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stringlengths 2
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| description
stringlengths 29
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| source
int64 1
7
| difficulty
int64 0
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| solution
stringlengths 7
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| language
stringclasses 4
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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 | #include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
long long int n;
cin >> n;
for (long long int i = 0; i < n; i++) {
long long int a, x;
cin >> a;
x = a;
vector<int> v;
for (long long int i = 2; i <= sqrt(a); i++) {
if (a % i == 0) {
if (a / i == i)
v.push_back(i);
else {
v.push_back(i);
v.push_back(a / i);
}
a = a / i;
}
}
if (v.size() < 2)
cout << "NO" << '\n';
else {
sort(v.begin(), v.end());
if (x / (v[0] * v[1]) > 1 && x / (v[0] * v[1]) != v[0] &&
x / (v[0] * v[1]) != v[1]) {
cout << "YES"
<< "\n";
cout << v[0] << ' ' << v[1] << ' ' << x / (v[0] * v[1]) << '\n';
} else
cout << "NO" << '\n';
}
}
}
| CPP |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | for _ in range(int(input())):
n=int(input())
a=b=c=1
for i in range(2,int((n**0.5)+1)):
if n%i==0:
a=i
break
if a==1:
print('NO')
continue
x=n//a
for i in range(2,int(((x)**0.5)+1)):
if x%i==0 and i!=a:
b=i
break
if b==1:
print('NO')
continue
c=int(n//(a*b))
if c!=1 and c!=a and c!=b:
print('YES')
print(a,b,c)
else:
print('NO') | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | """
This template is made by Satwik_Tiwari.
python programmers can use this template :)) .
"""
#===============================================================================================
#importing some useful libraries.
import sys
import bisect
import heapq
from math import *
from collections import Counter as counter # Counter(list) return a dict with {key: count}
from itertools import combinations as comb # if a = [1,2,3] then print(list(comb(a,2))) -----> [(1, 2), (1, 3), (2, 3)]
from itertools import permutations as permutate
from bisect import bisect_left as bl #
from bisect import bisect_right as br
from bisect import bisect
#===============================================================================================
#some shortcuts
mod = pow(10, 9) + 7
def inp(): return sys.stdin.readline().strip() #for fast input
def out(var): sys.stdout.write(str(var)) #for fast output, always take string
def lis(): return list(map(int, inp().split()))
def stringlis(): return list(map(str, inp().split()))
def sep(): return map(int, inp().split())
def strsep(): return map(str, inp().split())
def graph(vertex): return [[] for i in range(0,vertex+1)]
def zerolist(n): return [0]*n
def nextline(): out("\n") #as stdout.write always print sring.
def testcase(t):
for p in range(t):
solve()
def printlist(a) :
for p in range(0,len(a)):
out(str(a[p]) + ' ')
def lcm(a,b): return (a*b)//gcd(a,b)
#===============================================================================================
# code here ;))
def sieve(a): #O(n loglogn) nearly linear
#all odd mark 1
for i in range(3,((10**5)+1),2):
a[i] = 1
#marking multiples of i form i*i 0. they are nt prime
for i in range(3,((10**5)+1),2):
for j in range(i*i,((10**5)+1),i):
a[j] = 0
a[2] = 1 #special left case
return (a)
a = [0]*((10**5)+1)
a = sieve(a)
# print(a[2])
# print(12345**0.5)
def solve():
n = int(inp())
i = 3
f = 0
while(i*i < n):
if(n%(i) == 0):
f = 1
break
i +=2
if(n%2 == 0):
f = 1
if(f == 0):
print('NO')
else:
i = 2
div = []
cnt = 0
while((i*i) <n):
if(n%i==0):
div.append(i)
n = n//i
cnt +=1
if(cnt == 2):
break
i+=1
if(cnt == 2):
if(n>div[1]):
print('YES')
print(div[0],div[1],n)
else:
print('NO')
else:
print('NO')
testcase(int(inp()))
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible 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())
abc = set()
for i in range(2, int(n**0.5)+1):
if n % i == 0:
if len(abc) == 0:
abc.add(i)
n = n//i
else:
if i != n//i and n//i != 1:
abc.add(i)
abc.add(n//i)
break
if len(abc) == 3:
print("YES")
for num in abc:
print(num,end=" ")
print("")
else:
print("NO") | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | t = int(input())
for _ in range(t):
n= int(input())
import math
temp = int(math.sqrt(n))
done = 0
for i in range(2,temp+1):
if n%i == 0:
a = i
for j in range(2,int(math.sqrt(a))+1):
if a%j == 0 and a/j != a:
x = n//i
y = j
z = a//j
if x!=y and z!=y and z!=x:
done = 1
break
if done == 1:
break
if i != n/i :
b = n//i
for j in range(2,int(math.sqrt(b))+1):
if b%j == 0 and b!=b//j:
x = i ; y = b//j;z = j
if x!=y and z!=y and z!=x:
done = 1
break
if done == 1:
break
if done:
print("YES")
print(x,y,z)
else:
print('NO')
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 |
# import libraries for input/ output handling
# on generic level
import atexit, io, sys
# A stream implementation using an in-memory bytes
# buffer. It inherits BufferedIOBase.
buffer = io.BytesIO()
sys.stdout = buffer
# print via here
@atexit.register
def write():
sys.__stdout__.write(buffer.getvalue())
import math
# A function to print all prime factors of
# a given number n
def primeFactors(n):
m={}
# Print the number of two's that divide n
while n % 2 == 0:
if 2 not in m:
m[2]=1
else:
m[2]+=1
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:
if i not in m:
m[i]=1
else:
m[i]+=1
n = n / i
# Condition if n is a prime
# number greater than 2
if n > 2:
m[n]=1
return m
for _ in range(input()):
n=input()
v=primeFactors(n)
if len(v)>=3:
print"YES"
for k,g in v.items():
t=pow(k,g);break
for k,g in v.items():
r=pow(k,g);
if r!=t:
break
print t,r,n/(t*r)
elif len(v)==2:
c=0;d=[];t=0;p,q,r=[],[],[]
for k,g in v.items():
d.append(pow(k,g))
t+=g
if g>2:
p.append(k)
if g==2:
q.append(k)
else:
r.append(k)
if t<=3:
print"NO"
else:
print"YES";
if len(p):
k=p[0]
print k,k*k,n/(k*k*k)
elif len(q):
print q[0],q[1],n/(q[0]*q[1])
else:
for k,g in v.items():
if g>5:
print"YES";print k,k*k,n/(k*k*k)
else:
print"NO"
| PYTHON |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
long long n, m, ni, mi;
long long cn;
string s;
long long num[100005];
long long ok = false;
deque<long long> v;
map<string, bool> M;
map<string, bool> MC;
long long sch(long long a, long long b, long long c, long long i) {
string s = to_string(a) + " " + to_string(b) + " " + to_string(c) + " " +
to_string(i);
if (M[s]) return MC[s];
if (a > 1 && b > 1 && c > 1) {
if (a != b && a != c && b != c) {
if (a * b * c == cn) {
cout << "YES" << endl;
cout << a << " " << b << " " << c << endl;
return 1;
}
}
}
long long na, nb, nc;
if (i == v.size()) return 0;
na = a * v[i];
nb = b * v[i];
nc = c * v[i];
M[s] = 1;
MC[s] = sch(na, b, c, i + 1) || sch(a, nb, c, i + 1) || sch(a, b, nc, i + 1);
return MC[s];
}
void solve() {
cin >> n;
cn = n;
v.clear();
M.clear();
MC.clear();
for (long long i = 2; i * i < n + 5; i++) {
if (n % i == 0) {
while (n % i == 0) {
v.push_back(i);
n /= i;
}
}
}
if (n != 1) {
v.push_back(n);
}
if (!sch(1, 1, 1, 0)) {
cout << "NO" << endl;
}
}
int32_t main() {
cin.tie(0);
ios_base::sync_with_stdio(NULL);
long long t = 1;
cin >> t;
while (t--) solve();
return 0;
}
| CPP |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | ////////////////////---------------------------SHUBHAM CHAUDHARI-------------------------------///////////////////////
import java.util.*;
public class Main {
public static void main(String[] args) {
Scanner in=new Scanner(System.in);
int t=in.nextInt();
StringBuilder res=new StringBuilder();
while (t-->0)
{
int n=in.nextInt();
ArrayList<Integer>primes=primeFactors(n);
HashSet<Integer>set=new HashSet<>();
for(int i=0;i<primes.size();i++)
{
set.add(primes.get(i));
}
primes=new ArrayList<>();
for(int i:set)
{
primes.add(i);
}
int tmp=n;
int size=0;
for(int i=0;i<primes.size();i++)
{
while (tmp%primes.get(i)==0)
{
tmp/=primes.get(i);
size++;
}
}
//System.out.println(primes.size());
if(primes.size()==1)
{
if(size>=6)
{
res.append("YES\n");
//System.out.println("ji");
res.append(primes.get(0)+" "+(primes.get(0)*primes.get(0))+" "+(n/(primes.get(0)*primes.get(0)*primes.get(0)))+"\n");
}
else
res.append("NO\n");
}
else if(primes.size()==2)
{
if(size>3)
{
res.append("YES\n");
res.append(primes.get(0)+" "+primes.get(1)+" "+(n/(primes.get(0)*primes.get(1)))+"\n");
}
else
{
res.append("NO\n");
}
}
else
{
res.append("YES\n");
res.append(primes.get(0)+" "+primes.get(1)+" "+(n/(primes.get(0)*primes.get(1)))+"\n");
}
}
System.out.println(res);
}
public static ArrayList<Integer> primeFactors(int n)
{
ArrayList<Integer>list=new ArrayList<>();
while (n%2==0)
{
list.add(2);
n /= 2;
}
for (int i = 3; i <= Math.sqrt(n); i+= 2)
{
while (n%i == 0)
{
list.add(i);
n /= i;
}
}
if (n > 2)
list.add(n);
return list;
}
static class Node implements Comparable<Node>
{
int x;
int y;
public Node(int x,int y)
{
this.x=x;
this.y=y;
}
@Override
public int compareTo(Node o) {
if(x>o.x)
{
return 1;
}
if(x==o.x)
{
if(y>o.y)
return 1;
else if(y<o.y)return -1;
return 0;
}
return -1;
}
}
} | JAVA |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import collections
import itertools
import sys
from collections import defaultdict, Counter
from math import sqrt, ceil
input = sys.stdin.readline
############ ---- Input Functions ---- ############
def inp():
return (int(input()))
def inlt():
return (list(map(int, input().split())))
def insr():
s = input()
return (list(s[:len(s) - 1]))
def invr():
return (map(int, input().split()))
def ds(d, s):
for i in range(s, d + 1):
if i*i>d:
break
if d % i == 0:
return i
def main():
n = inp()
for _ in range(n):
d = inp()
d1 = ds(d, 2)
if d1 is not None:
d2 = ds(d // d1, d1 + 1)
if d2 is not None:
d3 = d // d1 // d2
if d3 != 1 and d3 != d1 and d3 != d2:
print("YES")
print(" ".join([str(v) for v in [d1, d2, d3]]))
else:
print("NO")
else:
print("NO")
else:
print("NO")
