and j != n // j:
n, b = n // j, j
break
if a != 0 and b != 0:
print("YES")
print(a, b, n)
else:
print("NO") | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import math
def p(n) :
l=[]
i = 1
while i <= math.sqrt(n):
if (n % i == 0) :
if (n / i == i) :
if(i!=1):
l.append(i)
else :
if(i!=1):
l.append(i)
if(n//i!=1):
l.append(n//i)
i = i + 1
return l
for _ in range(int(input())):
n=int(input())
a=p(n)
#l.extend(l)
a.sort()
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 | def factorization(n):
arr = []
temp = n
for i in range(2, int(-(-n ** 0.5 // 1)) + 1):
if temp % i == 0:
cnt = 0
while temp % i == 0:
cnt += 1
temp //= i
for j in range(cnt):
arr.append(i)
if temp != 1:
arr.append(temp)
if arr == []:
arr.append(n)
return arr
n = int(input())
for _ in range(n):
x = int(input())
dat = factorization(x)
datlen = len(dat)
#print("--")
#print(x)
#print(l)
if len(dat) <= 2:
print("NO")
else:
f = False
a = 1
for i in range(datlen):
a *= dat[i]
b = 1
for j in range(i+1, datlen):
b *= dat[j]
c = 1
for k in range(j+1, datlen):
c *= dat[k]
if a != b and b != c and a != c and a != 1 and b != 1 and c != 1:
f = True
break
if f:
break
if f:
print("YES")
print("{0} {1} {2}".format(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 sys
import math
import collections
def get_ints(): return map(int, sys.stdin.readline().strip().split())
def get_list(): return list(map(int, sys.stdin.readline().strip().split()))
def get_string(): return sys.stdin.readline().strip()
for t in range(int(input())):
n=int(input())
p=0
for i in range(2,math.ceil(math.sqrt(n))):
count=0
ans=[]
if n%i==0:
count+=1
no=n//i
ans.append(i)
for j in range(2,math.ceil(math.sqrt(no))):
if i==j:
continue
if no%j==0:
val=no//j
if val!=i and val!=j:
ans.append(val)
ans.append(j)
if len(ans)==3:
print("YES")
print(*ans)
p+=1
count=3
break
if count==3:
break
if p==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;
import java.util.Scanner;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
import java.util.HashMap;
import java.util.TreeMap;
import java.math.BigInteger;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Stack;
public class q_1 {
static Scanner scn = new Scanner(System.in);
static long cunt = 0;
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;
}
}
static FastReader s=new FastReader();
public static void main(String[] args) {
// TODO Auto-generated method stub
// System.out.format("%.10f", ans);char c = sc.next().charAt(0);
int t=s.nextInt();
while(t-->0)
{
int n=s.nextInt();
int a=0,b=0,c=1,d=0;
for(int i=2;i<Math.sqrt(n);i++)
{
if(n%i==0&&d==0)
{
a=i;
n=n/i;
d=1;
}
else if(n%i==0&&d==1)
{
b=i;
c=n/i;
d=10;
break;
}
}
if(d==10)
{ System.out.println("YES");
System.out.println(a+" "+b+" "+c);
}
else
System.out.println("NO");
}
}
public static String factorial(int n) {
BigInteger fact = new BigInteger("1");
for (int i = 1; i <= n; i++) {
fact = fact.multiply(new BigInteger(i + ""));
}
return fact.toString();
}
public static int bs(int arr[], int k) {
int high = 0, low = Integer.MIN_VALUE, ans = 0;
for (int i = 0; i < arr.length; i++) {
high += arr[i];
if (arr[i] > low)
low = arr[i];
}
while (low <= high) {
int mid = (high + low) / 2;
if (valid(arr, mid, k)) {
ans = mid;
high = mid - 1;
} else
low = mid + 1;
}
return ans;
}
public static boolean valid(int arr[], int mid, int k) {
int sum = 0, cs = 1;
for (int i = 0; i < arr.length; i++) {
if (sum + arr[i] > mid) {
cs++;
if (cs > k)
return false;
sum = arr[i];
} else
sum += arr[i];
}
return true;
}
public static long gcd(long a, long n) {
if (a == 0)
return n;
return gcd(n % a, a);
}
}
| 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())
x=int(n)
v=2
k=0
t=[]
while n>v*v and k<2:
if n%v==0:
t.append(v)
k+=1
n=n//v
v+=1
if len(t)<2:
print("NO")
else:
k=x//(t[0]*t[1])
if k==t[0] or k==t[1]:
print("NO")
else:
print("YES")
print(t[0],t[1],x//(t[0]*t[1]))
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | from math import sqrt
t=int(input())
while t:
n=int(input())
nn=n
nnn=n
num=[]
n=int(sqrt(n))+1
cou=0
if(nn%2==0):
cou+=1
num.append(2)
while nn%2==0:
nn=nn//2
for i in range(3,n,2):
if(cou==3):
break
if(nn%i==0):
cou+=1
num.append(i)
while nn%i==0:
nn=nn//i
if(nn!=1):
cou+=1
num.append(nn)
if(cou>=3):
print("YES")
print(num[0],num[1],nnn//(num[0]*num[1]))
else:
if(cou==1):
a=num[0]
b=num[0]*num[0]
c=nnn//(a*b)
if(a!=b and a!=c and c>b):
print("YES")
print(a,b,c)
else:
print("NO")
else:
a=num[0]
b=num[1]
c=nnn//(num[0]*num[1])
if(c!=a and c!=b and c>1):
print("YES")
print(num[0],num[1],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 | #include <bits/stdc++.h>
using namespace std;
const int mod = 1000000007;
void add(long long int& a, long long int b) {
a += b;
while (a >= mod) a -= mod;
while (a < 0) a += mod;
}
void mul(long long int& a, long long int b) {
a *= b;
while (a >= mod) a -= mod;
while (a < 0) a += mod;
}
long long int binexpomodulo(long long int x, long long int y) {
long long int res = 1;
x %= mod;
if (!x) return 0;
while (y) {
if (y & 1) {
mul(res, x);
}
mul(x, x);
y >>= 1;
}
return res;
}
long long int nCrInOr(long long int n, long long int r) {
long long int res = 1;
if (r > n - r) r = n - r;
long long int rin = 1;
for (long long int i = 1; i <= r; i++) rin = (rin * i) % mod;
rin = binexpomodulo(rin, mod - 2);
for (long long int i = 1; i <= r; i++) res = (res * (n - i + 1)) % mod;
res = (res * rin) % mod;
return res;
}
int main() {
int t;
cin >> t;
while (t--) {
int n;
cin >> n;
vector<pair<int, int>> fac;
for (int i = 2; i * i <= n; i++) {
if (n % i == 0) {
int c = 0;
while (n % i == 0) {
c++;
n /= i;
}
fac.push_back({i, c});
}
}
if (n >= 2) fac.push_back({n, 1});
if (fac.size() == 1) {
if (fac[0].second >= 6) {
int f = fac[0].first;
int p = fac[0].second;
cout << "YES" << endl;
cout << f << " " << f * f << " " << (int)(pow(f, p - 3) + 0.5) << endl;
} else
cout << "NO" << endl;
} else if (fac.size() == 2) {
if (fac[0].second > 2 || fac[1].second > 2) {
cout << "YES" << endl;
if (fac[0].second > 2) {
int f = fac[0].first;
int p = fac[0].second;
cout << f << " " << (int)(pow(f, p - 1) + 0.5) << " ";
cout << (int)(pow(fac[1].first, fac[1].second) + 0.5) << endl;
} else {
int f = fac[1].first;
int p = fac[1].second;
cout << f << " " << (int)(pow(f, p - 1) + 0.5) << " ";
cout << (int)(pow(fac[0].first, fac[0].second) + 0.5) << endl;
}
} else if (fac[0].second == 2 && fac[1].second == 2) {
cout << "YES" << endl;
cout << fac[0].first << " " << fac[1].first << " "
<< fac[0].first * fac[1].first << endl;
} else {
cout << "NO" << endl;
}
} else {
cout << "YES" << endl;
