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stringlengths 2
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| description
stringlengths 29
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| source
int64 1
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| difficulty
int64 0
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| solution
stringlengths 7
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| language
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1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | def solution(n):
for first_dis in range(2, int(n**0.5) + 1):
if n % first_dis == 0:
next_value = n // first_dis
for second_dis in range(first_dis + 1, int(next_value**0.5) + 1):
if next_value % second_dis == 0:
third = next_value // second_dis
if third != second_dis:
return ("YES", f"{first_dis} {second_dis} {third}")
return ("NO",)
def read():
cnt = int(input())
for _ in range(cnt):
value = int(input())
for item in solution(value):
print(item)
read() | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | from math import sqrt
def solve(n):
if isprime(n,2)==-1:return 'NO'
i=2
while i<=sqrt(n):
if n%i==0:
p=isprime(n//i,i+1)
if p!=-1:return i,p
i+=1
return 'NO'
def isprime(n,b):
i=b
while i<=sqrt(n):
if n%i==0 and i!=n//i:return i,n//i
i+=1
return -1
for _ in range(int(input())):
n=int(input());s=solve(n)
if s=='NO':print(s)
else:print('YES');print(s[0],*s[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())
for _ in range(t):
n = int(input())
sqn = int(n**(1/2))+10
ans = []
flag = False
for i in range(2,sqn):
if n%i==0:
ans.append(i)
n//=i
if len(ans)==2:
if n==ans[1] or n==ans[0] or n==1:
print("NO")
else:
print("YES")
print(*ans,n)
flag = True
break
if not flag:
print("NO")
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import java.io.*;
import java.util.*;
import java.lang.*;
// A U T H O R : s a n 1 d h y a
public class cf1294C {
static long Mod1 = 1000000007;
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 ni() {
return Integer.parseInt(next());
}
long nl() {
return Long.parseLong(next());
}
double nd() {
return Double.parseDouble(next());
}
String nln() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
/*
* public static void sortbyColumn(int arr[][], int col) { // Using built-in
* sort function Arrays.sort Arrays.sort(arr, new Comparator<int[]>() {
*
* @Override // Compare values according to columns public int compare(final
* int[] entry1, final int[] entry2) {
*
* // To sort in descending order revert // the '>' Operator if (entry1[col] >
* entry2[col]) return 1; else return -1; } }); // End of function call sort().
* }
*/
/* public static long modex(long x, long y) {
x = x % Mod1;
long ret = 1;
while (y > 0) {
if (y % 2 != 0) {
ret = ret * x % Mod1;
}
x = x * x % Mod1;
y >>= 1;
}
return ret;
}
*/
/* public static long mplr(long x, long y) {
long ret = 0;
while (y != 0) {
if (y % 2 != 0) {
ret = (ret + x) % Mod1;
}
x = (x + x) % Mod1;
y >>= 1;
}
return ret;
}
*/
public static void main(String[] args) {
FastReader sc = new FastReader();
int t=sc.ni();
while(t-->0) {
long n=sc.nl();
long q=n;
// long sq=(long)Math.ceil(Math.sqrt((double)n));
long a[]=new long[2];
int c=0; int f=0;
for(long i=2;i<(long)Math.sqrt(q);i++) {
if(n%i==0)
{
a[c++]=i;
n=n/i;
}
if(c==2)
{
f=1;
break;
}
}
if(f==0)
System.out.println("NO");
else
{ if(q%n==0&&(n!=a[0]&&n!=a[1]))
{ System.out.println("YES");
System.out.println(a[0]+" "+a[1]+" "+n);
}
else
System.out.println("NO");
}
}
}
} | JAVA |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | def d(i,n):
if(i==2 and n%2==0):
return 2;
while(i**2<=n):
if(n%i == 0):
return i
i+=1
return n;
t=int(input())
for _ in range(t):
n=int(input())
a = d(2,n);
b= d(a+1,n/a);