if __name__ == "__main__":
# sys.setrecursionlimit(10 ** 6)
# threading.stack_size(10 ** 8)
# t = threading.Thread(target=main)
# t.start()
# t.join()
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 | from math import sqrt
n = int(input())
for i in range(n):
f = int(input())
a = []
for j in range(2, int(sqrt(f))):
if f % j == 0:
a.append(j)
f = f // j
if len(a) == 2:
break
if len(a)==2 and a.count(f)==0:
print("YES")
print(a[0], a[1], f)
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 | '''input
3
1 2
3 4
5 6
'''
import math
# Method to print the divisors
def printDivisors(n) :
list = []
# List to store half of the divisors
a = []
x = n
for i in range(2, int(math.sqrt(n) + 1)) :
if (x % i == 0) :
a.append(i)
# i -= 1
x/=i
if len(a)==2:
p = 1
# for i in range(2,len(a)):
# p*=a[i]
if x==a[0] or x==a[1]:
print("NO")
return
print("YES")
print(a[0], end = " ")
print(a[1], end = " ")
print(int(x))
return
print("NO")
t = int(input())
# print(t)
for i in range(t):
a = int(input())
# print((60-m) + (23-h)*60)
printDivisors(a) | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | from math import sqrt, floor
t = int(input())
for _ in range(t):
n = int(input())
if n < 24:
print("NO")
else:
m = floor(sqrt(n)) + 1
d = 0
divs = []
for i in range(2, m):
if n % i == 0:
d += 1
divs.append(i)
n //= i
break
if d == 0:
print("NO")
else:
m = floor(sqrt(n)) + 1
for i in range(divs[0]+1, m):
if n % i == 0:
d += 1
divs.append(i)
n //= i
break
if d == 2:
if n in divs:
print("NO")
else:
divs.append(n)
ans = map(str, divs)
print("YES")
print(' '.join(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 _ in[0]*int(input()):
n=int(input());a=['YES'];i=j=2
while j<4and i*i<n:
if n%i<1:a+=i,;n//=i;j+=1
i+=1
print(*(a+[n],['NO'])[j<4][:4]) | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import math
t = int(input())
for j in range(t):
n = int(input())
a = 1
b = 1
c = 1
for i in range(2, int(math.sqrt(n)) + 1):
if n % i == 0:
a = i
break
m = n // a
for i in range(a + 1, int(math.sqrt(m)) + 1):
if m % i == 0:
b = i
c = m // i
break
if (a > 1 and b > 1) and c > b:
print("YES")
print(a, b, c)
else:
print("NO") | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
double squrt(int64_t num) {
double high = num, low = 0, mid;
while (high - low > 0.00001) {
mid = low + (high - low) / 2;
if (mid * mid > num) {
high = mid;
} else {
low = mid;
}
}
return mid;
}
void solve() {
int64_t num;
cin >> num;
for (int64_t i = 2; i * i < num; i++) {
if (num % i != 0) continue;
for (int64_t j = i + 1; j * j <= (num / i); j++) {
int64_t x;
if (num % (i * j) == 0) {
int64_t x = num / (i * j);
if (x != i && i != j && x != j) {
cout << "YES"
<< "\n";
cout << i << " " << j << " " << x << "\n";
return;
}
}
}
}
cout << "NO"
<< "\n";
}
int32_t main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
;
cout.precision(10);
cout << fixed;
;
int64_t test;
cin >> test;
while (test--) {
solve();
}
return 0;
}
| CPP |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | #include <bits/stdc++.h>
#pragma GCC optimize("O3")
using namespace std;
int add(int a, int b) {
long long x = a + b;
if (x >= 1000000007) x -= 1000000007;
if (x < 0) x += 1000000007;
return x;
}
long long mul(long long a, long long b) { return (a * b) % 1000000007; }
long long pw(long long a, long long b) {
long long ans = 1;
while (b) {
if (b & 1) ans = (ans * a) % 1000000007;
a = (a * a) % 1000000007;
b >>= 1;
}
return ans;
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
long long rand_seed() {
long long a = rng();
return a;
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int t;
cin >> t;
for (; t; --t) {
int n;
cin >> n;
int aa = 0, bb = 0, cc = 0;
for (int i = 2; i * i <= n; ++i)
if (n % i == 0) {
aa = i;
n /= i;
break;
}
for (int i = 2; i * i <= n; ++i)
if (n % i == 0 && i != aa) {
bb = i;
n /= i;
break;
}
cc = n;
if (aa && bb && cc && aa != bb && aa != cc && bb != cc) {
cout << "YES\n";
cout << aa << " " << bb << " " << cc << '\n';
} else
cout << "NO\n";
}
return 0;
}
| CPP |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const int MAX_N = -1;
const int inf = 2e9 + 19;
vector<long long> vec;
void primeDiv(long long n) {
for (int i = 2; i * i <= n; i++) {
if (n % i == 0) {
while (n % i == 0) vec.push_back(i), n /= i;
}
}
if (n) vec.push_back(n);
}
int main() {
ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
long long Q;
cin >> Q;
while (Q--) {
long long t, a = 1, b = 1, c = 1;
cin >> t;
vec.clear();
primeDiv(t);
if ((int)vec.size() < 3) {
cout << "NO" << endl;
continue;
}
a = vec[0];
b = vec[1];
bool flag = false;
if (a == b) b *= vec[2], flag = true;
for (int i = 2 + flag; i < (int)vec.size(); i++) {
c *= vec[i];
}
if (a < 2 || b < 2 || c < 2 || a == b || b == c || a == c)
cout << "NO" << endl;
else
cout << "YES" << endl << a << ' ' << b << ' ' << c << 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 | t = int(input())
for i in range(t):
n = int(input())
dvdrs = []
passs = True
if n < 24:
print("NO")
continue
for k in range(2,n):
if k*k*k > n:
passs = False
break
if n%k == 0:
dvdrs += [k]
n = int(n/k)
break
if passs:
for k in range(dvdrs[0]+1, n):
if k*k > n:
passs = False
break
if n%k == 0:
dvdrs += [k]
n = int(n/k)
break
if passs and n != dvdrs[0] and n != dvdrs[1]:
print("YES")
print(str(dvdrs[0]) + " " + str(dvdrs[1]) + " " + str(n))
else:
print("NO")
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import math
def div(n):
l = []
i = 2
while i <= math.sqrt(n):
if (n % i == 0):
if (n / i == i):
l.append(int(i))
else:
l.append(int(i))
l.append(int(n/i))
i = i + 1
l.sort()
return l
for _ in range(int(input())):
n = int(input())
y = n
l = div(n)
l = l[::-1]
#print(l)
o = []
if len(l) != 0:
y = y//l[-1]
x1 = l[-1]
o.append(l[-1])
l.pop()
else:
print("NO")
continue
if len(l) != 0:
f = 1
while(f == 1 and (y % l[-1] != 0)):
l.pop()
if len(l) == 0:
f = 0
if f == 0:
print("NO")
continue
y = y // l[-1]
o.append(l[-1])
x2 = l[-1]
l.pop()
if len(l) == 0:
print("NO")
continue
if (x2 == y) or (x1 == x2) or (x1 == y) or (x1 == 1) or (x2 == 1) or (y == 1):
print("NO")
continue
o.append(y)
else:
print("NO")
continue
cnt = 1
for i in o:
cnt = cnt*i
if cnt == n:
print("YES")
for i in o:
print(i,end = " ")
print("")
else:
print("NO") | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import java.util.*;
import java.lang.Math;
import java.io.* ;
public class Account {
class Pair {
int key ;
int val ;
Pair(int key ,int val) {
this.key = key ;
this.val = val ;
}
int getKey() {
return this.key ;
}
int getVal() {
return this.val ;
}
}
public static void main (String[] args) {
// Scanner sc= new Scanner(System.in) ;
FastReader sc= new FastReader() ;
// StringBuilder res= new StringBuilder() ;
int t= 1 ;
t= sc.nextInt() ;
while(t-- > 0) {
long n= sc.nextLong() ;
int rt= (int) Math.sqrt(n) ;
int x= -1 ;
boolean flg= false ;
int m= rt - ((rt*rt == n) ? 1 : 0) ;
for(int i=2 ;i<=m ;i++) {
if(x == -1 && n%i == 0) {
x= i ;
}else if(n%(x*i) == 0) {
long num= (n) / (x*i) ;
if(num != x && num != i) {
flg= true ;
System.out.println("YES");
System.out.println(x+" "+i+" "+num);
}
break ;
}
}if(flg == false) {
System.out.println("NO");
}
}
}
public static int gcd(int a, int b)
{
if (b == 0)
return a;
return gcd(b, a % b);
}
public 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;
}
int [] readintarray(int n) {
int res [] = new int [n];
for(int i = 0; i<n; i++)res[i] = nextInt();
return res;
}
long [] readlongarray(int n) {
long res [] = new long [n];
for(int i = 0; i<n; i++)res[i] = nextLong();
return res;
}
}
} | JAVA |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import math
def primeFactors(n):
li =[]
while n % 2 == 0:
li.append(2)
#print (2)
n = n//2
for i in range(3,int(math.sqrt(n))+1,2):
while n % i== 0:
##print (i)
li.append(i)
n = n//i
if n > 2:
li.append(n)
#print (n)
return li
for _ in range(int(input())):
n = int(input())
li = primeFactors(n)
d = dict()
for i in li:
if i not in d:
d[i] = 1
else:
d[i] += 1
#print(d)
if(len(d) == 1):
x = li[0]
if(d[x]>=6):
print("YES")
print(x,x*x,pow(x,d[x]-3))
else:
print("NO")
if(len(d)>=3):
x = 1
lee = []
for i in d:
lee.append((i,d[i]))
a = pow(lee[0][0],lee[0][1])
b = pow(lee[1][0],lee[1][1])
c = 1
for i in range(2,len(d)):
c = c*pow(lee[i][0],lee[i][1])
print("YES")
print(a,b,c)
if(len(d) ==2):
lee = []
for i in d:
lee.append((i,d[i]))
if(lee[0][1]>=3):
print("YES")
x= lee[0][1]
print(lee[0][0], pow(lee[0][0], lee[0][1]-1), pow(lee[1][0],lee[1][1]))
elif(lee[1][1]>=3):
print("YES")
x= lee[1][1]
print(pow(lee[0][0],lee[0][1]), lee[1][0], pow(lee[1][0], lee[1][1]-1))
elif(lee[0][1] == 2 and lee[1][1]==2):
print("YES")
print(lee[0][0],lee[0][0]*lee[1][0], lee[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 | #include <bits/stdc++.h>
using namespace std;
using ll = long long;
using lli = long long int;
using ld = long double;
const int N = 2e5 + 5, inf = 2e9;
bool valid(int a, int b, int c) {
if (a == b || b == c || a == c) return false;
return true;
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int t;
cin >> t;
while (t--) {
int n;
cin >> n;
int x = 0, y = 0;
vector<pair<int, int>> v;
set<int> st;
for (int i = 2; i * i <= n; i++) {
if (n % i == 0) {
v.push_back({i, n / i});
}
}
int a = 0, b = 0, c = 0;
int found = false;
for (auto p : v) {
for (int i = 2; i * i <= p.first; i++) {
if (p.first % i == 0) {
a = i;
b = p.first / i;
c = p.second;
if (valid(a, b, c)) {
found = true;
break;
}
}
}
if (found) break;
for (int i = 2; i * i <= p.second; i++) {
if (p.second % i == 0) {
a = i;
b = p.second / i;
c = p.first;
if (valid(a, b, c)) {
found = true;
break;
}
}
}
if (found) break;
}
if (found) {
cout << "YES"
<< "\n";
cout << a << " " << b << " " << c << "\n";
} else
cout << "NO"
<< "\n";
}
return 0;
}
| CPP |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import java.util.*;
public class geek {
public static void main(String[] args) {
Scanner s = new Scanner(System.in);