cout << (int)(pow(fac[0].first, fac[0].second) + 0.5) << " "
<< (int)(pow(fac[1].first, fac[1].second) + 0.5) << " ";
int final = 1;
for (int i = 2; i < fac.size(); i++)
final *= (int)(pow(fac[i].first, fac[i].second) + 0.5);
cout << final << endl;
}
}
return 0;
}
| CPP |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import math
t=int(input())
for i in range(t):
n=int(input())
flag=0
a=[]
count=0
for j in range(2,31623):
if(n%j==0):
count+=1
a.append(j)
n=n//j
if(count==2):
if(a[0]!=n and a[1]!=n and n!=1):
flag=1
a.append(n)
break
if(flag==0):
print("NO")
else:
print("YES")
for j in range(3):
print(a[j],end=" ")
print()
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
inline int read() {
char c = getchar();
int x = 0, f = 1;
while (c < '0' || c > '9') {
if (c == '-') f = -1;
c = getchar();
}
while (c >= '0' && c <= '9') {
x = x * 10 + c - '0';
c = getchar();
}
return x * f;
}
const int maxn = 1e6 + 10;
const int inf = 2147483647;
int main() {
vector<int> v, prim;
vector<bool> isprim;
int lim = sqrt(1.0 * 1000000000) + 10;
isprim.resize(lim);
isprim[0] = isprim[1] = 1;
for (int i = 2; i <= lim; ++i) {
if (isprim[i] == false) {
for (int j = i + i; j <= lim; j = j + i) isprim[j] = true;
}
}
for (int i = 2; i <= lim; ++i) {
if (isprim[i] == false) prim.push_back(i);
}
int sz = prim.size();
set<int> se;
set<int>::iterator it;
int T;
scanf("%d", &T);
while (T--) {
int n = read();
bool flag = false;
se.clear();
for (int i = 2; i * i <= n; ++i) {
if (n % i == 0 && isprim[i] == false) {
se.insert(i);
n /= i;
flag = true;
break;
}
}
if (!flag) {
puts("NO");
continue;
}
flag = false;
for (int i = sqrt(n); i >= 2; i--) {
if (n % i == 0 && se.count(i) == 0 && se.count(n / i) == 0 &&
i != n / i) {
flag = true;
se.insert(i);
se.insert(n / i);
break;
}
}
if (!flag) {
puts("NO");
continue;
}
puts("YES");
for (it = se.begin(); it != se.end(); ++it) {
printf("%d ", *it);
}
puts("");
}
return 0;
}
| CPP |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | from bisect import *
from collections import *
def factorize(n): # o(sqr(n))
c, ans = 2, []
while (c * c < n):
if n % c == 0:
ans.append(c)
ans.append(n // c)
c += 1
if c * c == n:
ans.append(c)
return sorted(ans)
for i in range(int(input())):
n = int(input())
a = factorize(n)
c, l, ans = defaultdict(int, Counter(a)), len(a), 'NO'
for i in range(l):
for j in range(i+1, l):
num = n / (a[i] * a[j])
if c[num] and num not in [a[i], a[j]]:
print('YES')
print(*[a[i], a[j], int(num)])
ans = 'YES'
break
if ans == 'YES':
break
if ans == 'NO':
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.lang.*;
import java.io.*;
import java.util.*;
import java.math.*;
public class Template implements Runnable{
public static void main(String args[] ) throws Exception {
new Thread(null, new Template(),"Main",1<<27).start();
}
public void run() {
InputReader in = new InputReader(System.in);
PrintWriter out = new PrintWriter(System.out);
int t=in.nextInt();
while(t--!=0){
int n=in.nextInt(),a=0,b=0,j=2,t1=0;
for(;j<Math.sqrt(n);j++) {
if(n%j==0) {
n=n/j;
if(a==0) {
a=j;
t1++;
}
else if(b==0) {
b=j;
t1++;
break;
}
}
}
out.print(t1==2?("YES\n"+a+" "+b+" "+n+"\n"):"NO"+"\n");
}
out.close();
}
static class InputReader {
private InputStream stream;
private byte[] buf = new byte[1024];
private int curChar;
private int numChars;
private SpaceCharFilter filter;
private BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
public InputReader(InputStream stream) {
this.stream = stream;
}
public int read() {
if (numChars==-1)
throw new InputMismatchException();
if (curChar >= numChars) {
curChar = 0;
try {
numChars = stream.read(buf);
}
catch (IOException e) {
throw new InputMismatchException();
}
if(numChars <= 0)
return -1;
}
return buf[curChar++];
}
public String nextLine() {
String str = "";
try {
str = br.readLine();
}
catch (IOException e) {
e.printStackTrace();
}
return str;
}
public int nextInt() {
int c = read();
while(isSpaceChar(c))
c = read();
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
int res = 0;
do {
if(c<'0'||c>'9')
throw new InputMismatchException();
res *= 10;
res += c - '0';
c = read();
}
while (!isSpaceChar(c));
return res * sgn;
}
public long nextLong() {
int c = read();
while (isSpaceChar(c))
c = read();
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
long res = 0;
do {
if (c < '0' || c > '9')
throw new InputMismatchException();
res *= 10;
res += c - '0';
c = read();
}
while (!isSpaceChar(c));
return res * sgn;
}
public double nextDouble() {
int c = read();
while (isSpaceChar(c))
c = read();
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
double res = 0;
while (!isSpaceChar(c) && c != '.') {
if (c == 'e' || c == 'E')
return res * Math.pow(10, nextInt());
if (c < '0' || c > '9')
throw new InputMismatchException();
res *= 10;
res += c - '0';
c = read();
}
if (c == '.') {
c = read();
double m = 1;
while (!isSpaceChar(c)) {
if (c == 'e' || c == 'E')
return res * Math.pow(10, nextInt());
if (c < '0' || c > '9')
throw new InputMismatchException();
m /= 10;
res += (c - '0') * m;
c = read();
}
}
return res * sgn;
}
public String readString() {
int c = read();
while (isSpaceChar(c))
c = read();
StringBuilder res = new StringBuilder();
do {
res.appendCodePoint(c);
c = read();
}
while (!isSpaceChar(c));
return res.toString();
}
public boolean isSpaceChar(int c) {
if (filter != null)
return filter.isSpaceChar(c);
return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
}
public String next() {
return readString();
}
public interface SpaceCharFilter {
public boolean isSpaceChar(int ch);
}
}
} | JAVA |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | from sys import stdin,stdout
from collections import Counter
from math import ceil
from bisect import bisect_left
from bisect import bisect_right
import math
ai = lambda: list(map(int, stdin.readline().split()))
ei = lambda: map(int, stdin.readline().split())
ip = lambda: int(stdin.readline().strip())
def primeFactors(N,li):
while N % 2 == 0:
li.append(2)
N//=2
for i in range(3,int(math.sqrt(N))+1,2):
while N%i == 0:
li.append(i)
N //= i
if N > 2:
li.append(N)
return li
#2 2 2 2 2 2
n = ip()
for i in range(n):
pf = primeFactors(ip(),[])
#print(pf)
if len(set(pf)) == 3 and len(pf) == 3:
print('YES')
print(*pf)
elif len(pf) < 3:
print('NO')
else:
# a,b,c= pf[0],pf[1]*pf[2],1
# for i in range(3,len(pf)):
# c *= pf[i]
a,b,c = pf[0],1,1
for i in range(1,len(pf)):
if b <= a:
b *= pf[i]
else:
c *= pf[i]
if c != b and c != a and c!= 1:
print('YES')
print(a,b,c)
else:
print('NO')
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | '''for j in range(int(input())):