c = n//(a*b)
if(c>=2 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 | # _1294c
##########
def productOfThreeNumbers(n, answer=(1, )):
lengthOfAnswer = len(answer)
if lengthOfAnswer == 3:
if n > answer[-1]:
return f'YES\n{answer[1]} {answer[2]} {int(n)}'
return 'NO'
for i in range(answer[-1]+1, int(n**(1/(4-lengthOfAnswer)))+1):
if not n%i:
return productOfThreeNumbers(n/i, answer+(i, ))
return 'NO'
nTestCases = int(input())
testCases = [int(input()) for x in range(nTestCases)]
[print(productOfThreeNumbers(testCase)) for testCase in testCases]
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible 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 nt in range(int(input())):
n = int(input())
if n<24:
print ("NO")
continue
p = []
curr = 2
while curr*curr<n:
if n%curr==0:
n=n//curr
p.append(curr)
if len(p)==2:
break
curr+=1
if len(p)<=1:
print ("NO")
continue
if n not in p:
print ("YES")
print (p[0],p[1],n)
else:
print ("NO") | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import math
def primeFactors(n):
prime = []
while n % 2 == 0:
prime.append(2)
n = n // 2
i = 3
while i <= math.sqrt(n):
while n % i == 0:
prime.append(i)
n = n//i
i += 2
if n > 2:
prime.append(n)
return prime
te = int(input())
while te > 0:
te -= 1
n = int(input())
factors = primeFactors(n)
# print(factors)
if len(factors) < 3:
print('NO')
else:
if len(factors) == 3:
x, y, z = factors
else:
x = factors[0]
j = 3
if factors[0] != factors[1]:
y = factors[1]
j = 2
else:
y = factors[1]*factors[2]
z = 1
while j < len(factors):
z = z*factors[j]
j += 1
if x == y or y == z or x == z:
print('NO')
else:
print('YES')
print(x, y, z)
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
long long t;
cin >> t;
while (t--) {
long long n;
cin >> n;
long long copy = n;
vector<pair<long long, long long>> v;
long long count = 0;
while (!(n % 2)) {
n >>= 1;
count++;
}
if (count) v.push_back({2, count});
for (long long i = 3; i <= sqrt(n); i += 2) {
count = 0;
while (n % i == 0) {
count++;
n = n / i;
}
if (count) v.push_back({i, count});
}
if (n > 2) v.push_back({n, 1});
if (v.size() < 3) {
if (v.size() == 2) {
long long f1 = v[0].first;
long long f2 = v[1].first;
long long f3 = copy / (v[0].first * v[1].first);
if (f1 != f2 && f1 != f3 && f2 != f3 && f1 >= 2 && f2 >= 2 && f3 >= 2) {
cout << "YES\n";
cout << f1 << " " << f2 << " " << f3 << '\n';
} else
cout << "NO\n";
} else {
if (v[0].second <= 5)
cout << "NO\n";
else {
cout << "YES\n";
cout << v[0].first << " " << v[0].first * v[0].first << " "
<< copy / (v[0].first * v[0].first * v[0].first) << "\n";
}
}
} else {
cout << "YES\n";
cout << v[0].first << " " << v[1].first << " ";
cout << copy / (v[0].first * v[1].first) << '\n';
}
}
}
| CPP |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import math
t=int(input())
for i in range(t):
n=int(input())
flag=0
for j in range(2,int(pow(n,0.5)+1)):
if n%j==0:
n1=n//j
for k in range(j+1,int(pow(n1,0.5)+1)):
n2=n1//k
if n1%k==0 and n2!=k:
flag=1
n2=n1//k
print("YES")
print(j,end=" ")
print(k,end=" ")
print(n2,end=" ")
break
break
if flag==0:
print("NO") | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 |
from __future__ import division
import sys
input = sys.stdin.readline
import math
from math import sqrt, floor, ceil
from collections import Counter, defaultdict
from copy import deepcopy as dc
############ ---- 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 insr2():
s = input()
return(s.split(" "))
def prime_factorization(n):
if n == 1:
return [1]
ans=[]
i = 2
cap = sqrt(n)
while i <= cap:
if n % i == 0:
ans.append(i)
n = n//i
cap=sqrt(n)
else:
i += 1
if n > 1:
ans.append(n)
return ans
def binomial(n, k):
if n == 1 or n == k:
return 1
if k > n:
return 0
else:
a = math.factorial(n)