int t = s.nextInt();
while (t-- > 0) {
int n = s.nextInt();
int a = 0;
int b = 0;
int c = 0;
boolean flag = true;
boolean flag1 = true;
boolean flag2 = true;
StringBuffer sb = new StringBuffer();
for (int i = 2; i <= Math.sqrt(n); i++) {
if (n % i == 0) {
a = i;
flag = false;
break;
}
}
if (flag) {
System.out.println("NO");
} else {
int k = n / a;
for (int i = a + 1; i <= Math.sqrt(k); i++) {
if (k % i == 0) {
b = i;
break;
}
}if(b<2){
System.out.println("NO");
}else {
int p = k / b;
if (p != a && p != b && a != b) {
System.out.println("YES\n" + "" + a + " " + b + " " + p);
} else {
System.out.println("NO");
}
}
}
}
}
}
| JAVA |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import sys
from collections import defaultdict as dc
import math
from bisect import bisect, bisect_left,bisect_right
input=sys.stdin.readline
for _ in range(int(input())):
n=int(input())
i=2
f=0
#p=math.sqrt(n)
while(i<=math.sqrt(n)):
if n%i==0:
f=1
break
i+=1
if f==0:
#print("c1")
print("NO")
else:
f=0
n=n//i
j=2
while(j<=math.sqrt(n)):
if j not in [1,i] and n%j==0:
f=1
break
j+=1
if f==0:
#print("c2")
print("NO")
else:
n=n//j
if i!=j and n!=i and n!=j:
print("YES")
print(i,j,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(n):
if n == 1:
return []
if n == 2:
return [2]
i = 2
while i*i <= n:
if n%i == 0:
return prime(n//i)+[i]
i += 1
return [n]
for _ in range(int(input())):
n = int(input())
a = sorted(prime(n))
if len(set(a)) >= 3:
print("YES")
prod = 1
for x in a[1:-1]:
prod *= x
print(a[0], prod, a[-1])
if len(set(a)) == 2:
if len(a) >= 4:
print("YES")
prod = 1
for x in a[1:-1]:
prod *= x
print(a[0], prod, a[-1])
else:
print("NO")
if len(set(a)) == 1:
if len(a) >= 6:
print("YES")
print(a[0], a[0]*a[0], a[0]**(len(a)-3))
else:
print("NO")
if a == []:
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 prod(li):
p = 1
for i in li: p*=i
return p
def getAns(n):
d = dict()
while n % 2 == 0:
if 2 in d.keys():
d[2] += 1
else:
d[2] = 1
n//=2
for i in range(3, int(math.sqrt(n)) + 3, 2):
while n % i == 0:
if i in d.keys():
d[i] += 1
else:
d[i] = 1
n//=i
if n > 2:
d[n] = 1
k = list(d.keys())
if len(k) >= 3:
a = k[0]**d[k[0]]
b = k[1]**d[k[1]]
c = prod([(k[i]**d[k[i]]) for i in range(2, len(k))])
elif len(k) == 2:
if d[k[0]] >= 3:
a = k[0]
b = k[0]**(d[k[0]] - 1)
c = k[1]**(d[k[1]])
else:
if d[k[1]] >= 3:
a = k[0]**d[k[0]]
b = k[1]
c = k[1]**(d[k[1]] - 1)
elif d[k[1]] == 2 and d[k[0]] == 2:
a = k[0]
b = k[1]
c = k[0]*k[1]
else:
return False, -1, -1, -1
else:
if d[k[0]] >= 6:
a = k[0]
b = k[0] * k[0]
c = k[0]**(d[k[0]] - 3)
else:
return False, -1, -1, -1
return True, a, b, c
for _ in range(int(input())):
n = int(input())
truth,a,b,c = getAns(n)
if truth:
print("YES")
print(*sorted([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 string
import math
import collections
def LeReseauLMD(n):
orin = n
a,b,c = 0,0,0
for a in range(2,math.floor(n**0.5)+1):
if n%a==0:
n //=a
for b in range(2,math.floor((n)**0.5)+1):
if n%b == 0 and b != a:
"""for c in range(2,math.floor(n**0.5)+1):
if n%c == 0 and c != b and c != a:
break"""
c = n//b
if c == a or c == b:
c = 0
break
break
if a*b*c == orin:
return f"{a} {b} {c}"
else:
return "NO"
def main():
q = int(input())
for _ in range(q):
n = int(input())
rep = LeReseauLMD(n)
if rep != "NO":
print("YES")
print(rep)
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 | #include <bits/stdc++.h>
using namespace std;
const int max6 = 1e6 + 6;
map<int, int> rem;
int e[max6], p[max6];
int main() {
int tests;
scanf("%d", &tests);
for (int i = 1; i <= 1000000; ++i) e[i] = i;
int sl = 0;
for (int i = 2; i <= 1000000; ++i)
if (e[i] == i) {
p[++sl] = i;
for (int j = 1; j * i <= 1000000; ++j) e[i * j] = i;
}
while (tests--) {
int n;
scanf("%d", &n);
int _n = n;
rem.clear();
sl = 0;
int j = 1;
while (n > 1000000) {
int t;
while (p[j] <= n / p[j] && n % p[j] != 0) j++;
if (p[j] > n / p[j])
t = n;
else
t = p[j];
while (n % t == 0) {
rem[t]++;
n = n / t;
}
}
while (n > 1) {
int p = e[n];
rem[p]++;
n = n / p;
}
auto v = rem.begin();
int a = (*v).first;
rem[a]--;
int b = 1;
bool ok = false;
int c;
while (b == 1 || b == a) {
if ((*v).second == 0) {
v++;
if (v == rem.end()) break;
}
(*v).second--;
b = b * (*v).first;
}
c = _n / (a * b);
if (a >= 2 && b >= 2 && c >= 2)
if (a != b && b != c && c != a) ok = true;
if (ok) {
cout << "YES\n";
cout << a << " " << b << " " << c << "\n";
} else
cout << "NO\n";
}
}
| CPP |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | for _ in range(int(input())):
i = 2
l = []
n = int(input())
size = 0
while(i*i<=n and size < 2):
if(n % i == 0):
l.append(i)
n //= i
size += 1
i += 1
if(size == 2 and (l[0]!=n and l[1]!=n)):
print('YES')
print(l[0],l[1],n)
else:
print('NO') | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | def read_ints(): return map(int, input().split())
def read_int(): return int(input())
def factors(n):
factors_p = dict()
twos = 0
while n % 2 == 0:
n //= 2
twos += 1
if twos > 0:
factors_p[2] = twos
d = 3
while d * d <= n:
p = 0
while n % d == 0:
p += 1
n //= d
if p > 0:
factors_p[d] = p
d += 2
if n > 1:
factors_p[n] = 1
return factors_p
t = read_int()
for _ in range(t):
n = read_int()
primes = factors(n)
p_sum = 0
abc = []
for k, v in primes.items():
p_sum += v
if len(primes) < 1:
print("No")
continue
elif len(primes) == 1:
if p_sum < 6:
print("No")
continue
else:
print("Yes")
d = list(primes.keys())[0]
abc = [d, d ** 2, n // (d ** 3)]
elif len(primes) == 2:
psum = 0
for k, v in primes.items():
psum += v
if psum < 4:
print("No")
continue
print("Yes")
p1, p2 = list(primes.keys())[:2]
abc = [p1, p2, n // (p1 * p2)]
elif len(primes) >= 3:
print("Yes")
p1, p2 = list(primes.keys())[:2]
abc = [p1, p2, n // (p1 * p2)]
print(*abc)
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible 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 floor, sqrt, ceil
t = int(input())
for _ in range(t):
a = -1
b = -1
n = int(input())
i = 2
while i*i < n:
if n % i == 0:
a = i
n //= a
break
i += 1
i = 2
while i*i < n:
if n % i == 0 and i != a:
b = i
n //= b
break
i += 1
if a == -1 or b == -1 or n == 1 or n == a or n == b:
print('NO')
continue
print('YES')
print(a, b, n)
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | m=int(input())
for l in range(m):
n=int(input())
q,t=int(n**.5)+1,0
for i in range(2,q):
if n%i==0:
a=n//i
w=int(a**.5)+1
for j in range(2,w):
if a%j==0:
r=a//j
if r!=j and j!=i and r!=i:
print("YES")
print(i,j,r)
t=1
break
if t:
break
if not t:
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
m=1000000007
def fact(n):
ans=1
for i in range(1,n+1):
ans=((ans%m)*(i%m))%m
return ans
def power_2(n):
ans=1
for i in range(n):
ans=((ans%m)*(2))%m
return ans
for z in range(int(input())):
n=int(input())
fin=int(math.ceil(n**(1/3)))
cnt,i=0,2
# to find the first no in three we check the
#value when it is less than cube root of n
# becouse left no(2nd and 3rd) must be greater
#than first becoz any no.less than first is not
#able to divide n
while(i<=fin):
if(n%i==0):
cnt+=1
n=int(n/i)
break
i+=1
fin=int(math.ceil(n**(1/2)))
j=i+1
while(j<=fin):
if(n%j==0 and j!=int(n/j) and int(n/j)!=i):
cnt+=1
break
j+=1
if(cnt==2):
print("YES")
print(i,j,int(n/j))
else:
print("NO") | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import sys
sys.setrecursionlimit(10 ** 6)
int1 = lambda x: int(x) - 1
p2D = lambda x: print(*x, sep=" ")
def II(): return int(sys.stdin.readline())
def MI(): return map(int, sys.stdin.readline().split())
def LI(): return list(map(int, sys.stdin.readline().split()))
def LLI(rows_number): return [LI() for _ in range(rows_number)]
def SI(): return sys.stdin.readline()[:-1]
g = II()
for j in range(g):
n = II()
fl = 0
i = 2
used = set()
while i*i*i<n:
if n % i == 0:
used.add(i)
n //= i
break
i += 1
if len(used) == 0:
print("NO")
continue
i += 1
while i * i < n:
if n % i == 0:
used.add(i)
n //= i
break
i += 1
if len(used) == 1:
print("NO")
continue
if n not in used:
used.add(n)
print("YES")
p2D(used)
else:
print("NO")
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import sys
input=sys.stdin.readline
t=int(input())
for i in range(t):
x=int(input())
ar=[]
l=0
a=1
b=1
c=1
ct=0
bl=False
for j in range(2,int(x**(0.5))):
if(x%j==0 and ct==0):
a=j
ct+=1
elif(x%j==0 and ct==1):
temp=j
if(x%(a*temp)==0 and x//(a*temp)!=a and x//(a*temp)!=temp):
b=temp
c=x//(a*temp)
bl=True
break
if(bl==True):
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 | from sys import stdin, stdout
import sys
INF=1e9
import bisect
def get_int(): return int(stdin.readline().strip())
def get_ints(): return map(int,stdin.readline().strip().split())
def get_array(): return list(map(int,stdin.readline().strip().split()))
def get_string(): return stdin.readline().strip()
def op(c): return stdout.write(c)
#from collections import defaultdict
import math
for _ in range(int(stdin.readline())):
n=get_int()
a,b,c=0,0,0
f=0
for i in range(2,int(math.sqrt(n))+1):
if n%i==0:
a=i
n=n//i
for j in range(i+1,int(math.sqrt(n))+1):
if n%j==0:
b=j
n=n//j
f=1
break
if f==1:
break
if f==1 and n not in [a,b] :
print("YES")
print(a,b,n)
continue
print("NO")
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | for _ in range(int(input())):
num = int(input())
list_of_nums = [1]
for i in range(2, int(pow(num, .5)+1)):
if num % i == 0:
num = num // i
list_of_nums.append(i)
break
# print(list_of_nums)
for i in range(2, int(pow(num, .5))+1):
if num % i == 0 and i not in list_of_nums:
num //= i
list_of_nums.append(i)
break
# print(list_of_nums)
if num not in list_of_nums and len(list_of_nums) > 2:
print("YES")
list_of_nums = [str(x) for x in list_of_nums]
print(num, ' '.join(list_of_nums[1:]))
else:
print("NO")
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | 'use strict'
const problem = (n) => {
for (let i = 2; i < Math.cbrt(n); i++) {
if (n % i === 0) {
n /= i;
for (let j = i + 1; j < Math.sqrt(n); j++) {
if (n % j === 0) return `YES\n${i} ${j} ${n/j}`;
}
}
}
return 'NO';