n=int(input())
flag = False
for i in range(2,int(n**(1/3))+1):
if n%i==0:
z=n//i
for k in range(i+1,int(z**0.5)+1):
if z%k==0 and k*k!=z:
flag=True
print("YES")
print(i,k,z//k)
if flag==True:
break
if flag==False:
print("NO")'''
for j in range(int(input())):
n=int(input())
c=0
l=[]
for i in range(2,int(n**0.5)+1):
if n%i==0:
l.append(i)
n=n//i
if len(l)==2:
break
if len(l)==2 and n>l[1]:
print("YES")
l.append(n)
print(*l)
else:
print("NO") | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | from math import *
for _ in range(int(input())):
n = int(input())
ch1 = 1
ch2 = 1
a1 = 0
a2 = 0
a3 = 0
g = 0
for i in range(2,int(sqrt(n)+1)):
if(n%i == 0):
a1 = i
g = n//i
break
else:
ch1 = 0
if(ch1 == 1):
for i in range(2,int(sqrt(g)+1)):
if(g%i == 0 and i!=a1):
a2 = i
a3 = g//i
if(a3 == a1 or a3 == a2):
continue
break
else:
ch2 = 0
if(ch1 ==0 or ch2 == 0):
print("NO")
else:
print("YES")
print(a1,a2,a3) | 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 spf(long long n) {
if (n % 2 == 0) {
return 2;
}
while (n % 2 == 0) {
n /= 2;
}
for (long long i = 3; i * i <= n; i++) {
if (n % i == 0) {
return i;
}
}
return n;
}
int32_t main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
long long t;
t = 1;
cin >> t;
while (t--) {
long long n;
cin >> n;
long long no = 1;
for (long long i = 2; i * i <= n && no; i++) {
if (n % i == 0 && (n != i * i)) {
long long next = n / i;
for (long long j = i + 1; j * j <= next && no; j++) {
if (next % j == 0 && (next / j) != j && (next / j) != i) {
cout << "YES"
<< "\n";
cout << i << " " << j << " " << next / j << "\n";
no = 0;
}
}
}
}
if (no) {
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 | import java.io.*;
import java.util.*;
public class Template {
public static void main(String[] args) {
FastScanner sc = new FastScanner();
PrintWriter pw = new PrintWriter(System.out);
int yo = sc.nextInt();
while (yo-- > 0) {
long n = sc.nextLong();
long a = 0, b = 0, c = 0;
boolean ok = false;
for(long a1 = 2; a1*a1*a1 <= n; a1++) {
if(n%a1 == 0) {
a = a1;
long[] bc = find(n/a1,b,c,a);
b = bc[0];
c = bc[1];
if(a != b && b != c && a != 1 && c != 1 && b != 1) {
ok = true;
break;
}
}
}
if(!ok) {
pw.println("NO");
}
else {
pw.println("YES");
pw.println(a + " " + b + " " + c);
}
}
pw.close();
}
private static long[] find(long k, long b, long c, long a) {
for(long i = 2; i <= k/i; i++) {
if(k%i == 0) {
b = i;
c = k/i;
if(b == a || c == a || b == c) {
continue;
}
else {
return new long[] {b,c};
}
}
}
return new long[]{-1,-1};
}
static class Pair {
int x;
int y;
public Pair(int x, int y) {
this.x = x;
this.y = y;
}
}
// Use this instead of Arrays.sort() on an array of ints. Arrays.sort() is n^2
// worst case since it uses a version of quicksort. Although this would never
// actually show up in the real world, in codeforces, people can hack, so
// this is needed.
static void ruffleSort(int[] a) {
//ruffle
int n = a.length;
Random r = new Random();
for (int i = 0; i < a.length; i++) {
int oi = r.nextInt(n), temp = a[i];
a[i] = a[oi];
a[oi] = temp;
}
//then sort
Arrays.sort(a);
}
static long gcd(long a, long b) {
if (b == 0)
return a;
return gcd(b, a % b);
}
static boolean[] sieve(int n) {
boolean isPrime[] = new boolean[n + 1];
for (int i = 2; i <= n; i++) {
if (isPrime[i])
continue;
for (int j = 2 * i; j <= n; j += i) {
isPrime[j] = true;
}
}
return isPrime;
}
static int mod = 1000000007;
static long pow(int a, long b) {
if (b == 0) {
return 1;
}
if (b == 1) {
return a;
}
if (b % 2 == 0) {
long ans = pow(a, b / 2);
return ans * ans;
} else {
long ans = pow(a, (b - 1) / 2);
return a * ans * ans;
}
}
static class FastScanner {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st = new StringTokenizer("");
String next() {
while (!st.hasMoreTokens())
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
int[] readArray(int n) {
int[] a = new int[n];
for (int i = 0; i < n; i++)
a[i] = nextInt();
return a;
}
long nextLong() {
return Long.parseLong(next());
}
}
// For Input.txt and Output.txt
// FileInputStream in = new FileInputStream("input.txt");
// FileOutputStream out = new FileOutputStream("output.txt");
// PrintWriter pw = new PrintWriter(out);
// Scanner sc = new Scanner(in);
}
| JAVA |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const long double INF = 1e18 + 7;
long long N = 1e6 * 5;
long double EPS = 1 / 1e18;
long long a[1005];
map<long long, long long> cnt;
signed main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
long long t;
cin >> t;
for (long long i = 0; i < t; i++) {
long long k = 0, err = 0, n;
cin >> n;
for (long long j = 2; j <= round(sqrt(n)); j++) {
if (!(n % j)) {
a[k++] = j;
cnt[j]++;
a[k++] = n / j;
cnt[n / j]++;
}
}
for (long long i = 0; i < k; i++) {
for (long long j = 0; j < k; j++) {
if (!(a[i] % a[j]) && cnt[a[i] / a[j]] != 0 && a[j] != n / a[i] &&
a[j] != a[i] / a[j] && n / a[i] != a[i] / a[j]) {
cout << "YES\n"
<< a[j] << ' ' << a[i] / a[j] << ' ' << n / a[i] << '\n';
err = 1;
break;
}
}
if (err) break;
}
if (!err) 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 | from math import sqrt
n=int(input())
s = 32000
lst=[2]
for i in range(3, s+1, 2):
if (i > 10) and (i%10==5):
continue
for j in lst:
if j*j-1 > i:
lst.append(i)
break
if (i % j == 0):
break
else:
lst.append(i)
for i in range(n):
m=int(input())
m1=m
p=[]
k=0
c=lst[k]
while len(p)<3 and c<=sqrt(m1):
if m%c==0:
m=m//c
p.append(c)
else:
k=k+1
c=lst[k]
if len(p)==3:
if p[0]==p[1]:
p[1]=p[1]*p[2]
p[2]=m1//p[0]//p[1]
if len(p)==2:
p.append(m1//p[0]//p[1])
if len(p)==3 and p[2]>1 and p[0]!=p[2] and p[1]!=p[2] and p[0]!=p[1]:
print ('YES')
print (*p)
else:
print ('NO') | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | t=int(input())
for i in range(t):
n=int(input())
k1=int(pow(n,0.333))
k2=k1**2
a=0
b=0
s=n
for j in range(2,k1+1):
if(n%j==0):
a=j
n=n//j
break
if(a==0):
print ('NO')
continue
else:
k=int(pow(n,0.5))
for j in range(a+1,k+1):
if(n%j==0):
b=j
n=n//j
break
if(b==0):
print ('NO')
continue
else:
c=s//(a*b)
if(c==1 or c==a or c==b):
print ('NO')
else:
print ('YES')
print (a,end=" ")
print (b,end=" ")
print (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 calc(n: int):
myStack: list = [n]
for i in range(2, int(math.sqrt(myStack[-1])) + 1):
x, rem = divmod(myStack[-1], i)
if rem == 0:
myStack.pop()
myStack.append(i)
myStack.append(x)
break
else:
print('NO')
return
for i in range(2, int(math.sqrt(myStack[-1])) + 1):
x, rem = divmod(myStack[-1], i)
if rem == 0 and i not in myStack and x not in myStack:
myStack.pop()
myStack.append(i)
myStack.append(x)
break
else:
print('NO')
return
if len(set(myStack)) == 3:
if (myStack[0] * myStack[1] * myStack[2]) == n:
print('YES')
print('{0} {1} {2}'.format(myStack[0], myStack[1], myStack[2]))
else:
print('NO')
else:
print('NO')
def main():
for _ in range(int(input())):
calc(int(input()))
if __name__ == '__main__':
main()
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import math
def primeFactors(n):
factors = []
while n % 2 == 0:
factors.append(2)
n = n // 2
for i in range(3, int(math.sqrt(n)) + 1, 2):
while n % i == 0:
factors.append(i)
n = n // i
if n > 2:
factors.append(n)
return factors
t=int(input())
for q in range(t):
n=int(input())
factor=primeFactors(n)
#print(factor)
if len(factor)<3:
print('NO')
continue
a,b,c=0,0,0
a=factor[0]
b=1
c=1
for i in range(1,len(factor)):
b*=factor[i]
if b!=a:
cin=i
break
for i in range(cin+1,len(factor)):
c *= factor[i]
if len(set([a,b,c]))==3 and a>=2 and b>=2 and c>=2:
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 java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.*;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.InputStream;
public class First {
public static void main(String[] args) {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
InputReader in = new InputReader(inputStream);
PrintWriter out = new PrintWriter(outputStream);
TaskA solver = new TaskA();
int t;
t = in.nextInt();
//t = 1;
while (t > 0) {
solver.call(in,out);
t--;
}
out.close();
}
static class TaskA {
public void call(InputReader in, PrintWriter out) {
int n;
n = in.nextInt();
Set<Integer> set = new HashSet<>();
for (int i = 2; i*i <= n; i++) {
if(n%i==0){
if(set.add(i)) {
n/=i;
i=1;
}
}
if(set.size()==2){
set.add(n);
break;
}
}
if(set.size()==3){
out.println("YES");
Iterator<Integer> iterator = set.iterator();
for (int i = 0; i < 3; i++) {
out.print(iterator.next()+" ");
}
out.println();
}
else{
out.println("NO");
}
}
}
static long gcd(long a, long b)
{
if (b == 0)
return a;
return gcd(b, a % b);
}
static void sort(int[] a) {
ArrayList<Integer> l=new ArrayList<>();
for (int i:a) l.add(i);
Collections.sort(l);
for (int i=0; i<a.length; i++) a[i]=l.get(i);
}
static final Random random=new Random();
static void shuffleSort(int[] arr) {
int n=arr.length;
for (int i=0; i<n; i++) {
int a=random.nextInt(n), temp=arr[a];
arr[a]=arr[i];
arr[i]=temp;
}
Arrays.sort(arr);
}
static class InputReader {
public BufferedReader reader;
public StringTokenizer tokenizer;
public InputReader(InputStream stream) {
reader = new BufferedReader(new InputStreamReader(stream), 32768);
tokenizer = null;
}
public String next() {
while (tokenizer == null || !tokenizer.hasMoreTokens()) {
try {
tokenizer = new StringTokenizer(reader.readLine());
} catch (IOException e) {
throw new RuntimeException(e);
}
}
return tokenizer.nextToken();
}
public int nextInt() {
return Integer.parseInt(next());
}
public long nextLong(){
return Long.parseLong(next());
}
public double nextDouble() {
return Double.parseDouble(next());
}
}
} | JAVA |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import math
def find(n):
count=0
flag=0
ori=n
for i in range(2,int(math.sqrt(n))+3):
if(n%i==0):
count+=1
n=n//i
liste.append(i)
if(count>=2):
r=ori//(liste[0]*liste[1])
if(r!=liste[0] and r!=liste[1] and r!=1):
liste.append(ori//(liste[0]*liste[1]))
flag=1
break
else:
count=1
del liste[len(liste)-1]
if(flag==1):
return "YES"
else:
return "NO"
t=int(input())
for i in range(t):
n=int(input())
liste=[]
y=find(n)
if(y=="YES"):
print(y)
print(*liste)
elif(y=="NO"):
print(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 |
for i in range(int(input())):
n = int(input())
p=[]
m=2
while m*m<=n:
if n%m==0:
n=n//m
p.append(m)
if len(p)==2:break
m+=1
if len(p)<2:
print("NO")
elif p[0]!=p[1]!=n and n>2:
print("YES")
print(*p, 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.io.*;
import java.util.*;
import java.lang.*;
public class Codechef {
PrintWriter out;
StringTokenizer st;
BufferedReader br;
class Pair implements Comparable<Pair>
{
int x;
int y;
Pair(int t, int r) {
x = t;
y = r;
}
public int compareTo(Pair p)
{
return this.x-p.x;
}
}
String ns() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
String nextLine() throws Exception {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
throw new Exception(e.toString());
}
return str;
}
int nextInt() {
return Integer.parseInt(ns());
}
long nextLong() {
return Long.parseLong(ns());
}
double nextDouble() {
return Double.parseDouble(ns());
}
public static void main(String args[]) throws IOException {
new Codechef().run();
}
void run() throws IOException {
br = new BufferedReader(new InputStreamReader(System.in));
out = new PrintWriter(System.out);
long s = System.currentTimeMillis();
solve();
out.flush();
out.close();
}
void solve() throws IOException{
int t=nextInt();
while(t-->0)
{
int n = nextInt();
ArrayList<Integer> list = new ArrayList<>();
int a=1,b=1,c=1;
int i=2,flag=0;
for(i=2;i*i<n;i++)
{
if(n%i==0)
{
list.add(i);
list.add(n/i);
}
}
if(i*i==n)
list.add(i);
Collections.sort(list);
int size=list.size();
for(int j=0;j<size/2;j++)
{
for(int k=j+1;k<size/2;k++)
{
int cn;
if(n%(list.get(j)*list.get(k))==0)
{
cn=n/(list.get(j)*list.get(k));
if(cn!=list.get(j)&&cn!=list.get(k)&&cn>1)
{
a=cn;
b=list.get(j);
c=list.get(k);
flag=1;
break;
}
}
}
}
if(flag==1)
{
out.println("YES");
out.println(a+" "+b+" "+c);
}
else
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())
X = set()
x = 1
for d in range(2, 1+int(n**.5)):
while not n%d:
x *= d
n //= d
if len(X)<2 and x not in X:
X.add(x)
x = 1
X.add(n*x)
if len(X)<3 or 1 in X:
print('NO')
else:
print('YES')
print(*X) | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import sys
import math
input=sys.stdin.readline
# A function to print all prime factors of
# a given number n
def primeFactors(n):
# Print the number of two's that divide n
while n % 2 == 0:
n = n / 2
return(2,n)
# n must be odd at this point
# so a skip of 2 ( i = i + 2) can be used
for i in range(3,int(math.sqrt(n))+1,2):
# while i divides n , print i ad divide n
while n % i== 0:
n = n / i
return(i,n)
# Condition if n is a prime
# number greater than 2
return(-1,-1)
t=int(input())
for _ in range(t):
n=int(input())
a,n=primeFactors(n)
b=-1
c=-1
flag=0
if(a==-1):
print("NO")
else:
i=2
while(i<=math.sqrt(n)):
if(n%i==0):
if(n//i>=2):
if(i!=a and(n//i!=a and i!=n//i)):
b=i
c=n//i
flag=1
i+=1
if(flag==1):
print("YES")
print(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 | # Unmesh Kumar
# IITR CSE '22
from math import *
for _ in range(int(input())):
n=int(input())
pos=False
for a in range(2,int(pow(n,1/3))+1):