b = math.factorial(k)
c = math.factorial(n-k)
div = a // (b * c)
return div
def computeGCD(x, y):
if x > y:
small = y
else:
small = x
for i in range(1, small+1):
if((x % i == 0) and (y % i == 0)):
gcd = i
return gcd
# r = raw_input()
for _ in range(inp()):
n = inp()
z = prime_factorization(n)
ans = set()
temp = 1
for i, a in enumerate(z):
if len(ans) >= 2:
temp *= a
else:
if a not in ans:
ans.add(a)
else:
temp *= a
if temp not in ans:
ans.add(temp)
temp = 1
if temp != 1 and temp not in ans and len(ans) == 2:
ans.add(temp)
print 'YES'
print ' '.join(str(a) for a in list(ans))
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 | # Problem C
T = int(input())
for _ in range(T):
n = int(input())
a = 0
b = 0
c = 0
for i in range(2, int(n**0.5)):
if n % i == 0:
a = i
break
else:
continue
if a != 0:
for i in range(a+1, int(n**0.5)):
if n/a % i == 0:
b = i
c = int(n/(a*b))
break
else:
continue
if c > 1 and b != c and a!= c:
print("YES")
print(a,b,c)
else:
print("NO") | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import math
for i in range(int(input())):
n=int(input())
l=[]
count=0
for i in range(2,int(math.sqrt(n))+1):
if n%i==0:
l.append(i)
n=n//i
break
for i in range(2,int(math.sqrt(n))+1):
if n%i==0 and len(l)==1 and l[0]!=i:
l.append(i)
n=n//i
count=1
break
flag=0
if count==1:
if n!=l[0] and n!=l[1]:
l.append(n)
else:
flag=1
if len(l)==3 and flag==0:
print("YES")
print(*l)
else:
print("NO") | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import java.util.*;
import java.io.*;
public class Aman {
public static void main(String args[]) {
Scanner in = new Scanner(System.in);
int t = in.nextInt();
for(int i = 0; i < t; i++) {
int n = in.nextInt();
int ans[] = new int[3];
int m = 0;
int temp = (int) Math.sqrt(n);
for(int j = 2; j < temp && m < 2; j++) {
if(n%j == 0) {
ans[m++] = j;
n = n/j;
}
}
if(m == 2) {
ans[m] = n;
if(ans[1] != ans[2] && ans[0] != ans[2] && n != 1) {
System.out.println("YES");
System.out.print(ans[0] + " " + ans[1] + " " + ans[2] + "\n");
} else
System.out.println("NO");
} else {
System.out.println("NO");
}
}
}
} | JAVA |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int main() {
int t = 1;
cin >> t;
while (t--) {
long long int i, j, k = 0, n, m, s, s1, l, r, x, y, a = 1, b = 1;
cin >> n;
for (i = 2; i * i <= n; i++) {
if (n % i == 0) {
a = i;
n = n / i;
break;
}
}
for (i = 2; i * i <= n; i++) {
if (n % i == 0 && i != a) {
b = i;
n = n / i;
break;
}
}
if (a > 1 && b > 1 && n > 1 && a != b && b != n && a != n) {
cout << "YES" << endl;
cout << a << " " << b << " " << n << endl;
} else
cout << "NO" << endl;
}
}
| CPP |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int main() {
int t;
cin >> t;
while (t--) {
int n, ans = 0, i;
cin >> n;
vector<int> v;
for (i = 2; i < pow(n, 0.5); i++) {
if (ans == 2) break;
if (n % i == 0) {
ans++;
n = n / i;
v.push_back(i);
}
}
if (v.size() == 2) {
v.push_back(n);
cout << "YES"
<< "\n";
for (i = 0; i < 3; i++) {
cout << v[i] << " ";
}
cout << "\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 | prime = []
l = []
for i in range(40000):
prime.append(True)
for i in range(2,40000):
if prime[i] == True:
l.append(i)
tmp = 2*i
while tmp < 40000:
prime[tmp] = False
tmp += i
t = int(input())
for i in range(t):
n = int(input())
flag = False
for j in l:
if j**2 > n:
break
if n%j == 0:
q = n//j
for k in range(j+1, q):
if k**2 > q:
break
if k != j and q%k == 0:
if q//k != k:
flag = True
print("YES")
print(j, k, q//k, sep = " ")
break
break
if flag == False:
print("NO")
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 |
from math import sqrt
t = int(input())
for i in range(t):
n = int(input())
l = []
temp = n
for j in range(2,int(sqrt(n))+1):
if temp%j == 0:
l.append(j)
temp = temp//j