}
let t = +readline();
while(t--) print(problem(+readline()));
| 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 my_def(x):
a, b, c = 1, 1, 1
for j in range(2, int(x ** 0.5) + 2):
while x % j == 0:
if a == 1:
a = j
elif b == 1 or b == a:
b *= j
else:
return (a, b, x)
x //= j
return (a, b, x)
n = int(input())
for i in range(n):
x = int(input())
a, b, c = my_def(x)
if a != b and b != c and a != c and a != 1 and b != 1 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 | from math import sqrt
def calcul(x,message):
for j in range (2,int(sqrt(x))):
if x%j==0:
m=x/j
s=int(sqrt(m))
if j+1>s:
s=int(m/2)
for k in range (j+1,s+1):
if (m%k==0) and (int(m/k)!=k) and (int(m/k)!=j):
message="YES"
print(message)
print(j,' ',k,' ',int(m/k))
return -1
n = int(input())
l=[]
for i in range (0,n):
l.append(int(input()))
for i in range (0,n):
message="NO"
if calcul(l[i],message)!=-1:
print(message) | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | from math import sqrt
t=int(input())
for w in range(t):
n=int(input())
c=0
ans=[]
for i in range(2,int(sqrt(n))+1):
if n==1 or c==2:
break
if n%i==0:
ans.append(i)
n=n//i
c+=1
if c<2 or n==ans[1] or n==ans[0]:
print("NO")
else:
print("YES")
print(ans[0],ans[1],n) | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import math
from collections import Counter,defaultdict
I =lambda:int(input())
M =lambda:map(int,input().split())
LI=lambda:list(map(int,input().split()))
for _ in range(I()):
n=I();i=2;a,b,c=-1,-1,-1
while i*i<=n:
while n%i==0:
n//=i
if a==-1:
a=i
elif b==-1 or a==b:
if b==-1:b=i
else:b*=i
else:
if c==-1:c=i
else:c*=i
i+=1
if n>1:
if c==-1:c=n
else:c*=n
if min(a,b,c)==-1 or b==c or a==b or a==c:
print("NO")
else:
print("YES")
print(a,b,c)
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | #include <bits/stdc++.h>
int main() {
int t;
scanf("%d", &t);
while (t--) {
int n;
scanf("%d", &n);
int flag, a, b, c;
flag = 1;
if (n % 2 == 0) {
a = 2;
flag = 0;
} else
for (a = 3; a * a * a <= n; a += 2)
if (n % a == 0) {
flag = 0;
break;
}
if (!flag) {
int flag2 = 0;
n /= a;
flag = 1;
if (n % 2 == 0) flag2 = 1;
for (b = a; b * b <= n;) {
if ((n % b == 0) && (a != b)) {
flag = 0;
break;
}
b++;
}
if (!flag) {
n /= b;
flag = 1;
if ((n != a) && (n != b) && (n >= 2)) flag = 0;
}
}
if (flag)
printf("NO\n");
else
printf("YES\n%d %d %d\n", a, b, 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 code {
static int sqrt(int x) {
for(int i=2; true; i++) {
if(i*i==x) {
return i;
}
if(i*i>x) {
return i-1;
}
}
}
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
for(int i=0; i<n; i++) {
int a= sc.nextInt();
int b= sqrt(a);
int c=0;
double[] arr= new double[2];
for(double j=2; j<=b; j++) {
if((a/j)%1==0) {
arr[c]=j;
c++;
}
if(c==2) {
double z=((a/arr[0])/arr[1]);
if(z%1==0 && z!=arr[1] && z!=arr[0]) {
break;
}
else {
c--;
}
}
}
if(c==2) {
System.out.println("YES");
System.out.println((int)arr[0]+" "+(int)arr[1]+" "+(int)((a/arr[0])/arr[1]));
}
else {
System.out.println("NO");
}
}
}
} | JAVA |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | def find_multipliers(n):
if n < 8:
return None
a = b = c = None
for i in range(2, int(n**0.5 + 1)):
if n % i == 0:
a = i
break
if a is None:
return None
n /= a
for i in range(2, int(n**0.5 + 1)):
if (i != a) and (n % i == 0):
b = i
break
if b is None:
return None
c = n / b
if (c == a) or (c == b):
return None
return a, b, c
t = int(input())
for _ in range(t):
n = int(input())
result = find_multipliers(n)
if result is None:
print("NO")
else:
print("YES")
print(" ".join(map(str, result)))
| PYTHON |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | def Sieve(n):
ret = []
divlis = [-1] * (n+1)
flag = [True] * (n+1)
flag[0] = False
flag[1] = False
ind = 2
while ind <= n:
if flag[ind]:
ret.append(ind)
ind2 = ind ** 2
while ind2 <= n:
flag[ind2] = False
divlis[ind2] = ind
ind2 += ind
ind += 1
return ret,divlis
t = int(input())
ret,divs = Sieve(10**6)
for loop in range(t):
n = int(input())
ans = []
now = 1
flag = False
for i in ret:
while n % i == 0:
n //= i
now *= i
if now > 1 and now not in ans:
ans.append(now)
now = 1
if len(ans) == 2:
if n not in ans and n > 1:
ans.append(n)
print ("YES")
print (" ".join(map(str,ans)))
flag = True
break
else:
print ("NO")
flag = True
break
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 | #include <bits/stdc++.h>
using namespace std;
int main() {
long long int T;
cin >> T;
while (T--) {
long long int n, k = 0, i, j, b, c, flag = 0, p = 0;
cin >> n;
vector<int> a;
for (int i = 1; i <= sqrt(n); i++) {
if (n % i == 0) {
if (n / i == i)
a.push_back(i);
else {
a.push_back(i);
a.push_back(n / i);
}
}
}
sort(a.begin(), a.end());
for (i = 0; i < a.size(); i++) {
if (a[i] < 2)
continue;
else {
for (j = 0; j < a.size(); j++) {
if (a[j] < 2)
continue;
else {
for (k = 0; k < a.size(); k++) {
if (a[k] < 2)
continue;
else if (a[i] * a[j] * a[k] == n && a[i] != a[j] &&
a[j] != a[k] && a[k] != a[i]) {
cout << "YES" << endl;
cout << a[i] << " " << a[j] << " " << a[k] << endl;
flag = 1;
break;
}
}
}
if (flag == 1) break;
}
}
if (flag == 1) break;
}
if (flag == 0) cout << "NO" << endl;
}
return 0;
}
| CPP |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | for _ in range(int(input())):
n=int(input())
l=[]
a=2
m=n
while a*a<=n:
if(n%a==0):
l.append(a)
n=n//a
break
a+=1
b=2
while b*b<=n:
if(n%b==0):
if b not in l:
l.append(b)
n=n//b
break
b+=1
if(len(l)==2):
c=m//(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 | t = int(input())
for i in range(t):
N = int(input())
answer = []
F = 0
for i in range(2,int(N**0.5)+1):
if N%i == 0:
X = i
Z = N//i
for j in range(2,int(Z**0.5)+1):
if Z%j==0:
ANS1=Z//j
if ANS1!=j and ANS1!=X and j!=X:
print("YES")
print(ANS1,j,X)
F=1
break
if F==1:
break
if F==0:
print("NO") | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | from bisect import bisect_left as bl
from bisect import bisect_right as br
import heapq
import math
from collections import *
from functools import reduce,cmp_to_key
import sys
input = sys.stdin.readline
# M = mod = 998244353
def factors(n):return sorted(set(reduce(list.__add__, ([i, n//i] for i in range(1, int(n**0.5) + 1) if n % i == 0))))
# def inv_mod(n):return pow(n, mod - 2, mod)
def li():return [int(i) for i in input().rstrip('\n').split()]
def st():return input().rstrip('\n')
def val():return int(input().rstrip('\n'))
def li2():return [i for i in input().rstrip('\n').split(' ')]
def li3():return [int(i) for i in input().rstrip('\n')]
def primefactors(n):
l = []
while not n%2:
l.append(2)
n//=2
for i in range(3,int(math.sqrt(n)) + 1,2):
while not n%i:
l.append(i)
n//=i
if n>2:
l.append(n)
return l
# # https://www.geeksforgeeks.org/print-all-prime-factors-of-a-given-number/
# 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(i)
# n = n // i
# # Condition if n is a prime
# # number greater than 2
# if n > 2:
# l.append(n)
# return l
for _ in range(val()):
l = sorted(primefactors(val()))
ans = 'YES'
if len(l) < 3:
print('NO')
continue
first = l[0]
second = 1
# print(l)
for i in range(1,len(l)-1):
secondlast = l[i-1]
second*=l[i]
last = l[i]
third = l[-1]
if second == third:
second//=last
third*=last
if third == first:
second//=secondlast
third*=secondlast
if second == first or second == third or first == third or 1 in set([first,second,third]):
print('NO')
continue
print('YES')
print(first,second,third)
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 |
import java.io.*;
import java.util.*;
import static java.lang.Math.max;
public class Main {
public static void main(String[] args) throws IOException {
run();
end();
}
static void run() throws IOException {
int t = nextInt();
while (t-- > 0) {
solve(nextInt());
}
}
private static void solve(int n) {
for (int i = 2; i * i <= n; ++i) {
if (n % i == 0) {
for (int j = 2; j * j < n / i; ++j) {
if ((n / i) % j == 0 && j != i && n / i / j != i) {
out.println("YES");
out.println(n / i / j + " " + i + " " + j);
return;
}
}
}
}
out.println("NO");
}
static class mem {
int x, y;
}
static class comparator implements Comparator<mem> {
public int compare(mem o1, mem o2) {
if (o1.x != o2.x)
return Integer.compare(o1.x, o2.x);
return Integer.compare(o1.y, o2.y);
}
}
static long stepenb(long a, long b) {
long h = 1;
while (b > 0) {
if (b % 2 != 0) {
h *= a;
b--;
}
b /= 2;
a *= a;
}
return h;
}
static long lcm(long a, long b) {
return a * b / gcd(a, b);
}
private static long gcd(long a, long b) {
return b == 0 ? a : gcd(b, a % b);
}
static void end() {
out.flush();
out.close();
}
static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
static PrintWriter out = new PrintWriter(System.out);
static StringTokenizer in = new StringTokenizer("");
public static String nextToken() throws IOException {
while (in == null || !in.hasMoreTokens()) {
String s = br.readLine();
if (s == null) return null;
in = new StringTokenizer(s);
}
return in.nextToken();
}
public static int nextInt() throws IOException {
return Integer.parseInt(nextToken());
}
public static double nextDouble() throws IOException {
return Double.parseDouble(nextToken());
}
public static long nextLong() throws IOException {
return Long.parseLong(nextToken());
}
} | 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 factors(x):
l=[]
while(x>1):
a=0
for i in range(2,int(math.sqrt(x)+1)):
if(x%i==0):
l.append(i)
a=1
break
if(a==0):
l.append(x)
break
x=x//l[-1]
return l
t=int(input())
while(t):
n=int(input())
l1 = factors(n)
if(len(l1)<3):
print("NO")
elif(len(l1)==3):
if((l1[0]!=l1[-1])and(l1[0]!=l1[1])and(l1[1]!=l1[-1])):
print("YES")
s=""
for i in l1:
s+=str(i)+" "
print(s)
else:
print("NO")
else:
if(l1[0]==l1[1]):
l1[1] = l1[1]*l1[2]
l1.pop(2)
i=3
while(i<len(l1)):
l1[2]*=l1[i]
l1.pop(i)
if((l1[0]!=l1[-1])and(l1[0]!=l1[1])and(l1[1]!=l1[-1])):
print("YES")
s=""
for i in l1:
s+=str(i)+" "
print(s)
else:
print("NO")
else:
i=3
while(i<len(l1)):
l1[2]*=l1[i]
l1.pop(i)
if((l1[0]!=l1[-1])and(l1[0]!=l1[1])and(l1[1]!=l1[-1])):