if n%a==0:
z=n//a
for b in range(a+1,int(sqrt(z))+1):
if z%b==0 and b*b!=z:
pos=True
print('YES')
print(a,b,z//b)
break
if pos:
break
if not pos:
print('NO')
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import java.util.*;
import java.io.*;
public class Main {
static InputReader scn = new InputReader(System.in);
static PrintWriter out = new PrintWriter(System.out);
public static void main(String[] HastaLaVistaLa) {
// Running Number Of TestCases (t)
int t = scn.nextInt();
while(t-- > 0)
solve();
out.close();
}
public static void dummy(){
int n = scn.nextInt();
out.println(n);
}
static int max = 0;
static int[] dp;
static int MOD = 100;
public static void solve() {
// Main Solution (AC)
int n = scn.nextInt();
int k = (int) Math.sqrt(n);
for(int i = 2; i <= k; i++) {
if(n % i == 0) {
int fact1 = n / i;
int[] factor23 = Factors(fact1);
if(factor23[0] != -1 && factor23[0] != i && factor23[1] != i) {
out.println("YES");
out.println(i + " " + factor23[0] + " " + factor23[1]);
return;
}
}
}
out.println("NO");
}
public static int[] Factors(int a) {
int k = (int) Math.sqrt(a);
int[] arr = new int[2];
Arrays.fill(arr, -1);
for(int i = 2; i <= k; i++) {
if(a % i == 0) {
int fact1 = a / i;
if(fact1 != i) {
arr[0] = i;
arr[1] = fact1;
}
}
}
return arr;
}
public static long binpow(long n, long m, long mod) {
long res = 1;
while (m > 0) {
if (m % 2 == 1) {
res *= n;
res %= mod;
m--;
}
n *= n;
n %= mod;
m /= 2;
}
return res;
}
public static HashMap<Integer, Integer> CountFrequencies(int[] arr) {
HashMap<Integer, Integer> map = new HashMap<>();
for (int i : arr) {
if (map.containsKey(i)) map.put(i, map.get(i) + 1);
else map.put(i, 1);
}
return map;
}
public static long gcd(long a, long b) {
if (a == 0)
return b;
return gcd(b % a, a);
}
public static long sum(int[] arr) {
long sum = 0;
for(int i : arr) sum += i;
return sum;
}
// Number of Divisor for Range [l, r]
public static int[] NumberOFDivisors(int l, int r) {
int[] arr = new int[l - r + 1];
for(int i = l; i <= r; i++) {
for(int j = i; j <= r; j += i)
arr[j]++;
}
return arr;
}
public static int LowerBound(int a[], int x) { // x is the target value or key
int l = -1,r = a.length;
while(l + 1 < r) {
int m = (l + r) >>> 1;
if(a[m] >= x) r = m;
else l = m;
}
return r;
}
static boolean[] prime;
public static void sieveOfEratosthenes(int n) {
prime = new boolean[n+1];
for(int i=0;i<n;i++)
prime[i] = true;
for(int p = 2; p*p <=n; p++) {
if(prime[p] == true) {
for(int i = p*p; i <= n; i += p)
prime[i] = false;
}
}
}
public static int UpperBound(long a[], long x) {// x is the key or target value
int l = -1,r = a.length;
while(l + 1 < r) {
int m = (l + r) >>> 1;
if(a[m] <= x) l = m;
else r = m;
}
return l + 1;
}
public static long lcm(long a, long b) {
return (a * b) / gcd(a, b);
}
public static void sort(int[] arr) {
Random rand = new Random();
int n = arr.length;
for(int i = 0; i<n; i++) {
swap(arr, i, rand.nextInt(n));
}
Arrays.sort(arr);
}
public static void swap(int[] arr, int i, int j) {
if(i!=j) {
arr[i] ^= arr[j];
arr[j] ^= arr[i];
arr[i] ^= arr[j];
}
}
public static void sortbyColumn(int[][] arr, int col) {
Arrays.sort(arr, new Comparator<int[]>() {
public int compare(final int[] entry1, final int[] entry2) {
if (entry1[col] > entry2[col])
return 1;
else
return -1;
}
});
}
public static void ArraySort2D(double[][] arr, int xy) {
// xy == 0, for sorting wrt X-Axis
// xy == 1, for sorting wrt Y-Axis
Arrays.sort(arr, Comparator.comparingDouble(o -> o[xy]));
}
public static int binarySearch(int arr[], int l, int r, int x) {
if (r >= l) {
int mid = l + (r - l) / 2;
if (arr[mid] == x)
return mid;
if (arr[mid] > x)
return binarySearch(arr, l, mid - 1, x);
return binarySearch(arr, mid + 1, r, x);
}
return -1;
}
static class InputReader {
private InputStream stream;
private byte[] buf = new byte[1024];
private int curChar;
private int numChars;
private SpaceCharFilter filter;
public InputReader(InputStream stream) {
this.stream = stream;
}
public int[] readIntArray(int n) {
int[] a = new int[n];
for (int i = 0; i < n; i++) {
a[i] = scn.nextInt();
}
return a;
}
public long[] readLongArray(int n) {
long[] a = new long[n];
for (int i = 0; i < n; i++) {
a[i] = scn.nextLong();
}
return a;
}
public int read() {
if (numChars == -1)
throw new InputMismatchException();
if (curChar >= numChars) {
curChar = 0;
try {
numChars = stream.read(buf);
} catch (IOException e) {
throw new InputMismatchException();
}
if (numChars <= 0)
return -1;
}
return buf[curChar++];
}
public int nextInt() {
int c = read();
while (isSpaceChar(c))
c = read();
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
int res = 0;
do {
if (c < '0' || c > '9')
throw new InputMismatchException();
res *= 10;
res += c - '0';
c = read();
} while (!isSpaceChar(c));
return res * sgn;
}
public long nextLong() {
int c = read();
while (isSpaceChar(c))
c = read();
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
long res = 0;
do {
if (c < '0' || c > '9')
throw new InputMismatchException();
res *= 10;
res += c - '0';
c = read();
} while (!isSpaceChar(c));
return res * sgn;
}
public String nextLine() {
int c = read();
while (isSpaceChar(c))
c = read();
StringBuilder res = new StringBuilder();
do {
res.appendCodePoint(c);
c = read();
} while (!isSpaceChar(c));
return res.toString();
}
public boolean isSpaceChar(int c) {
if (filter != null)
return filter.isSpaceChar(c);
return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
}
public String next() {
return nextLine();
}
public interface SpaceCharFilter {
public boolean isSpaceChar(int ch);
}
}
} | JAVA |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import math
t=int(input())
def f(a):
b=2
a=int(a)
fac=[]
c=math.sqrt(a)
while b<=c and a>1:
if a%b==0:
while a%b==0:
fac.append(b)
a/=b
b+=1
else:
b+=1
if a!=1:
fac.append(a)
return fac
for i in range(t):
n=int(input())
list=f(n)
list2=[]
list2.clear()
while list:
i=list[0]
b=list.count(i)
list2.append([i,b])
for j in range(b):
list.remove(i)
c=len(list2)
if c>=3:
print("YES")
print(list2[0][0],list2[1][0],int(n/list2[0][0]/list2[1][0]))
elif c==1:
if(list2[0][1]>=6):
print("YES")
print(list2[0][0],list2[0][0]*list2[0][0],int(n/list2[0][0]/list2[0][0]/list2[0][0]))
else:
print("NO")
elif c==0:
print("NO")
else:
if(list2[0][1]>=2 and list2[1][1]>=2):
print("YES")
print(list2[0][0],list2[1][0],int(n/list2[0][0]/list2[1][0]))
elif list2[0][1]>=3:
print("YES")
print(list2[0][0],int(list2[1][0]),int(n/list2[0][0]/list2[1][0]))
elif list2[1][1]>=3:
print("YES")