if len(l) == 2:
break
if len(l) < 2:
print("NO")
continue
l.append(n//(l[0]*l[1]))
# print(l)
l.sort()
a,b,c = l[0],l[1],l[2]
if a*b*c != n or a == b or b == c or c == a or a == 1 or b == 1 or c == 1:
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 | from math import sqrt
def primes(n):
y, count, result = n, 0, list()
for i in range(2, int(sqrt(n)) + 1):
if y % i == 0:
count += 1
y = y // i
result.append(i)
if count >= 2:
break
print("YES" if count >= 2 and y not in result else "NO")
if count >= 2 and y not in result:
print(*(result + [y]))
for _ in range(int(input())):
primes(int(input()))
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import math
for i in range(int(input())):
n=int(input())
r=[]
q=2
i=0
while i<2 and q<=math.sqrt(n):
if n%q==0:
r.append(q)
i+=1
n=n//q
q+=1
if n!=1:
r+=[n]
if len(r)<3 or r[1]==r[2] or r[0]==r[2] or r[0]==r[1]:
print('NO')
else:
print('YES')
print(*r)
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible 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 int j, pri[100005], i;
vector<long long int> prime;
int main() {
long long int t;
cin >> t;
while (t--) {
long long int n, tmp, cnt = 0, prod = 1;
cin >> n;
tmp = n;
unordered_set<long long int> st;
for (i = 2; i <= ceil(sqrt(tmp)); i++) {
if (n % i == 0) {
st.insert(i);
n /= i;
cnt++;
prod *= i;
if (cnt == 2) break;
}
}
if (cnt < 2) {
cout << "NO\n";
continue;
}
long long int lst = tmp / (prod);
if (lst != 1) {
st.insert(lst);
if (st.size() == 3) {
cout << "YES\n";
for (auto it = st.begin(); it != st.end(); it++) cout << *it << " ";
cout << "\n";
} else
cout << "NO\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 | '''
* Author : Ayushman Chahar #
* About : II Year, IT Undergrad #
* Insti : VIT, Vellore #
'''
import os
import sys
import math
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
def pf(n):
ans = []
while(n % 2 == 0):
ans.append(2)
n //= 2
for i in range(3, int(math.sqrt(n)) + 1, 2):
while(n % i == 0):
ans.append(i)
n //= i
if(n > 2):
ans.append(n)
return ans
def main():
for _ in range(int(input())):
n = int(input())
res = pf(n)
# print(res)
if(len(res) <= 2):
print("NO")
else:
if(len(res) == 3):
if(len(set(res)) == 3):
print("YES")
print(*res)
else:
print("NO")
else:
a1 = res[0]
a2 = 1
idx = -69
for i in range(1, len(res)):
if a2 != a1 and a2 != 1:
idx = i
break
else:
a2 *= res[i]
a3 = 1
for i in range(idx, len(res)):
a3 *= res[i]
if(a3 != a2 and a3 != a1):
print("YES")
print(a1, a2, a3)
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 | #Ashish Sagar
import math
q=int(input())
for _ in range(q):
n=int(input())
l=[]
for i in range(2,int(math.sqrt(n))):
if n%i==0:
n=n//i
l.append(i)
break
for i in range(2,int(math.sqrt(n))+1):
if n%i==0 and len(l)==1 and i!=l[0]:
n=n//i
l.append(i)
break
if n!=1 and len(l)==2 and n!=l[0] and n!=l[1]:
l.append(n)
print("YES")
print(*l)
else:
print("NO") | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import sys
# sys.setrecursionlimit(10**6)
from sys import stdin, stdout
import bisect #c++ upperbound
import math
import heapq
def modinv(n,p):
return pow(n,p-2,p)
def cin():
return map(int,sin().split())
def ain(): #takes array as input
return list(map(int,sin().split()))
def sin():
return input()
def inin():
return int(input())
import math
def Divisors(n) :
l = []
for i in range(1, int(math.sqrt(n) + 1)) :
if (n % i == 0) :
if (n // i == i) :
l.append(i)
else :
l.append(i)
l.append(n//i)
return l
"""*******************************************************"""
def main():
t=inin()
for _ in range(t):
n=inin()
a=[]
c=0
for i in range(2,int(math.sqrt(n))+5):
if(c==2):
break
if(n%i==0):
a.append(i)
n=n//i
c+=1
if(n>1):
a.append(n)
a=list(set(a))
if(len(a)==3):
print("YES")
print(*a)
else:
print("NO")