print("YES")
s=""
for i in l1:
s+=str(i)+" "
print(s)
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 | t = int(input())
while(t):
n = int(input())
l = []
i = 2
c = 0
while(c < 2 and i*i < n):
if(n % i == 0):
l.append(i)
n //= i
c += 1
i += 1
#print(*l)
if(c == 2 and n not in l):
print('YES')
print(*l, n)
else:
print("NO")
t -= 1
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | # -*- coding: utf-8 -*-
"""
Created on Tue Feb 4 23:28:21 2020
@author: User
"""
import math
n = int(input())
num = []
for i in range(n):
num.append(int(input()))
for x in num:
flag = 0
counter = 0
y = int(math.sqrt(x))
for j in range(2,y+1) :
if flag == 1 : break
elif x % j == 0 :
a = int(j)
nu = int(x/j)
z=int(math.sqrt(nu))
for k in range(j+1,z+1) :
if nu % k == 0 :
b = int(k)
c = int(nu/k)
if b==c or a==c :
counter = 0
else :
print('YES')
print(a,b,c)
flag = 1
counter = 1
break
if counter == 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 whatever //do not write package name here */
import java.util.*;
import java.lang.*;
import java.io.*;
public class GFG {
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();mm(n);}
}
static int mm (int n){
int nsqr=(int)Math.sqrt(n);
for(int j=2;j<=nsqr;j++)
{
if(n%j==0)
{int nbyjsqr=(int)Math.sqrt(n/j);
for(int k=j+1;k<=nbyjsqr;k++)
{
if((n/j)%k==0)
{if(j!=k&&k!=(n/(j*k))&&(n/(j*k))!=j)
{System.out.println("YES");
System.out.println(j+" "+k+" "+n/(k*j));
return 1;
}}
}
}
}
System.out.println("NO");
return -1;
}
}
| JAVA |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | #-------------Program--------------
#----Kuzlyaev-Nikita-Codeforces----
#-------------Round615-------------
#----------------------------------
t=int(input())
for i in range(t):
n=int(input())
a=[]
for i in range(2,int(n**0.5)+2):
if len(a)==2:
a.append(n)
break
if n%i==0:
a.append(i)
n//=i
a=list(set(a))
if len(a)==3 and a.count(1)==0:
print('YES')
a.sort()
print(a[0],a[1],a[2])
else:
print('NO')
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import sys
def minp():
return sys.stdin.readline().strip()
def mint():
return int(minp())
def mints():
return map(int, minp().split())
def solve():
n = mint()
a = []
nn = n
for i in range(2,n+1):
if i*i > n:
break
if n % i == 0:
c = 0
while n % i == 0:
n //= i
if len(a) == 2:
break
c += 1
a.append((i, c))
if n != 1:
a.append((n, 1))
if len(a) == 1:
if a[0][1] < 6:
print('NO')
else:
print('YES')
print(a[0][0],a[0][0]**2,nn // (a[0][0]**3))
elif len(a) == 2:
if a[0][1] > 1 and a[1][1] > 1:
print('YES')
print(a[0][0],a[1][0],nn // (a[0][0]*a[1][0]))
elif a[0][1] > 2:
print('YES')
print(a[0][0],a[0][0]**2,nn // (a[0][0]**3))
elif a[1][1] > 2:
print('YES')
print(a[1][0],a[1][0]**2,nn // (a[1][0]**3))
else:
print('NO')
else:
print('YES')
print(a[0][0],a[1][0],nn // (a[0][0]*a[1][0]))
for i in range(mint()):
solve()
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
inline void setPrecision(int n) { cout.precision(n); }
long long int INF = 1e10;
int MOD = 1e9 + 7;
inline bool same2(int x, int y, int z) {
return (x == y) || (y == z) || (z == x);
}
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
int tt;
cin >> tt;
while (tt--) {
int n;
cin >> n;
int first = -1;
for (int i = 2; i * i <= n; ++i) {
if (n % i == 0) {
first = i;
break;
}
}
if (first == -1) {
cout << "NO\n";
continue;
}
n /= first;
int second = -1;
for (int i = first + 1; i * i <= n; ++i) {
if (n % i == 0) {
second = i;
break;
}
}
if (second == -1 || same2(first, second, n / second)) {
cout << "NO\n";
continue;
}
cout << "YES\n";
cout << first << " " << second << " " << n / second << "\n";
}
return 0;
}
| CPP |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import math
t=int(input())
for _ in range(t):
n=int(input())
raj=set()
for i in range(2, math.ceil(math.sqrt(n))):
if(n%i==0):
raj.add(i)
raj.add(n//i)
flg=0
for i in raj:
a=i
dup=n//a
andro=set()
for j in range(2, math.ceil(math.sqrt(dup))):
if(dup%j==0):
b=j
c=dup//b
if(b!=a and b!=c and c>1 and a!=c):
ans=[a,b,c]
ans=sorted(ans)
print("YES")
print(*ans)
flg=1
break
if(flg==1):
break
if(not flg):
print("NO")
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import math
def div(n, r):
sq = math.sqrt(n)
d = 1
for x in range(2, int(sq) + 1):
if n % x == 0 and x != r:
d = x
break
return d
t = int(input())
for i in range(0, t):
y = int(input())
a = div(y, 1)
if a == 1:
print("NO")
continue
else:
z = int(y / a)
b = div(z, a)
if b == 1 or b == a:
print("NO")
continue
else:
c = int(y / (a * b))
if c == 1 or c == a or c == b:
print("NO")
continue
else:
print("YES")
print(a, b, c)
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import math
def deliteli_chisla(cur):
mas = []
ot = "NO"
n = cur
for i in range(2, math.ceil(cur ** 0.5)):
if n % i == 0:
mas.append(i)
for i in range(len(mas)):
del1 = mas[i]
nn = n // del1
s = nn
for j in range(i, len(mas)):
nn = s
if nn % mas[j] == 0:
nn = nn // mas[j]
del2 = mas[j]
if del1 * del2 != n and n % (del1 * del2) == 0:
if del1 != del2 and del2 != (n // del1 // del2) and del1 != n // del1 // del2:
ot = "YES\n" + str(del1) + " " + str(del2) + " " + str(n // del1 // del2)
return ot
for t in range(int(input())):
n = int(input())
if deliteli_chisla(n) is None:
print("NO")
else:
print(deliteli_chisla(n))
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int main() {
int t;
cin >> t;
while (t--) {
int n;
cin >> n;
int a, b, c, c1 = 0;
for (int i = 2; i < sqrt(n) + 1; i++) {
if (n % i == 0) {
if (c1 == 0) {
a = i;
c1++;
n = n / i;
} else if (c1 == 1) {
b = i;
c = n / i;
c1++;
break;
}
}
}
if (c1 < 2 || b == c || a == c)
cout << "NO" << endl;
else {
cout << "YES" << endl << a << " " << b << " " << c << endl;
}
}
}
| CPP |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | for _ in range(int(input())):
n = int(input())
a = []
i=2
while(len(a)<2 and i*i <n):
if n%i ==0:
n=n//i
a.append(i)
i+=1
if len(a)==2 and n not in a:print("YES");print(n,*a)
else:print("NO") | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 |
t=int(input())
for i in range(t):
n=int(input())
a=[]
for j in range(2,int(n**0.5)+1):
if(n%j==0):
a.append(j)
n=n/j
break
if(len(a)==1):
for k in range(j+1,int(n**0.5)+1):
if(n%k==0):
a.append(k)
n=int(n/k)
break
if(len(a)==2):
if(n!=a[0] and n!=a[1]):
print("YES")
print(a[0],end=" ")
print(a[1],end=" ")
print(int(n))
else:
print("NO")
else:
print("NO")
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | t = int(input())
for i in range(t):
n = int(input())
ans=list()
for i in range(2,int(n**(2/3))+1):
if len(ans) == 2:
break
if n%i==0:
ans.append(i)
n//=i
if len(ans) == 2 and ans[1] < n:
print('YES')
print(*ans, 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
T=int(input())
for i in range((T)):
x=input()
x=int(x)
d=0
a,b,c=-1,-1,-1
p=x
for j in range(2,int(math.sqrt(x))+1):
if x%j==0:
a=j
x=x//j
break
if a!=-1:
for j in range(a+1,int(math.sqrt(x))+1):
if x%j==0:
b=j
x=x//j
break
if p%x==0:
c=x
if a==-1 or b==-1 or c==-1 or b==c or a==c:
print("NO")
else:
print("YES")
print(a,b,c,sep=" ")
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | t = int(input())
for _ in range(t):
n = int(input())
f = 0
for x in range(2, n):
if x * x * x >= n: break
for y in range(x + 1, n):
#print(x, y)
if x * y * y >= n: break
if n % (x * y) == 0:
f = 1
print("YES")
print(x, y, n // (x * y))
break
if f > 0: break
if f == 0: print("NO")
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | #include <bits/stdc++.h>
int main() {
int t;
std::cin >> t;
for (; t > 0; t--) {
long n;
std::cin >> n;
int i = 2;
std::vector<int> primes;
while (i * i <= n) {
if (n % i == 0) {
while (n % i == 0) {
primes.push_back(i);
n /= i;
}
}
i++;
}
if (n != 1) {
primes.push_back(n);
}
if (primes.size() < 3) {
std::cout << "NO\n";
continue;
}
long a = primes[0];
long b = primes[1];
int next_prime = 2;
if (a == b) {
b *= primes[next_prime++];
}
if (primes.size() <= next_prime) {
std::cout << "NO\n";
continue;
}
long c = primes[next_prime++];
while (next_prime < primes.size()) {
c *= primes[next_prime++];
}
if (c == a or c == b) {
std::cout << "NO\n";
} else {
std::cout << "YES\n";
std::cout << a << " " << b << " " << c << "\n";
}
}
}
| CPP |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 |
import java.util.Arrays;
import java.util.Comparator;
import java.util.HashSet;
import java.util.LinkedList;
import java.util.Queue;
import java.util.Scanner;
import java.util.Set;
public class Main {
private static long gcd(long a, long b) {
return (a == 0 ? b : gcd(b % a, a));
}
public static void main(String[] args) {
Scanner cin = new Scanner(System.in);
int t = cin.nextInt();
while (t-- != 0) {
int n = cin.nextInt();
int a = 0,b = 0,c = 0;
int flag = 0;
for(int i = 2;i * i <= n;i++) {
if(n % i == 0) {
for(int j = 2;j * j <= n / i;j++) {
if((n / i) % j == 0 && ((n / i) / j) >= 2 && (i) != (n / i / j) && i != j && j != (n / i / j)) {
a = i;
b = j;
c = n / i / j;
flag = 1;
break;
}
}
}
if(flag == 1)break;
}
if(flag == 1)System.out.println("YES" + "\n" + 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 | from math import sqrt
def check(n):
k=0
for i in range(2,int(sqrt(n+1))):
if k==2 and n>=i:
return a,b,n
if k==3:
return a,b,c
if k<3 and n<i:
return 'NO'
if n%i==0:
k+=1
n //= i
if k==1:
a=i
elif k==2:
b=i
elif k==3:
c=i
return 'NO'
t = int(input())
for i in range (t):
n = int(input())
out = check(n)
if type(out)==str:
print(out)
else:
print('YES')
for x in list(out):
print(x,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 | # Python3 implementation of the approach
from math import sqrt
# Function to find the required triplets
def findTriplets(x):
# To store the factors
fact = [];
factors = set();
# Find factors in sqrt(x) time
for i in range(2, int(sqrt(x))):
if (x % i == 0):
fact.append(i);
if (x / i != i):
fact.append(x // i);
factors.add(i);
factors.add(x // i);
found = False;
k = len(fact);
for i in range(k):
# Choose a factor
a = fact[i];
for j in range(k):