print(list2[0][0], list2[1][0],int(n/list2[0][0]/list2[1][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 | import sys
from functools import reduce
reader = (s.rstrip() for s in sys.stdin)
input = reader.__next__
def factors(n):
return list(set(reduce(list.__add__,
([i, n//i] for i in range(1, int(n**0.5) + 1) if n % i == 0))))
def product():
for _ in range(t):
n=int(input())
yn=factors(n)
yn.sort()
#print(yn)
if len(yn)>=4:
a,b=yn[1],yn[3]
c=n//yn[1]//yn[3]
d=n//yn[1]%yn[3]
a1,b1=yn[1],yn[2]
e=n//yn[1]//yn[2]
f=n//yn[1]%yn[2]
if c>1 and c!=a and c!=b and d==0:
yield 'YES'
yield str(a)+' '+str(b)+' '+str(c)
elif e>1 and e!=a1 and e!=b1 and f==0:
yield 'YES'
yield str(a1)+' '+str(b1)+' '+str(e)
else:
yield'NO'
else:
yield 'NO'
if __name__ == '__main__':
t= int(input())
ans = product()
print(*ans,sep='\n')
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import math
t = int(input())
while t>0:
# print("t: ",t)
t-=1
n = int(input())
num_factors=0
first_factor = -1
second_factor = -1
if n==1 or n==2:
print("NO")
continue
n_copy=n
s = int(math.sqrt(n))
for i in range(2,s):
if n_copy%i==0:
if first_factor==-1:
first_factor=i
elif second_factor==-1:
second_factor=i
else:
break
n_copy= n_copy//i
if first_factor==-1 or second_factor==-1 or first_factor==n_copy or second_factor==n_copy or n_copy==1:
print("NO")
else:
print("YES")
print(first_factor,second_factor,n_copy)
| 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(2)
#pragma GCC optimize(3)
using namespace std;
const int N = 1e5 + 7;
const int M = 1e6 + 7;
const int INF = 1e9 + 8;
int p[N];
int c[N];
int m;
void divide(int n) {
m = 0;
for (int i = 2; i * i <= n; i++) {
if (n % i == 0) {
p[++m] = i, c[m] = 0;
while (n % i == 0) {
n /= i;
c[m]++;
}
}
}
if (n > 1) {
p[++m] = n;
c[m] = 1;
}
return;
}
int main() {
int t;
cin >> t;
while (t--) {
int n;
cin >> n;
divide(n);
int a = 1;
int b = 1;
int cc = 1;
int tt = 1;
for (int i = 1; i <= m; i++) {
int g = c[i];
for (int j = 1; j <= g; j++) {
if (tt <= 1) {
a *= p[i];
tt++;
} else {
if (b == 1 || b == a)
b *= p[i];
else
cc *= p[i];
}
}
}
if (cc == 1 || b == 1 || a == 1) {
cout << "NO" << endl;
continue;
}
if (cc != a && a != b && b != cc) {
cout << "YES" << endl;
cout << a << " " << b << " " << cc << endl;
} else
cout << "NO" << endl;
}
return 0;
}
| CPP |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 7;
int a[N];
void solve() {
int n;
cin >> n;
set<int> s;
for (long long i = 2; i * i <= n; i++) {
if (n % i == 0 && s.find(i) == s.end()) {
n /= i;
s.insert(i);
break;
}
}
for (long long i = 2; i * i <= n; i++) {
if (n % i == 0 && s.find(i) == s.end()) {
n /= i;
s.insert(i);
break;
}
}
if ((int)s.size() < 2 || n == 1 || s.find(n) != s.end())
cout << "NO\n";
else {
cout << "YES\n";
s.insert(n);
for (auto it : s) cout << it << " ";
cout << "\n";
}
}
int main() {
ios_base ::sync_with_stdio(false);
cin.tie(NULL);
cout.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 math
t=int(input())
for i in range(t):
n=int(input())
a=math.sqrt(n)
a=int(a)
c=[]
f=0
d=0
b=int(a)
for j in range(2,a+1,1):
if(n%j==0):
f=1
x=j
y=n//j
break
if(f==0):
print("NO")
else:
f=0
for j in range(2,int(y**(1/2))+1):
if(y%j==0):
if(x!=j and y//j!=x and y//j!=j):
z=j
y=y//j
f=1
break
if(f==0):
print("NO")
else:
print("YES")
print(x,y,z)
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import java.util.*;
public class Main
{
public static void main(String args[]){
Scanner sc = new Scanner(System.in);
int t=sc.nextInt();
while(t-->0) {
int n = sc.nextInt();
int[] arr = new int[3];
int index=0, divisor=2;
while(n%divisor==0 && index<2) {
arr[index++]=divisor;
n/=divisor;
divisor*=divisor;
}
for (int i = 3; i <= Math.sqrt(n) && index<2; i+= 2) {
divisor=i;
while(n%divisor==0 && index<2) {
arr[index++]=divisor;
n/=divisor;
divisor*=divisor;
}
}
if(n==1 || n==0 || n==arr[0] || n==arr[1] || arr[0]==0 || arr[1]==0)
System.out.println("NO");
else {
System.out.println("YES");
System.out.println(arr[0]+" "+arr[1]+" "+n);
}
}
}
}
| JAVA |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | #include <bits/stdc++.h>
#pragma 03
using namespace std;
vector<pair<long long, long long> > pfac;
long long binpow(long long x, long long y) {
if (x == 0) return 0;
if (y == 0) return 1;
long long res = binpow(x, y / 2);
if (y % 2 == 0)
return res * res;
else
return res * res * x;
}
signed main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
long long t;
cin >> t;
long long n, x, num;
while (t--) {
cin >> n;
pfac.clear();
if (n < 24) {
cout << "NO\n";
continue;
}
x = n;
for (long long i = 2; i < sqrt(n); i++) {
if (x % i == 0) {
num = 0;
while (x % i == 0) {
num++;
x /= i;
}
pfac.push_back(pair<long long, long long>(i, num));
}
if (x == 1) break;
}
if (x != 1) pfac.push_back(pair<long long, long long>(x, 1));
if (pfac.empty()) {
cout << "NO\n";
continue;
} else if (pfac.size() == 1) {
if (pfac[0].second >= 6) {
cout << "YES\n";
cout << pfac[0].first << " " << pfac[0].first * pfac[0].first << " "
<< binpow(pfac[0].first, pfac[0].second - 3) << "\n";
} else {
cout << "NO\n";
}
} else if (pfac.size() == 2) {
if (pfac[0].second <= 2 && pfac[1].second <= 2) {
if (pfac[0].second < 2 or pfac[1].second < 2) {
cout << "NO\n";
} else {
cout << "YES\n";
cout << pfac[0].first << " " << pfac[1].first << " "
<< pfac[0].first * pfac[1].first << "\n";
}
} else {
cout << "YES\n";
cout << pfac[0].first << " " << pfac[1].first << " "
<< binpow(pfac[0].first, pfac[0].second - 1) *
binpow(pfac[1].first, pfac[1].second - 1)
<< "\n";
}
} else {
long long neu = 1;
for (long long i = 2; i < pfac.size(); i++) {
neu *= binpow(pfac[i].first, pfac[i].second);
}
cout << "YES\n";
cout << binpow(pfac[0].first, pfac[0].second) << " "
<< binpow(pfac[1].first, pfac[1].second) << " " << neu << "\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
t = int(input())
for i in range(t):
n = int(input())
f = 1
for a in range(2, int(math.sqrt(n)) + 1):
if f and n % a == 0:
curn = n // a
for b in range(2, int(math.sqrt(curn)) + 1):
if f and b != a and curn % b == 0:
c = curn // b
if c != b and c != a:
print('YES')
print(a, b, c)
f = 0
if f:
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 facts(n):
l=[]
for i in range(2,int(pow(n,0.5))+1):
if n%i==0:
l.append(i)
l.append(n//i)
return list(set(l))
for _ in range(int(input())):
n=int(input())
l=facts(n)
l.sort()
d={}
for i in l:
d[i]=0
flag = 0
for i in range(len(l)):
for j in range(i+1,len(l)):
a,b=l[i],l[j]
c=n//(a*b)
try:
if d[c]==0 and a!=c and b!=c:
print("YES")
print(a,b,c)
flag = 1
except:
continue
if flag:
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 | # cook your dish here
import math