######## Python 2 and 3 footer by Pajenegod and c1729
# Note because cf runs old PyPy3 version which doesn't have the sped up
# unicode strings, PyPy3 strings will many times be slower than pypy2.
# There is a way to get around this by using binary strings in PyPy3
# but its syntax is different which makes it kind of a mess to use.
# So on cf, use PyPy2 for best string performance.
py2 = round(0.5)
if py2:
from future_builtins import ascii, filter, hex, map, oct, zip
range = xrange
import os, sys
from io import IOBase, BytesIO
BUFSIZE = 8192
class FastIO(BytesIO):
newlines = 0
def __init__(self, file):
self._file = file
self._fd = file.fileno()
self.writable = "x" in file.mode or "w" in file.mode
self.write = super(FastIO, self).write if self.writable else None
def _fill(self):
s = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.seek((self.tell(), self.seek(0,2), super(FastIO, self).write(s))[0])
return s
def read(self):
while self._fill(): pass
return super(FastIO,self).read()
def readline(self):
while self.newlines == 0:
s = self._fill(); self.newlines = s.count(b"\n") + (not s)
self.newlines -= 1
return super(FastIO, self).readline()
def flush(self):
if self.writable:
os.write(self._fd, self.getvalue())
self.truncate(0), self.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
if py2:
self.write = self.buffer.write
self.read = self.buffer.read
self.readline = self.buffer.readline
else:
self.write = lambda s:self.buffer.write(s.encode('ascii'))
self.read = lambda:self.buffer.read().decode('ascii')
self.readline = lambda:self.buffer.readline().decode('ascii')
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip('\r\n')
# Cout implemented in Python
import sys
class ostream:
def __lshift__(self,a):
sys.stdout.write(str(a))
return self
cout = ostream()
endl = '\n'
# Read all remaining integers in stdin, type is given by optional argument, this is fast
def readnumbers(zero = 0):
conv = ord if py2 else lambda x:x
A = []; numb = zero; sign = 1; i = 0; s = sys.stdin.buffer.read()
try:
while True:
if s[i] >= b'0' [0]:
numb = 10 * numb + conv(s[i]) - 48
elif s[i] == b'-' [0]: sign = -1
elif s[i] != b'\r' [0]:
A.append(sign*numb)
numb = zero; sign = 1
i += 1
except:pass
if s and s[-1] >= b'0' [0]:
A.append(sign*numb)
return A
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 | #include <bits/stdc++.h>
using namespace std;
const long double PI = acos(-1.0L);
const long double EPS = 1e-12L;
mt19937 rng((uint32_t)chrono::steady_clock::now().time_since_epoch().count());
namespace task {
int t, n, k;
vector<int> p;
int main() {
cin >> t;
while (t--) {
cin >> n;
k = n;
p = {};
for (int i = 2; i < min(n, 100000); ++i) {
if (n % i == 0) {
p.push_back(i);
n /= i;
}
}
if (p.size() < 2) {
cout << "NO\n";
} else if (p.size() == 2) {
if (n <= p.back()) {
cout << "NO\n";
} else {
cout << "YES\n" << p[0] << ' ' << p[1] << ' ' << n << '\n';
}
} else {
cout << "YES\n" << p[0] << ' ' << p[1] << ' ' << k / p[0] / p[1] << '\n';
}
}
return 0;
}
} // namespace task
int main() {
ios_base::sync_with_stdio(false);
cin.tie(nullptr);
cout.precision(11);
cout.setf(ios::fixed);
return task::main();
}
| CPP |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
void find(vector<long long int>& v, long long int n) {
for (long long int i = 2; i <= sqrt(n); i++) {
while (n > 0 && n % i == 0) {
v.push_back(i);
n = n / i;
}
}
}
int main() {
int t;
cin >> t;
while (t--) {
long long int n, a, b, c;
cin >> n;
vector<long long int> v;
find(v, n);
if (v.size() < 2) {
cout << "NO" << endl;
continue;
}
a = v[0];
if (v.size() > 2 && v[1] == v[0]) {
b = v[1] * v[2];
} else {
b = v[1];
}
if (n % (a * b) != 0) {
cout << "NO" << endl;
} else {
c = n / (a * b);
if (a > 1 && b > 1 && c > 1 && a != b && a != c && b != c) {
cout << "YES" << endl;
cout << a << " " << b << " " << c << endl;
} else {
cout << "NO" << endl;
}
}
}
return 0;
}
| CPP |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | # -*- coding: utf-8 -*-
"""
Created on Wed Feb 12 15:07:02 2020
@author: Milán
"""
import math
t = int(input())
L = []
for i in range(t):
L += [int(input())]
def multiples(n):
i = 2
while i <= math.floor(n ** (1./3.)):