# Choose another factor
b = fact[j];
# These conditions need to be
# met for a valid triplet
if ((a != b) and (x % (a * b) == 0)
and (x / (a * b) != a)
and (x / (a * b) != b)
and (x / (a * b) != 1)):
# Print the valid triplet
print("YES")
print(a, b, x // (a * b));
found = True;
break;
# Triplet found
if (found):
break;
# Triplet not found
if (not found):
print("NO");
# Driver code
if __name__ == "__main__":
t=int(input())
while(t):
t-=1
n=int(input())
findTriplets(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 | #!/usr/bin/env python3
# -*- coding: utf-8 -*-
"""
Created on Wed Jan 22 08:37:17 2020
@author: huynguyen
"""
def findTripple(n):
res = []
for i in range(2,int(n**.5)+1):
if n%i == 0:
res.append(i)
n//=i
if len(res) ==2:
break
if n!=1:
if res:
if n not in res:
res.append(n)
return res
t = int(input())
for test in range(t):
n = int(input())
res = findTripple(n)
if len(res)==3:
print("YES")
print (" ".join([str(item) for item in res]))
else:
print ("NO") | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | from math import sqrt
q = int(input())
for _ in range(q):
n = int(input())
cnt = 0
save = [0] * 6
for i in range(2, int(sqrt(n)) + 2):
if n % i == 0:
cnt += 1
save[cnt] = i
n = int(n/i)
if cnt == 2:
break
if cnt < 2:
print("NO")
elif n <= save[cnt]:
print("NO")
else:
print("YES")
print(save[1], save[2], n, ' ')
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int main() {
int t, i;
long n, d, arr[2];
cin >> t;
while (t--) {
cin >> n;
i = 0;
for (d = 2; d * d < n && i < 2; d++) {
if (n % d == 0) {
arr[i++] = d;
n /= d;
}
}
if (i == 2)
cout << "YES\n" << arr[0] << ' ' << arr[1] << ' ' << n << '\n';
else
cout << "NO\n";
}
return 0;
}
| CPP |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | t=int(input())
for x in range(t):
n=int(input())
l=[]
i=2
while len(l)<2 and n>i**2:
if n%i==0:
l.append(i)
n=n/i
i+=1
if len(l)==2 and n not in l:
print('YES')
print(*l,int(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 sys
import os
import time
import collections
from collections import Counter, deque
import itertools
import math
import timeit
import random
#########################
# imgur.com/Pkt7iIf.png #
#########################
def sieve(n):
if n < 2: return list()
prime = [True for _ in range(n + 1)]
p = 3
while p * p <= n:
if prime[p]:
for i in range(p * 2, n + 1, p):
prime[i] = False
p += 2
r = [2]
for p in range(3, n + 1, 2):
if prime[p]:
r.append(p)
return r
def divs(n, start=1):
divisors = []
for i in range(start, int(math.sqrt(n) + 1)):
if n % i == 0:
if n / i == i:
divisors.append(i)
else:
divisors.extend([i, n // i])
return divisors
def divn(n, primes):
divs_number = 1
for i in primes:
if n == 1:
return divs_number
t = 1
while n % i == 0:
t += 1
n //= i
divs_number *= t
def flin(d, x, default=-1):
left = right = -1
for i in range(len(d)):
if d[i] == x:
if left == -1: left = i
right = i
if left == -1:
return (default, default)
else:
return (left, right)
def ceil(n, k): return n // k + (n % k != 0)
def ii(): return int(input())
def mi(): return map(int, input().split())
def li(): return list(map(int, input().split()))
def lcm(a, b): return abs(a * b) // math.gcd(a, b)
def prr(a, sep=' '): print(sep.join(map(str, a)))
def dd(): return collections.defaultdict(int)
def ddl(): return collections.defaultdict(list)
# input = sys.stdin.readline
for _ in range(ii()):
n = ii()
d = divs(n, 2)
f = 1
if len(d):
x = d[0]
for y in divs(n // x, 2):
z = n // (x * y)
if x != y != z != x and z != 1:
print('YES')
print(x, y, z)
f = 0
break
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 | #!/usr/bin/env python3
import math
primes = [2]
for n in range(3,math.ceil(math.sqrt(1e9)),2):
is_prime = True
sqrt_n = math.sqrt(n)
for p in primes:
if p > sqrt_n:
break
if n%p == 0:
is_prime = False
break
if is_prime:
primes.append(n)
t = int(input())
for _ in range(t):
n = int(input())
_n = n
factors = {}
for p in primes:
while _n % p == 0:
if p not in factors:
factors[p] = 0
factors[p] += 1
_n /= p
if _n != 1:
factors[_n] = 1
f1 = None
f2 = None
if len(factors) == 1:
[f] = list(factors)
if factors[f] >= 6:
f1 = f
f2 = f*f
elif len(factors) == 2:
[_f1, _f2] = list(factors)
if (factors[_f1] + factors[_f2]) >= 4:
f1 = _f1
f2 = _f2
elif len(factors) >= 3:
[_f1,_f2] = list(factors)[:2]
f1 = _f1
f2 = _f2
if f1 != None and f2 != None:
print("YES")
print(f"{int(f1)} {int(f2)} {int(n/(f1*f2))}")
else:
print("NO")
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | # -*- coding: utf-8 -*-
"""
Created on Wed Jan 22 19:40:08 2020
@author: DELL
"""
import math
# A function to print all prime factors of
# a given number n
def primeFactors(n):
li=[]
# Print the number of two's that divide n
while n % 2 == 0:
li.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:
li.append(i)
n = n / i
# Condition if n is a prime
# number greater than 2
if n > 2:
li.append(int(n))
return li
t=int(input())
for i in range(t):
n=int(input())
p = primeFactors(n)
k=set(p)
w=list(k)
if len(w)==1:
if p.count(w[0])>5:
print("YES")
a=p[0]
b=p[1]*p[2]
c=1
for i in range(3,len(p)):
c=c*p[i]
print(str(a)+" "+str(b)+" "+str(c))
else:
print("NO")
elif len(w)==2:
if len(p)<=3:
print("NO")
else:
if len(p)==5:
print("YES")
a=p[0]
b=p[1]*p[2]*p[3]
c=p[4]
print(str(a)+" "+str(b)+" "+str(c))
else:
print("YES")
a=p[0]
b=p[1]*p[2]
c=1
for i in range(3,len(p)):
c=c*p[i]
print(str(a)+" "+str(b)+" "+str(c))
elif len(w)==0:
print("NO")
else:
if len(p)==3:
print("YES")
a=p[0]
b=p[1]
c=p[2]
print(str(a)+" "+str(b)+" "+str(c))
else:
print("YES")
a=p[0]
b=p[1]*p[2]
c=1
for i in range(3,len(p)):
c=c*p[i]
print(str(a)+" "+str(b)+" "+str(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 | for _ in range(int(input())):
n=int(input())
t=pow(n,0.5)+3
l=[]
co=0
i=2
while i<t:
if n%i==0:
n=n//i
l.append(i)
else:
i+=1
# print(l)
if n!=1:
l.append(n)
flag="YES"
t=len(l)
if t>=3:
a=l[-1]
l.pop(-1)
b=l[-1]
l.pop(-1)
if a==b:
b=b*l[-1]
l.pop(-1)
c=1
for i in l:
c=c*i
if t>=3:
if c>1 and a>1 and b>1 and b!=c and a!=b and a!=c:
print("YES")
print(a,b,c)
else:
print("NO")
else:
print("NO") | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import java.util.*;
public class Product_of_Three_Numbers
{
public static void main(String args[])
{
Scanner ob=new Scanner(System.in);
int t=ob.nextInt();
while(t-->0)
{
int n=ob.nextInt();
int a=0,b=0;
for(int i=2;i*i<n;i++)
{
if(n%i==0)
{
n/=i;
if(a==0)
a=i;
else{
b=i;
break;
}
}
}
if(a!=0 && b!=0 && n>b) {
System.out.println("YES");
System.out.println(a+" "+b+" "+n+" ");}
else
System.out.println("NO");
}
}
} | JAVA |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | t = int(input())
for _ in range(t):
n = int(input())
li = []
i = 2
while i*i<n:
if n%i==0:
li.append(i)
n/=i
if len(li)==2:
break
i+=1
if len(li)!=2 or n in li:
print('NO')
else:
print("YES")
print(li[0],li[1], int(n)) | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | from math import factorial
from collections import Counter
from sys import stdin
def RL(): return map(int, stdin.readline().rstrip().split() )
def comb(n, m): return factorial(n)/(factorial(m)*factorial(n-m)) if n>=m else 0
def perm(n, m): return factorial(n)//(factorial(n-m)) if n>=m else 0
def mdis(x1, y1, x2, y2): return abs(x1-x2) + abs(y1-y2)
mod = 1000000007
from math import gcd
# ------------------------------
from math import sqrt
N = int(input())
for _ in range(N):
n = int(input())
tag = False
for i in range(2, int(sqrt(n))+1):
if n%i==0:
for j in range(2, int(sqrt(n//i))+1):
if i!=j and (n//i)%j==0 and n//i//j!=j and n//i//j!=i and n//i//j!=1:
print("YES")
print(i, j, n//i//j)
tag = True
if tag: break
if tag: 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 | # Python program to print prime factors
import math
# a given number n
def prime(n):
a= []# Print the number of two's that divide n
while n % 2 == 0:
a.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:
a.append(int(i))
n = n / i
# Condition if n is a prime
# number greater than 2
if n > 2:
a.append(int(n))
return a
# Driver Program to test above function
t= int(input())
while(t>0):
n = int(input())
a = prime(n)
l = len(a)
if l<3:
print('NO')
elif l==3 and len(set(a))==3:
print('YES')
for i in a:
print(i,end=' ')
print()
else:
i = 0
k = 1
temp = []
temp.append(a[i])
i+=1
if a[i]!=a[i-1]:
temp.append(a[i])
i+=1
else:
temp.append(a[i]*a[i+1])
i+=2
for j in range(i,l):
k*=a[j]
if k!=1 and k not in temp:
temp.append(k)
s= set(temp)
if len(s)==3 and temp[0]*temp[1]*temp[2]==n:
print('YES')
for i in s:
print(i,end=' ')
print()
else:
print('NO')
t-=1 | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import math
t = int(input())
for _ in range(t):
n = int(input())
d = []
while n % 2 == 0:
d.append(2)
n //= 2
for i in range(3, (int(math.sqrt(n)) + 1), 2):
while n % i == 0:
d.append(i)
n //= i
if n > 2:
d.append(n)
if len(d) < 3:
print('NO')
elif len(d) == 3 and (d[0] == d[1] or d[1] == d[2]):
print('NO')
else:
# print(d)
a = d[0]
if d[1] == d[0]:
b = d[1] * d[2]
c = math.prod(d[3:])
else:
b = d[1]
c = math.prod(d[2:])
if a == b or a == c or b == c:
print('NO')
else:
print('YES')
print(a, b, c)
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import math as mt
n = int(input())
i = 0
while i < n:
case = int(input())
n1 = case
a, b, c = 1, 1, 1
divisor = []
for p in range(2, mt.ceil(mt.sqrt(case + 1))):
if len(divisor) == 2:
divisor.append(int(n1 / (divisor[0] * divisor[1])))
break
elif case % p == 0:
divisor.append(p)
case /= p
try:
a, b, c = tuple(divisor)
except ValueError:
a, b, c = 1, 1, 1
if a == 1 or b == 1 or c == 1 or a == c or a == b or b == c:
ans = False
else:
ans = True
if ans:
print("YES")
print(a, b, c)
else:
print("NO")
i += 1 | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int n, q;
cin >> q;
while (q--) {
cin >> n;
vector<int> res;
for (int i = 2; res.size() < 2 && i * i < n; ++i) {
if (n % i == 0) {
res.push_back(i);
n /= i;
}
}
res.push_back(n);
if (res.size() != 3) {