for t in range(int(input())):
n=int(input())
flag=False
for i in range(2,int(math.sqrt(n))+1):
if n%i==0:
x=i
yy=n//i
for j in range(i+1,int(math.sqrt(yy))+1):
if yy%j==0:
y=j
z=yy//j
if z>=2 and z!=y and z!=x:
flag=True
l=[x,y,z]
print("YES")
print(*l)
break
if flag:
break
if flag==False:
print("NO") | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import math
t=int(input())
for i in range(t):
n=int(input())
k=n
l=[]
for j in range(2,int(math.sqrt(n))+1):
if n%j==0:
l.append(j)
break
if len(l)!=0:
n=n//l[0]
for j in range(2,int(math.sqrt(n))+1):
if n%j==0:
if j not in l:
l.append(j)
break
if len(l)==2:
if k%(l[0]*l[1])==0:
c=k//(l[0]*l[1])
if c not in l:
print("YES")
print(l[0],l[1],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 | #include <bits/stdc++.h>
int n, a[105], m;
int main() {
scanf("%d", &n);
for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
for (int i = 1; i <= n; i++) {
int s1 = 0, s2 = 0, s3 = 0;
m = a[i];
for (int j = 2; j * j <= m; j++) {
if (m % j == 0) {
if (s1 == 0)
s1 = j;
else
s2 = j;
m /= j;
}
if (s2 != 0 && m != 1) {
s3 = a[i] / (s1 * s2);
if (s2 == s3) s3 = 0;
break;
}
}
if (s1 * s2 * s3 != 0)
printf("YES\n%d %d %d", s1, s2, s3);
else
printf("NO");
if (i != n) printf("\n");
}
scanf("\n");
}
| CPP |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | from math import *
from operator import *
def isPrime(x):
if(x in [1,4]):
return False
elif(x in [2,3,5]):
return True
else:
for i in range(2,int(sqrt(x))+1):
if(x%i == 0):
return False
return True
def Sieve(x):
checkPrime = [1 for i in range(x+1)]
zz = 2
L = []
while(zz**2 <= x):
if(checkPrime[zz] == 1):
for i in range(zz*2,x+1,zz):
checkPrime[i] = 0
zz += 1
for i in range(2,x):
if(checkPrime[i] == 1):
L.append(i)
return checkPrime
checkPrime = Sieve(100001)
def primeFactors(x):
d = {}
i = 0
X = x
while(x%2 == 0):
x //= 2
i += 1
if(i > 0):
d[2] = i
for i in range(3,int(sqrt(x))+1,2):
zz = 0
while(x%i == 0):
x //= i
zz += 1
if(zz > 0):
d[i] = zz
if(x > 2):
d[x] = 1
return d
T = int(input())
for tt in range(T):
x = int(input())
if(x%2 != 0 and x%3 != 0 and isPrime(x)):
print('NO')
elif(isPrime(x)):
print('NO')
else:
d = primeFactors(x)
# print('d:',d)
ans = [0,0,0]
if(len(d) > 2):
print('YES')
for i in d:
if(ans[0] == 0):
ans[0] = i
elif(ans[1] == 0):
ans[1] = i
else:
break
ans[2] = x//(ans[0]*ans[1])
print(*ans)
elif(len(d) == 1):
for i in d:
if(d[i] < 6):
print('NO')
else:
print('YES')
print(i,i**2,x//(i*i**2))
else:
ans = [0]*3
d = sorted(d.items(), key=itemgetter(1))
if(d[0][1] >= 6):
print('YES')
i = d[0][0]
print(i,i**2,x//(i*i**2))
elif(d[0][1] + d[1][1] >= 4):
print('YES')
print(d[0][0],d[1][0],x//(d[0][0]*d[1][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 P(n):
i = 2
factors = []
while i * i <= n:
if n % i:
i += 1
else:
n //= i
factors.append(i)
if n > 1:
factors.append(n)
return factors
t=int(input())
for i in range(t):
x=int(input())
p=P(x)
if len(p)>2:
a=p[0]
b=p[1]
c=1
for i in range(2,len(p)):
if a==b:
b=b*p[i]
else:
c*=p[i]
if c!=1 and a!=b and b!=c and c!=a:
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 | #include <bits/stdc++.h>
using namespace std;
void solve() {
long long int n;
cin >> n;
long long int ans[3];
long long int count = 0;
for (long long int i = 2; i * i <= n; i++) {
if (n % i == 0) {
n = n / i;
ans[count] = i;
count++;
}
if (count == 3) {
break;
}
}
if (count == 3 && n > 2) {
ans[2] = ans[2] * n;
} else if (count < 3 && n > 2) {
ans[count] = n;
count++;
}
if (count == 3 && ans[0] != ans[1] && ans[1] != ans[2] && ans[2] != ans[0]) {
cout << "YES" << endl;
cout << ans[0] << " " << ans[1] << " " << ans[2] << endl;
} else
cout << "NO" << endl;
}
int main() {
int t;
cin >> t;
while (t--) {
solve();
}
}
| CPP |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int main() {
long long int t, n, a, b, i, ra, rb, rc;
cin >> t;
while (t--) {
ra = rb = rc = 1;
cin >> a;
b = sqrt(a);
for (i = 2; i <= b + 1; i++) {
if (a % i == 0) {
a = a / i;
ra = i;
break;
}
}
b = sqrt(a);
for (i = 2; i <= b + 1 && ra != 1; i++) {
if (a % i == 0 && i != ra) {
a = a / i;
rb = i;
break;
}
}
if (a != ra && a != rb) rc = a;
if (ra == 1 || rb == 1 || rc == 1)
cout << "NO\n";
else {
cout << "YES\n";
cout << ra << " " << rb << " " << rc << endl;
}
}
}
| CPP |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import math
def isPrime(n) :
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
t=int(input())
for i in range(t):
n=int(input())
x=isPrime(n)
if x==True:
print("NO")
else:
c=[]
count=0
vv=math.sqrt(n)
vv=int(vv)
i=2
while i*i<n:
if n%i==0:
count+=1
n=n//i
c.append(i)
if count==2:
c.append(n)
break
i+=1
if len(c)<3 or c[0]==c[1] or c[1]==c[2]:
print("NO")
else:
print("YES")
print(*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 | # your code goes here
import math
t=int(input())
while(t>0):
n=int(input())
l=[]
c=0
s=int(math.sqrt(n))
for i in range(2,s+1):
if(n%i==0):
l.append(i)
c+=1
n=n//i
break
for i in range(2,s+1):
if(i not in l and n%i==0):
l.append(i)
c+=1
n=n//i
break
if(c<2 or n==1 or n in l):
print("NO")
else :
print("YES")
for i in l:
print(i,end=' ')
print(n)
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())
a=0
b=0
c=0
for i in range(2,int(n**0.5)+1):
if n%i==0:
a=i
t=n//i
for j in range(a+1,int(t**0.5)+1):
if t%j==0:
b=j
c=t//j
if b==c:
b=0
c=0
continue
break
break
if a and b and c:
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 java.math.BigInteger;
import java.util.Arrays;
import java.util.HashMap;
import java.util.Scanner;
public class T {
public static void main(String args[]) {
Scanner sc = new Scanner(System.in);
int x = sc.nextInt();
int c=0;
int c1=0;
int c2=0;
int c3=0;
boolean t ;
for(int i =0 ; i<x;i++){
int k = sc.nextInt();
t=false;
int v=k;
c=0;
for(int j =2 ; j<Math.sqrt(v);j++){
if(k%j==0 && c==0){
c++;
k/=j;
c1=j;
continue;
}
if(k%j==0 && c==1){
c++;
k/=j;
c2=j;
if(k<v &&k>c2){
c3=k;
t=true;
c++;
break;
}
}
}
if(t){
System.out.println("YES");
System.out.println(c1+" "+c2+" "+c3+" ");
}else{
System.out.println("NO");
}
c1=0;
c2=0;
c3=0;
}
}} | 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 miller_rabin_primalitytest(n,a):
z=n-1
d=z
while(d%2==0):
d=d//2
if(pow(a,d,n)!=1):
x=d
while(x<n):
r=pow(a,x,n)
if(r==n-1):
return "Probable Prime"
else:
x=2*x
return "Composite"
else:
return "Probable Prime"
def isprimeoptimized(n):
if(n%2==0 and n>2):
return False
elif(n==2):
return True
elif(n==1):
return False
else:
if(n<2047):
L=[2]
else:
L=[2,3,5,7]
i=0
while(i<len(L)):
if(miller_rabin_primalitytest(n,L[i])=="Probable Prime"): #WORKS FOR ALL NUMBERS LESS THAN 2^64.