if n%i == 0:
j = i + 1
k = n/i
while j <= math.sqrt(k):
if k%j == 0 and j != math.sqrt(k):
return "\n" + "YES" + "\n" + str(i) + " " + str(j) + " " + str(int(k/j))
j += 1
i += 1
return "\n" + "NO"
for l in L:
print(multiples(l)) | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int n, a, b, c;
bool solve() {
cin >> n;
int i = 2;
for (i = 2; i * i < n; i++) {
if (n % i == 0) {
n /= i;
a = i;
break;
}
}
for (int j = 2; j * j < n; j++) {
if (j == i) continue;
if (n % j == 0) {
b = j;
c = n / j;
return 1;
}
}
return 0;
}
int main() {
int tc;
cin >> tc;
while (tc--) {
if (solve()) {
cout << "YES\n";
cout << a << ' ' << b << ' ' << c << endl;
} 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 | from math import sqrt
test = int(input())
while test > 0:
test -= 1
n = int(input())
first = []
sz = 0
for i in range (2, int(sqrt(n)+1)):
if(n % i == 0):
first.append(i)
sz += 1
found = False
for i in range (sz):
for j in range (i+1, sz):
if first[i] != first[j] and n % (first[i]*first[j]) == 0\
and n / (first[i]*first[j]) != first[i] and n / (first[i]*first[j]) != first[j]:
print("YES")
print(first[i], end = " ")
print(first[j], end = " ")
print(int(n / (first[i]*first[j])))
found = True
break
if(found): break
if(found == False): print("NO")
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import sys
input = lambda:sys.stdin.readline()
int_arr = lambda: list(map(int,input().split()))
str_arr = lambda: list(map(str,input().split()))
get_str = lambda: map(str,input().split())
get_int = lambda: map(int,input().split())
get_flo = lambda: map(float,input().split())
mod = 1000000007
def solve(n):
ans = []
i = 2
while len(ans) < 2 and i*i < n:
if n % i == 0:
ans.append(i)
n //= i
i += 1
if len(ans) == 2:
print("YES")
print(ans[0],ans[1],n)
else:
print("NO")
for _ in range(int(input())):
n = int(input())
solve(n)
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | #include <bits/stdc++.h>
int main() {
int x, i, j, k, n, c, m[3];
scanf("%d", &x);
while (x--) {
j = 0;
c = 0;
scanf("%d", &n);
for (i = 2; i * i <= n; i++) {
if (n % i == 0) {
k = n / i;
if (k > i) {
for (j = 2; j * j <= k; j++) {
if (k % j == 0 && j != i && j * j != k && k / j != i) {
m[0] = i;
m[1] = j;
m[2] = k / j;
c++;
break;
}
}
if (c > 0) break;
}
}
}
if (c == 0)
printf("NO\n");
else {
printf("YES\n");
for (i = 0; i < 3; i++) printf("%d ", m[i]);
printf("\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.Scanner;
public class task_c {
public String answer;
public boolean correct;
public task_c() {
Scanner scanner = new Scanner(System.in);
int number = scanner.nextInt();
for (int i = 0; i<number; i++) {
int zahl = scanner.nextInt();
answer = "";
correct = false;
recursion(zahl, 3, 2);
if (correct) {
System.out.println("YES");
System.out.println(answer);
}
else {
System.out.println("NO");
}
}
scanner.close();
}
public void recursion(int number, int amount, int last) {
if (amount == 1) {
if (number>=last) {
answer = answer + number;
correct = true;
}
}
else if (last>Math.pow(number, 0.5)) {
correct = false;
}
else {
for (int j = last; j<(int)Math.pow(number, 0.5)+1; j++) {
if (number%j==0) {
answer = answer+ j + " ";
recursion(number/j, amount-1, j+1);
break;
}
}
}
}
public static void main(String[] args) {
new task_c();
}
}
| JAVA |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
#pragma comment(linker, "/STACK:10240000000,10240000000")
char ch;
int bo;
inline bool blank(char ch) {
return ch == ' ' || ch == '\n' || ch == '\r' || ch == '\t';
}
inline void rd(int &x) {
x = bo = 0;
for (ch = getchar(); ch < '0' || ch > '9'; ch = getchar())
if (ch == '-') bo = 1;
for (; ch >= '0' && ch <= '9';
x = (x << 1) + (x << 3) + ch - '0', ch = getchar())
;
if (bo) x = -x;
}
inline void RD(long long &x) {
x = bo = 0;
for (ch = getchar(); ch < '0' || ch > '9'; ch = getchar())
if (ch == '-') bo = 1;
for (; ch >= '0' && ch <= '9';
x = (x << 1) + (x << 3) + ch - '0', ch = getchar())
;
if (bo) x = -x;
}
inline long long RD() {
long long x;
RD(x);
return x;
}
inline int rd() {
int x;
rd(x);
return x;
}
inline void RD(char *s) {
for (ch = getchar(); blank(ch); ch = getchar())
;