cout << "NO" << endl;
} else {
cout << "YES" << endl;
cout << res[0] << " " << res[1] << " " << res[2] << endl;
}
}
}
| CPP |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | def IsPrime(n):
d = 2
while d * d <= n and n % d != 0:
d += 1
return d * d > n
t = int(input())
ans = []
for j in range(t):
n = int(input())
divs = []
i = 2
while i*i <= n:
if n % i == 0:
x = n // i
u = i + 1
while u*u <= x:
if x % u == 0:
if x//u == u:
break
ans.append(['YES', i, u, x//u])
u = x
i = n
u += 1
i +=1
if len(ans) == j:
ans.append(['NO'])
for i in range(len(ans)):
if len(ans[i]) == 1:
print(*ans[i])
else:
print(ans[i][0])
print(*ans[i][1::])
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import java.io.*;
import java.util.*;
import java.lang.Math.*;
public class CodeForces615C {
public static void main (String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int t = Integer.valueOf(br.readLine());
for (int i = 0;i<t ;i++ ){
Number n = new Number(Integer.valueOf(br.readLine()));
System.out.println(n.tutorialSolution());
}
}
}
class Number{
int input ;
public Number(int input){
this.input = input;
}
public String tutorialSolution(){
int a = findSmallestFactor(2,input);
if(a<=0){
return "NO";
}
int b = findSmallestFactor(a+1, input/a);
if(b<0){
return "NO";
}
//System.out.println(a+" "+b+" "+input/(a*b));
StringBuilder sb = new StringBuilder();
if(input/(a*b)==1){
sb.append("NO");
} else{
if(a!=b && b!= input/(a*b) && a!=input/(a*b)){
sb.append("YES\n");
sb.append(a+" "+b+" "+input/(a*b));
}else {
sb.append("NO");
}
}
return sb.toString();
}
public int findSmallestFactor(int start, int input){//complexity root(n) same as primes
int ret = -1;
double a = (double)input;
for (int i = start; i<=java.lang.Math.sqrt(a); i++) {
//System.out.println(i+" "+ i*i);
if(input%i==0){
ret= i;
break;
}
}
return ret;
}
public String findThreeFactors(){
int [] ret = new int[6];
int k = input;
boolean noMoreFactors = false;
Factors f = null;
int i = 0;
for ( i = 0;i<ret.length ;i++ ) {
f = returnFactors(k);
//System.out.println("k " + k + "("+f.a+" "+f.b +")");
ret[i] = f.a;
if(f.b<0|| f.b == 1){
i++;
noMoreFactors = true;
break;
}
k = f.b;
}//what happens when 1 is given
String retString = null;
Set<Integer> s = new HashSet<Integer>();
if(!noMoreFactors){
s.add(ret[0]);
s.add(ret[1]* ret[2]);
s.add(ret[3]* ret[4]* ret[5]*f.b);
}else{
if(i<3){
retString = "NO";
}
if(i==3){
s.add(ret[0]);
s.add(ret[1]);
s.add(ret[2]);
} else if(i==4){
s.add(ret[0]);
s.add(ret[1]* ret[2]);
s.add(ret[3]);
} else if(i==5){
s.add(ret[0]);
s.add(ret[1]* ret[2]);
s.add(ret[3]* ret[4]);
}else if(i==6) {
s.add(ret[0]);
s.add(ret[1]* ret[2]);
s.add(ret[3]* ret[4]* ret[5]);
}
}
if(s.size()==3){
StringBuilder sb = new StringBuilder();
sb.append("YES\n");
Iterator iter = s.iterator();
while (iter.hasNext()) {
sb.append(iter.next());
if(iter.hasNext()) {
sb.append(" ");
}
}
retString = sb.toString();
} else{
retString = "NO";
}
return retString ;
}
public Factors returnFactors(int i){
if(i==1)
return new Factors(-1) ;
if(i==3)
return new Factors(3) ;
if(i==2)
return new Factors(2) ;
if(i%2==0){
Factors f = new Factors(i/2, 2);
return f;
}
if(i%3 ==0){
Factors f = new Factors(i/3, 3);
return f;
}
int j =5;
while(j<= java.lang.Math.sqrt((double)i) ){// ignore ones where j * j = i
if(i%j == 0){
Factors f = new Factors(i/j, j);
return f;
} else if(i%(j+2)==0){
Factors f = new Factors(i/(j+2), j+2);
return f;
}
// System.out.println(j+" "+ (j+2));
j = j+6;
}
return new Factors(i);
}
}
class Factors{
int a;
int b;
public Factors(int i,int j){
//make sure a<b and dont add 1
if(i==1){
a = j;
b = -1;
}else if(j==1){
a = i;
b = -1;
}else if(i<j){
a = i;
b = j;
}else{
a = j;
b = i;
}
}
public Factors(int i){
a = i;
b = -1;
}
}
| JAVA |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int t = Integer.parseInt(sc.next());
for(int i = 0; i < t; i++){
int n = Integer.parseInt(sc.next());
List<Long> list = primeFactorization(n);
if(list.size() < 3){
System.out.println("NO");
}else{
long a = list.get(0);
long b = list.get(list.size()-1);
long c = 1;
for(int j = 1; j < list.size()-1; j++){
c *= list.get(j);
}
if(a == b || b == c || c == a){
if(list.size() < 6){
System.out.println("NO");
}else{
b = list.get(1) * list.get(2);
c = 1;
for(int j = 3; j < list.size(); j++){
c *= list.get(j);
}
System.out.println("YES");
System.out.println(a + " " + b + " " + c);
}
}else{
System.out.println("YES");
System.out.println(a + " " + b + " " + c);
}
}
}
}
static List<Long> primeFactorization(long n){
List<Long> list = new ArrayList<>();
while(n%2 == 0){
n /= 2;
list.add(2L);
}
double sqrt = Math.sqrt(n);
for(long l = 3; l <= sqrt; l += 2){
while(n%l == 0){
n /= l;
list.add(l);
}
}
if(n > 1) list.add(n);
return list;
}
} | 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 defaultdict
import math
rl = lambda: stdin.readline()
rll = lambda: stdin.readline().split()
def main():
cases = int(input())
for line in stdin:
n = int(line)
ans = []
f1 = 2
while f1 <= math.sqrt(n):
if n % f1 == 0:
ans.append(f1)
break
f1 += 1
if len(ans) == 0:
print("NO")
continue
m = n//f1
f2 = f1 + 1
while f2 <= math.sqrt(m):
if m % f2 == 0:
ans.append(f2)
break
f2 += 1
if len(ans) == 1:
print("NO")
continue
f3 = n//(f1*f2)
if f3 not in {f1, f2}:
print("YES")
print(*[f1, f2, f3])
else:
print("NO")
if __name__ == "__main__":
main() | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | from math import ceil, sqrt
n = int(input())
for i in range(n):
num = int(input())
ans = []
for mult in range(2, ceil(sqrt(num))):
#print('Try ', mult,' for ', num)
if num < mult:
break
if num % mult == 0:
ans.append(mult)
num = num / mult
if (len(ans) == 2) and (ans[0] != num) and (ans[1] != num) and (num != 1):
ans.append(int(num))
break
if len(ans) == 3:
print('YES')
print(' '.join([str(x) for x 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 | import java.util.*;
import java.lang.*;
import java.io.*;
public class Codeforces {
static FastReader sc=new FastReader();
static PrintWriter out=new PrintWriter(System.out);
static long mod=1000000007;
public static void main (String[] args) throws java.lang.Exception
{
// your code goes here
// StringBuffer sb=new StringBuffer();
int t=I();
while(t-->0)
{
int n=I();
int x=checkPrime(n,2);
if(x==-1)out.println("NO");
else{
int y=n/x;
int p=checkPrime(x,2);
if(p==-1){
int q=checkPrime(y,x+1);
if(q==-1)out.println("NO");
else{
out.println("YES");
out.println(x+" "+q+" "+(y/q));
}
}else{
out.println("YES");
out.println(p+" "+(x/p)+" "+y);
}
}
}
// out.print(sb.toString());
out.close();
}
public static int checkPrime(int n,int p)
{
for(int i=p;i*i<=n;i++){
if(n%i==0 && n/i!=i){
return i;
}
}
return -1;
}
public static void printArray(int a[])
{
for(int i=0;i<a.length;i++){
out.print(a[i]+" ");
}
out.println();
}
public static void printArray(ArrayList<Integer> arr)
{
for(int i=0;i<arr.size();i++){
out.print(arr.get(i)+" ");
}
out.println();
}
public static void DFS(ArrayList<Integer> arr[],int s,boolean visited[])
{
visited[s]=true;
for(int i:arr[s]){
if(!visited[i]){
DFS(arr,i,visited);
}
}
}
public static int BS(long a[],long x,int ii,int jj)
{
// int n=a.length;
int mid=0;
int i=ii,j=jj;
while(i<=j)
{
mid=(i+j)/2;
if(a[mid]>x)
j=mid-1;
else if(a[mid]<=x)
i=mid+1;
}
if(a[mid]>x && mid>ii)
return mid-1;
else
return mid;
}
public static ArrayList<Integer> prime(int n)
{
ArrayList<Integer> arr=new ArrayList<>();
boolean 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;
}
}
for (int i = 2; i <= n; i++)
{
if (prime[i] == true)
arr.add(i);
}
return arr;
}
public static HashSet<Integer> primeSet(int n)
{
HashSet<Integer> arr=new HashSet<>();
boolean 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;
}
}
for (int i = 2; i <= n; i++)
{
if (prime[i] == true)
arr.add(i);
}
return arr;
}
// Fenwick / BinaryIndexed Tree USE IT - FenwickTree ft1=new FenwickTree(n);
public static class FenwickTree
{
long farr[];
int n;
public FenwickTree(int c)
{
n=c+1;
farr=new long[n];
}
public void update_range(int l,int r,long p)
{
update(l,p);
update(r+1,(-1)*p);
}
public void update(int x,long p)
{
for(;x<n;x+=x&(-x))
{
farr[x]+=p;
}
}
public long get(int x)
{
long ans=0;
for(;x>0;x-=x&(-x))
{
ans=ans+farr[x];
}
return ans;
}
}
//Disjoint Set Union
public static class DSU
{
static int par[],rank[];
public DSU(int c)
{
par=new int[c+1];
rank=new int[c+1];
for(int i=0;i<=c;i++)
{
par[i]=i;
rank[i]=0;
}
}
public static int find(int a)
{
if(a==par[a])
return a;
return par[a]=find(par[a]);
}
public static void union(int a,int b)
{
int a_rep=find(a),b_rep=find(b);
if(a_rep==b_rep)
return;
if(rank[a_rep]<rank[b_rep])
par[a_rep]=b_rep;
else if(rank[a_rep]>rank[b_rep])
par[b_rep]=a_rep;
else
{
par[b_rep]=a_rep;
rank[a_rep]++;
}
}
}
//SEGMENT TREE CODE
// public static void segmentUpdate(int si,int ss,int se,int qs,int qe,long x)
// {
// if(ss>qe || se<qs)return;
// if(qs<=ss && qe>=se)
// {
// seg[si][0]+=1L;
// seg[si][1]+=x*x;
// seg[si][2]+=2*x;
// return;
// }
// int mid=(ss+se)/2;
// segmentUpdate(2*si+1,ss,mid,qs,qe,x);
// segmentUpdate(2*si+2,mid+1,se,qs,qe,x);
// }
// public static long segmentGet(int si,int ss,int se,int x,long f,long s,long t,long a[])
// {
// if(ss==se && ss==x)
// {
// f+=seg[si][0];
// s+=seg[si][1];
// t+=seg[si][2];
// long ans=a[x]+(f*((long)x+1L)*((long)x+1L))+s+(t*((long)x+1L));
// return ans;
// }
// int mid=(ss+se)/2;
// if(x>mid){
// return segmentGet(2*si+2,mid+1,se,x,f+seg[si][0],s+seg[si][1],t+seg[si][2],a);
// }else{
// return segmentGet(2*si+1,ss,mid,x,f+seg[si][0],s+seg[si][1],t+seg[si][2],a);
// }
// }
public static class pair
{
long a;
long b;
public pair(long val,long index)
{
a=val;
b=index;
}
}
public static class myComp implements Comparator<pair>
{
//sort in ascending order.
public int compare(pair p1,pair p2)
{
if(p1.a==p2.a)
return 0;
else if(p1.a<p2.a)
return -1;
else
return 1;
}
//sort in descending order.