i+=1
else:
return False
return True
t=int(input())
for i in range(0,t):
n=int(input())
if(isprimeoptimized(n)):
print("NO")
elif(n<24):
print("NO")
else:
L=[]
for i in range(2,int(n**0.5)+1):
if(n%i==0):
L.append(i)
n=n//i
break
for i in range(2,int(n**0.5)+1):
if(n%i==0 and L[0]!=i and L[0]!=n//i and i*i!=n):
L.append(i)
L.append(n//i)
break
if(len(L)==3):
print("YES")
print(*L)
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 | q = int(input())
def div_by_primes(n):
p = []
s = []
for d in range(2, int(n**(1/2))+1):
if n%d == 0:
p.append(d)
s.append(0)
while n%d == 0:
s[-1] += 1
n //= d
if n > 1:
p.append(n)
s.append(1)
return p, s
def s_mult(s, p):
ans = 1
for i in range(len(s)):
ans *= p[i]**s[i]
return ans
for _ in range(q):
n = int(input())
p, s = div_by_primes(n)
if len(p) == 1:
if s[0] < 6:
print("NO")
else:
print("YES\n", p[0], p[0]**2, p[0]**(s[0]-3))
elif len(p) == 2:
if s[0]+s[1] < 4:
print("NO")
elif s[0] == 1:
print("YES\n", p[0], p[1], p[1]**(s[1]-1))
elif s[1] == 1:
print("YES\n", p[0], p[1], p[0]**(s[0]-1))
else:
print("YES\n", p[0], p[1], p[0]**(s[0]-1)*p[1]**(s[1]-1))
else:
print("YES\n", p[0]**s[0], p[1]**s[1], s_mult(s[2:], p[2:]))
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import java.io.*;
import java.util.*;
public class C {
public static boolean[] isPrime;
public static void main(String[] args) {
FastScanner sc=new FastScanner();
int t = sc.nextInt();
isPrime = new boolean[1000007];
fillPrime();
while(t-->0) {
int n = sc.nextInt();
if(n< 1000007 && isPrime[n]) {
System.out.println("NO");
}else {
int ar[] = new int[3];
int i = 0;
int cur = 2;
while(i<2 && cur*cur <= n) {
if(n%cur == 0) {
//System.out.println(ar[i]);
ar[i++] = cur;
n/=cur;
//System.out.println(n+" "+ar[i-1]);
}
cur++;
}
if(i==2) {
ar[2] = n;
}
//System.out.println();
if(i==2 && ar[2] > 1 && ar[2] != ar[1] && ar[2] !=ar[0]) {
System.out.println("YES");
for(int val : ar) {
System.out.print(val + " ");
}
System.out.println();
}else {
System.out.println("NO");
}
}
}
}
public static void fillPrime() {
int n = 1000007;
Arrays.fill(isPrime, true);
isPrime[0] = 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;
}
}
}
}
static final Random random = new Random();
static void sort(int[] a) {
int n = a.length;
for(int i =0; i<n; i++) {
int val = random.nextInt(n);
int cur = a[i];
a[i] = a[val];
a[val] = cur;
}
Arrays.sort(a);
}
static void sortl(long[] a) {
int n = a.length;
for(int i =0; i<n; i++) {
int val = random.nextInt(n);
long cur = a[i];
a[i] = a[val];
a[val] = cur;
}
Arrays.sort(a);
}
static class FastScanner {
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st=new StringTokenizer("");
String next() {
while (st == null || !st.hasMoreTokens()) {
try {
st=new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
String nextLine() {
String str = "";
try
{
str = br.readLine();
}
catch (IOException e)
{
e.printStackTrace();
}
return str;
}
int nextInt() {
return Integer.parseInt(next());
}
int[] readArray(int n) {
int[] a=new int[n];
for (int i=0; i<n; i++) a[i]=nextInt();
return a;
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
}
}
| JAVA |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import math
class Read:
@staticmethod
def int():
return int(input())
@staticmethod
def list(sep=' '):
return input().split(sep)
@staticmethod
def list_int(sep=' '):
return list(map(int, input().split(sep)))
def main():
n = Read.int()
status = False
for a in range(2, int(math.sqrt(n) + 1)):
if n % a == 0:
k = n / a
for b in range(a+1, int(math.sqrt(k) +1)):
if k % b == 0:
c = int(k / b)
if b != c and c > 2:
print('YES')
print('{} {} {}'.format(a, b, c))
status = True
break;
if status:
break;
if status == False:
print('NO')
query_count = Read.int()
while query_count:
query_count -= 1
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 i in range(t):
n = int(input())
Ans = []
d = 2
while d * d <= n:
if n % d == 0:
Ans.append(d)
break
else:
d += 1
if len(Ans) == 0:
Ans.append(n)
n //= Ans[-1]
while d * d <= n:
if n % d == 0 and Ans[-1] != d:
Ans.append(d)
break
else:
d += 1
if len(Ans) == 1:
Ans.append(1)
if len(Ans) >= 2:
c = n // Ans[-1]
if c >= 2 and Ans[0] >= 2 and Ans[1] >= 2 and c != Ans[-1] != Ans[-2]:
Ans.append(c)
print('YES')
print(*Ans)
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
ts = int(input( ))
while ts > 0:
ts = ts - 1
n = int(input( ))
flag = 0
a, b, c = 0, 0, 0
for i in range( 2, int(math.sqrt(n) + 1 ) ):
if n % i == 0:
a = i
n = int( n / a )
break
for i in range( 2, int(math.sqrt(n) + 1 ) ):
if n % i == 0 and i != a:
b = i
n = int( n / b )
break
if a != 0 and b != 0:
c = n
if c != a and c != b and c != 1 and c != 0:
print("YES")
print(a,b,c)
else:
print('NO') | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int T;
int main(int argc, char const *argv[]) {
cin >> T;
for (int kase = 1; kase <= T; kase++) {
int n;
cin >> n;
map<int, int> mp;
int tmp = n;
int cnt = 3;
while (cnt) {
if (cnt == 1) {
mp[tmp] = 1;
break;
}
for (int i = 2; i <= sqrt(tmp); i++) {
if (tmp % i == 0 && mp[i] == 0) {
mp[i] = 1;
tmp /= i;
break;
}
}
cnt--;
}
int ans[5];
int tot = 0;
for (map<int, int>::iterator it = mp.begin(); it != mp.end(); it++) {
if ((*it).second == 1 && (*it).first >= 2) {
ans[++tot] = (*it).first;
}
}
if (tot != 3) {
cout << "NO" << endl;
} else {
cout << "YES" << endl;
cout << ans[1] << " " << ans[2] << " " << ans[3] << endl;
}
}
return 0;
}
| CPP |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | for _ in " "*int(input()):
n=int(input());a=[];i=2
while(len(a)<2 and i*i<n):
if n%i==0:n=n//i;a.append(i)
i+=1
if len(a)==2 and n not in a:print("YES");print(n,*a)
else:print("NO")
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | def factors(x):
result = []
i = 1
while i*i <= x:
if x % i == 0:
result.append(i)
if x//i != i: #
result.append(x//i)
i += 1
return(result)
for _ in range(int(input())):
n = int(input())
lst = sorted(factors(n))
a = lst[1]
x = n//a
lst2 = sorted(factors(x))
if len(lst2) <= 2:
b = -1
elif lst2[1] != a:
b = lst2[1]
elif lst2[1] == a and len(lst2) > 2:
b = lst2[2]
yo = a*b
res = n//yo
if res != a and res!= b and res != 1 and b > 1:
print("YES")
print(a,b,res)
else:
print("NO")
#print(lst2) | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import math
def div(n, m):
i = m+1
while i*i < n:
if(n%i == 0):
return(i, n // i)
i += 1
return(-1, -1)
t = int(input())
for _ in range(t):
n = int(input())
a = 0
i = 2
while i * i < n:
if (n % i == 0):
a = i
break
i += 1
if(a == 0):
print("NO")
else:
b, c = div(n//a, a)
if(b == -1):
print("NO")
else:
print("YES\n",a," ",b," ",c)
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int gcd(int r, int p) {
if (p == 0) return r;
return gcd(p, r % p);
}
struct Test {
int x, y, z;
};
long long int fact(long long int x) {
long long int p = (fact(x - 1) * x) % 1000000007;
return p;
}
void sieveOfEratosthenes(long long int N, long long int s[]) {
vector<bool> prime(N + 1, false);
for (long long int i = 2; i <= N; i += 2) s[i] = 2;
for (long long int i = 3; i <= N; i += 2) {
if (prime[i] == false) {
s[i] = i;
for (long long int j = i; j * i <= N; j += 2) {
if (prime[i * j] == false) {
prime[i * j] = true;
s[i * j] = i;
}
}
}
}
}
map<long long int, long long int> poww;
vector<long long int> vec;
void generatePrimeFactors(long long int N) {
long long int s[N + 1];
sieveOfEratosthenes(N, s);
long long int curr = s[N];
int cnt = 1;
while (N > 1) {
N /= s[N];
if (curr == s[N]) {
cnt++;
continue;
}
poww[curr] = cnt;
vec.push_back(curr);
if (vec.size() > 2) {
return;
}
curr = s[N];
cnt = 1;
}
}
int main() {
cin.tie(NULL);
int t;
cin >> t;
while (t--) {
long long int n;
int f = 0;
cin >> n;
long long int t, tt, ttt;
for (int i = 2; i < sqrt(n); i++) {
if (n % i == 0) {
for (int j = 2; j < sqrt(n); j++) {
if (n / i % j == 0 && i != j && n / (i * j) != j &&
n / (i * j) != i) {
f = 1;
t = i;
tt = j;
ttt = n / (i * j);
break;
}
}
}
}
if (f == 1) {
cout << "YES" << endl;
cout << t << " " << tt << " " << ttt << endl;
} else {
cout << "NO" << endl;
}
}
}
| CPP |
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