for (; !blank(ch); ch = getchar()) *s++ = ch;
*s = 0;
}
inline void RD(char &c) {
for (ch = getchar(); blank(c); ch = getchar())
;
}
template <class T>
inline void OT(T x) {
static char buf[20];
char *p1 = buf;
if (!x) *p1++ = '0';
if (x < 0) putchar('-'), x = -x;
while (x) *p1++ = x % 10 + '0', x /= 10;
while (p1-- != buf) putchar(*p1);
}
const int maxn = 1e3 + 10;
const int mod = 1e9 + 7;
vector<int> v;
int main() {
for (int _ = rd(), Case = 1; Case <= _; Case++) {
int n = rd();
int rn = n;
int num = 0;
v.clear();
for (int i = 2; i * i <= n; i++) {
if (n % i == 0) {
num++;
n /= i;
v.push_back(i);
if (v.size() >= 3) break;
}
}
if (v.size() < 3 && n != 1) v.push_back(n);
if (v.size() < 3)
puts("NO");
else {
int fl = 0;
for (int i = 0; i < v.size(); i++)
for (int j = i + 1; j < v.size(); j++)
if (v[i] == v[j]) fl = 1;
if (fl)
puts("NO");
else {
puts("YES");
printf("%d %d %d\n", v[0], v[1], rn / v[0] / v[1]);
}
}
}
}
| CPP |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import java.io.*;
import java.math.BigInteger;
import java.util.*;
public class Main {
static class FastReader
{
final private int BUFFER_SIZE = 1 << 16;
private DataInputStream din;
private byte[] buffer;
private int bufferPointer, bytesRead;
public FastReader()
{
din = new DataInputStream(System.in);
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public FastReader(String file_name) throws IOException
{
din = new DataInputStream(new FileInputStream(file_name));
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public String readLine() throws IOException
{
byte[] buf = new byte[64]; // line length
int cnt = 0, c;
while ((c = read()) != -1)
{
if (c == '\n')
break;
buf[cnt++] = (byte) c;
}
return new String(buf, 0, cnt);
}
public int nextInt() throws IOException
{
int ret = 0;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do
{
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public long nextLong() throws IOException
{
long ret = 0;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
}
while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public double nextDouble() throws IOException
{
double ret = 0, div = 1;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
}
while ((c = read()) >= '0' && c <= '9');
if (c == '.')
{
while ((c = read()) >= '0' && c <= '9')
{
ret += (c - '0') / (div *= 10);
}
}
if (neg)
return -ret;
return ret;
}
private void fillBuffer() throws IOException
{
bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
if (bytesRead == -1)
buffer[0] = -1;
}
private byte read() throws IOException
{
if (bufferPointer == bytesRead)
fillBuffer();
return buffer[bufferPointer++];
}
public void close() throws IOException
{
if (din == null)
return;
din.close();
}
}
public static PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
public static long mod = (long) (1e9 + 7);
public static int N = (int) 1e5;
public static long[][] dp;
public static List<Integer>[] edges ;
public static void main(String[] args) throws IOException {
// TODO Auto-generated method stub
FastReader sc = new FastReader();
int t = sc.nextInt();
while ((t--) != 0)
{
int n = sc.nextInt();
Map<Integer,Integer> map = new HashMap<>();
find(map, n);
}
out.close();
}
private static void find(Map<Integer, Integer> map, int n)
{
int t = n;
while ((n % 2) == 0)
{
map.put(2,map.getOrDefault(2,0)+1);
n /= 2;
}
int l = (int) Math.sqrt(n);
for (int i = 3 ; i <= l ; i += 2)
{
while (n % i == 0)
{
map.put(i,map.getOrDefault(i,0)+1);
n /= i;
}
}
if (n > 1)
{
map.put(n,map.getOrDefault(n,0)+1);
}
List<Integer> list = new ArrayList<>();
for (int i : map.keySet())
{
for (int j = 0 ; j < map.get(i) ; j++)
{
list.add(i);
}
}
long a = list.get(0);
long b = 1;
long c = 1;
for (int i = 1; i < list.size() ; i++)
{
b *= list.get(i);
long x = (t / (a * b));
if (a == b || b == x || x == a)
{
continue;
}
else
{
c = x;
break;
}
}
if (c == 1)
{
out.println("NO");
return;
}
out.println("YES");
out.println(a+" "+b+" "+c);
}
}
| JAVA |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
void solve() {
long long n;
cin >> n;
set<long long> s;
for (int i = 2; i * i <= n; i++) {
if (n % i == 0) {
n = n / i;
s.insert(i);
break;
}
}