// public int compare(pair p1,pair p2)
// {
// if(p1.a==p2.a)
// return 0;
// else if(p1.a<p2.a)
// return 1;
// else
// return -1;
// }
}
public static class myComp1 implements Comparator<pair1>
{
//sort in ascending order.
public int compare(pair1 p1,pair1 p2)
{
if(p1.a==p2.a)
return 0;
else if(p1.a<p2.a)
return -1;
else
return 1;
}
//sort in descending order.
// public int compare(pair p1,pair p2)
// {
// if(p1.a==p2.a)
// return 0;
// else if(p1.a<p2.a)
// return 1;
// else
// return -1;
// }
}
public static class pair1
{
long a;
long b;
public pair1(long val,long index)
{
a=val;
b=index;
}
}
public static ArrayList<pair1> mergeIntervals(ArrayList<pair1> arr)
{
//****************use this in main function-Collections.sort(arr,new myComp1());
ArrayList<pair1> a1=new ArrayList<>();
if(arr.size()<=1)
return arr;
a1.add(arr.get(0));
int i=1,j=0;
while(i<arr.size())
{
if(a1.get(j).b<arr.get(i).a)
{
a1.add(arr.get(i));
i++;
j++;
}
else if(a1.get(j).b>arr.get(i).a && a1.get(j).b>=arr.get(i).b)
{
i++;
}
else if(a1.get(j).b>=arr.get(i).a)
{
long a=a1.get(j).a;
long b=arr.get(i).b;
a1.remove(j);
a1.add(new pair1(a,b));
i++;
}
}
return a1;
}
public static boolean palindrome(String s,int n)
{
for(int i=0;i<=n/2;i++){
if(s.charAt(i)!=s.charAt(n-i-1)){
return false;
}
}
return true;
}
public static long countDigit(long n)
{
long sum=0;
while(n!=0)
{
sum++;
n=n/10;
}
return sum;
}
public static long digitSum(long n)
{
long sum=0;
while(n!=0)
{
sum=sum+n%10;
n=n/10;
}
return sum;
}
public static long gcd(long a,long b)
{
if(b==0)
return a;
else
return gcd(b,a%b);
}
public static int gcd(int a,int b)
{
if(b==0)
return a;
else
return gcd(b,a%b);
}
public static long pwr(long m,long n)
{
long res=1;
m=m%mod;
if(m==0)
return 0;
while(n>0)
{
if((n&1)!=0)
{
res=(res*m)%mod;
}
n=n>>1;
m=(m*m)%mod;
}
return res;
}
public static void sort(int[] A)
{
int n = A.length;
Random rnd = new Random();
for(int i=0; i<n; ++i)
{
int tmp = A[i];
int randomPos = i + rnd.nextInt(n-i);
A[i] = A[randomPos];
A[randomPos] = tmp;
}
Arrays.sort(A);
}
public static void sort(long[] A)
{
int n = A.length;
Random rnd = new Random();
for(int i=0; i<n; ++i)
{
long tmp = A[i];
int randomPos = i + rnd.nextInt(n-i);
A[i] = A[randomPos];
A[randomPos] = tmp;
}
Arrays.sort(A);
}
public static int I(){return sc.I();}
public static long L(){return sc.L();}
public static String S(){return sc.S();}
public static double D(){return sc.D();}
}
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 I(){
return Integer.parseInt(next());
}
long L(){
return Long.parseLong(next());
}
double D(){
return Double.parseDouble(next());
}
String S(){
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 | #!/usr/bin/env python
# coding: utf-8
# In[7]:
import math
def divisor(n):
for i in range(2,int(math.sqrt(n))+1):
if n%i==0:
k=n//i
for j in range(2,int(math.sqrt(k))+1):
if j!=i and k%j==0:
t=k//j
if t!=i and t!=j:
return ["YES",i,j,t]
return "NO"
t=int(input())
for i in range(t):
n=int(input())
if divisor(n)=="NO":
print("NO")
else:
print(divisor(n)[0])
print(*divisor(n)[1:])
# In[ ]:
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import java.io.*;
import java.util.*;
import java.math.*;
public class S{
static class Pair implements Comparable<Pair>{
int a;
int b;
public Pair(int x,int y){a=x;b=y;}
public Pair(){}
public int compareTo(Pair p1){
if(a == p1.a)
return b - p1.b;
return a - p1.a;
}
}
static class TrieNode{
TrieNode[]child;
int w;
boolean term;
TrieNode(){
child = new TrieNode[26];
}
}
public static int gcd(int a,int b)
{
if(a<b)
return gcd(b,a);
if(b==0)
return a;
return gcd(b,a%b);
}
//static long ans = 0;
static long mod =(long)( 1e9 + 7);
public static void main(String[] args) throws Exception {
new Thread(null, null, "Anshum Gupta", 99999999) {
public void run() {
try {
solve();
} catch(Exception e) {
e.printStackTrace();
System.exit(1);
}
}
}.start();
}
static void insert(TrieNode head,String str){
int n = str.length();
TrieNode cur = head;
for(int i=0;i<n;i++){
int ch = str.charAt(i)-'A';
if(cur.child[ch] == null)
cur.child[ch]=new TrieNode();
cur = cur.child[ch];
}
cur.term = true;
}
static boolean isRoot(TrieNode cur){
if(cur==root)return true;
return false;
}
static int f(TrieNode head){
int ans = 0;
for(int i=0;i<26;i++){
if(head.child[i]!=null){
ans += f(head.child[i]);
}
}
if(head.term)ans++;
if(ans >=2 && !isRoot(head))
ans-=2;
return ans;
}
static boolean done(int[]arr){
for(int i=1;i<arr.length;i++){
if(arr[i-1] > arr[i] )
return false;
}
return true;
}
static TrieNode root;
public static void solve() throws Exception {
// solve the problem here
MyScanner s = new MyScanner();
out = new PrintWriter(new BufferedOutputStream(System.out), true);
int t = s.nextInt();
int tc = 0;
while(tc++<t){
int n = s.nextInt();
int a=0,b=0,c=0;
for(int i = 2;i*i <= n;i++){
if(n%i == 0){
a = i;
break;
}
}
if(a==0)
{
out.println("NO");
continue;
}
for(int i=a+1;i*i<=n;i++){
if(n%(i*a) == 0 ){
b = i;
break;
}
}
if(b==0)
{
out.println("NO");
continue;
}
c = n/(a*b);
if(a == 0 || b==0 || c==0 || a==b || b==c || c==a)
out.println("NO");
else {
out.println("YES");
out.println(a+" "+b+" "+c);
}
}
out.flush();
}
//-----------PrintWriter for faster output---------------------------------
public static PrintWriter out;
//-----------MyScanner class for faster input----------
public static class MyScanner {
BufferedReader br;
StringTokenizer st;
public MyScanner() {
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() {
int q;
cin >> q;
for (int i = 0; i < q; ++i) {
int n;
cin >> n;
set<int> used;
for (int i = 2; i * i <= n; ++i) {
if (n % i == 0 && !used.count(i)) {
used.insert(i);
n /= i;
break;
}
}
for (int i = 2; i * i <= n; ++i) {
if (n % i == 0 && !used.count(i)) {
used.insert(i);
n /= i;
break;
}
}
if (int(used.size()) < 2 || used.count(n) || n == 1) {
cout << "NO" << endl;
} else {
cout << "YES" << endl;
used.insert(n);
for (auto it : used) cout << it << " ";
cout << 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 | def pm(n):
arr = []
while(n%2==0):
arr.append(2)
n//=2
for i in range(3,int(pow(n,0.5))+1,2):
while n%i==0:
arr.append(i)
n//=i
if n>2:
arr.append(n)
return arr
tc = int(input())
for ii in range(tc):
n = int(input())
arr = pm(n)
a = arr[0]
b = 1
c = 1
index = 0
for i in range(1,len(arr)):
b*=arr[i]
if b!=a:
index = i
break
for j in range(index+1,len(arr)):
c*=arr[j]
if a!=b and b!=c and a!=c and a!=1 and b!=1 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 _ in range(int(input())):
n=int(input())
l=[];i=2
while len(l)<2 and i*i<n:
if n%i==0:l.append(i);n//=i;
i+=1
if len(l)==2 and n not in l:print("YES");print(*l,n);
else: print("NO")
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import math
from functools import reduce
t = int(input())
for _ in range(t):
n = int(input())
m = n
sqrt_n = int(math.sqrt(n))
res = []
for i in range(2, sqrt_n):
if n == 1:
break
while n > 1 and n % i == 0:
res.append(i)
n //= i
if n > 1:
res.append(n)
if len(res) < 3:
print('NO')
elif len(res) == 3:
if len(set(res)) < 3:
print('NO')
else:
print('YES')
print(res[0], res[1], res[2])
elif len(res) == 4:
if len(set(res)) == 1:
print('NO')
else:
print('YES')
print(res[0], res[1] * res[2], res[3])
elif len(res) == 5:
if len(set(res)) == 1:
print('NO')
else:
print('YES')
print(res[0], res[1] * res[2] * res[3], res[4])
else:
print('YES')
print(res[0], res[1] * res[2], reduce(lambda x, y: x * y, res[3:])) | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | # from sys import stdin, stdout
# print = lambda x: stdout.write(x)
# input = stdin.readline
from math import sqrt
#Ref: Geeksforgeeks
def primeFactors(n):
arr = []
# Print the number of two's that divide n
while n % 2 == 0:
arr.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(sqrt(n))+1,2):
# while i divides n , print i ad divide n
while n % i== 0:
arr.append(i)
n = n // i
# Condition if n is a prime
# number greater than 2
if n > 2:
arr.append(n)
return arr
def solve(a,n,res):
for i in res:
b = n//i
if a!=b:
return a,b
return False
for t in range(int(input())):
n = int(input())
res = primeFactors(n)
if len(res) <= 2:
print('NO')
continue
resset = list(set(res))
ans = []
if len(resset) >= 2:
ans.append(resset[0])
ans.append(resset[1])
ans.append(n//(ans[-1]*ans[-2]))
ans = set(ans)
else:
ans.append(resset[0])
ans.append(resset[0]**2)
ans.append(n//(ans[-1]*ans[-2]))
ans = set(ans)
ans = list(ans)
if len(ans) == 3 and min(ans)>=2:
print('YES')
for i in ans:
print(i, end = ' ')
print()
else:
print('NO')
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | from math import sqrt
def factor(n):
s = int(sqrt(n))
rLst = [1, n]
for i in xrange(2, s+1):
if n%i==0:
rLst.append(i)
if n//i != i:
rLst.append(n//i)
rLst = list(set(rLst))
rLst.sort()
return rLst
from sys import stdin
raw_input = lambda: stdin.readline().rstrip()
input = lambda: int(raw_input())
I=lambda: map(int, raw_input().split())
t = input()
for _ in xrange(t):
n = input()
facts = factor(n)
q1 = facts[1]
facts1 = factor(n//q1)
q2 = -1
for x in facts1:
if x>1 and x != q1:
q2 = x
break
if q2==-1:
print 'NO'
continue
q3 = n//(q1*q2)
if q3==1 or q3==q1 or q3==q2:
print 'NO'
else:
print 'YES'
print q1, q2, q3
# if len(facts)<7:
# print 'NO'
# else:
# print 'YES'
# q1 = facts[1]
# q2 = facts[2]
# print q1, q2, n//(q1*q2)
| PYTHON |
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