for (int i = 2; i * i <= n; i++) {
if (n % i == 0 && !s.count(i)) {
n = n / i;
s.insert(i);
break;
}
}
if (s.size() < 2 || s.count(n) || n == 1)
cout << "NO" << endl;
else {
s.insert(n);
cout << "YES" << endl;
for (auto e : s) cout << e << " ";
cout << endl;
}
}
int main() {
int t;
cin >> t;
while (t--) {
solve();
}
return 0;
}
| CPP |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | t = int(input())
for _ in range(t):
n = int(input())
if n < 2*3*4:
print("NO")
else:
r = []
i = 2
j = 3
while j > 1:
l = len(r)
for k in range(i, int(n**(1/j) + 2)):
if n % k == 0:
r.append(k)
n //= k
i = k + 1
j -= 1
break
if len(r) == l or len(r) == 2:
break
if len(r) == 2 and n > int(r[-1]):
r.append(n)
print('YES')
print(' '.join(map(str, r)))
else:
print('NO')
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import math
t = int(input())
for i in range(t):
n = int(input())
li = []
n_ = math.ceil(n** 0.5)
for j in range(2,n_):
if n % j == 0:
li.append(j)
n = n // j
if n != 1:
li.append(n)
if len(li) <= 2:
print("NO")
else:
w1 = li[0]
w2 = li[1]
w3 = 1
for k in range(2,len(li)):
w3 = w3 * li[k]
if w1 != w2 and w2 != w3 and w3 != w1:
print("YES")
print(str(w1) + " " + str(w2) + " " + str(int(w3)))
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 find_factors(n):
l = []
while n % 2 == 0:
l.append(2)
n //= 2
for i in range(3,int(math.sqrt(n)+1),2):
while n % i == 0:
l.append(i)
n //= i
if n > 2:
l.append(n)
return l
for _ in range(int(input())):
n = int(input())
k0 = find_factors(n)
k1 = list(dict.fromkeys(k0))
# print(k0)
# print(k1)
if len(k1) == 1:
if len(k0) <= 5:
print("NO")
else:
print('YES')
pro = 1
for i in range(3,len(k0)):
pro *= k0[i]
print(k0[0],k0[1]*k0[2],pro)
elif len(k1) == 2:
if len(k0) <= 3:
print("NO")
else:
pro = 1
print("YES")
for i in range(1,len(k0)-1):
pro *= k0[i]
print(k0[0],k0[-1],pro)
elif len(k1) == 3:
b = [1,1,1]
print("YES")
for i in range(len(k0)):
if k0[i] == k1[0]:
b[0] *= k0[i]
elif k0[i] == k1[1]:
b[1] *= k0[i]
else:
b[2] *= k0[i]
print(*b)
else:
m = k1[0]
n = k1[1]
k0.remove(m)
k0.remove(n)
pro = 1
print("YES")
for i in range(len(k0)):
pro *= k0[i]
print(m,n,pro)
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import java.util.* ;
import java.math.*;
import java.io.*;
public class javaTemplate {
// public static final int M = 1000000007 ;
public static final int M = 1000003 ;
static FastReader sc = new FastReader();
// static Scanner sc = new Scanner(System.in) ;
public static void main(String[] args) {
int t= sc.nextInt() ;
while(t -- != 0) {
int n = sc.nextInt() ;
int a = n , b = n , c = n ;
for(int i = 2 ; i*i <= n ; i++) {
if(n % i == 0) {
a = i ;
break ;
}
}
n = n/a ;
for(int i = 2; i*i <= n ; i++) {
if(n%i == 0) {
if(i != a) {
b = Math.min(b, i) ;
break ;
} else {
if((n/i) != i && (n/i) != a) {
b = Math.min(b,(n/i)) ;
}
}
}
}
c = n/b ;
if(a!= b && b!= c && c != a && c >= 2) {
System.out.println("YES");
System.out.println(a + " " + b + " " + c);
} else {
System.out.println("NO");
}
}
}
//_________________________//Template//_____________________________________________________________________
// FAST I/O
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;
}
}
// Check Perfect Squre
public static boolean checkPerfectSquare(double number) {
double sqrt=Math.sqrt(number);
return ((sqrt - Math.floor(sqrt)) == 0);
}
static boolean isPrime(int n) {
if (n <= 1) return false;
for (int i = 2; i < n; i++) {
if (n % i == 0) return false;
}
return true;
}
// Modulo
static int mod(int a)
{
return (a%M + M) % M;
}
// Palindrom or Not
static boolean isPalindrome(StringBuilder str, int low, int high) {
while (low < high) {
if (str.charAt(low) != str.charAt(high)) return false;
low++;
high--;
}
return true;
}
// Euler Tortient fx
static int Phi(int n) {
int count = 0 ;
for(int i = 1 ; i< n ; i++) {
if(GCD(i,n) == 1) count ++ ;
}
return count ;
}
// Integer Sort
static void sort(int[] a)
{
ArrayList<Integer> l=new ArrayList<>();
for (int i:a) l.add(i);
Collections.sort(l);
for (int i=0; i<a.length; i++) a[i]=l.get(i);
}
// Long Sort
static void sort(long[] a)
{
ArrayList<Long> l=new ArrayList<>();
for (long i:a) l.add(i);
Collections.sort(l);
for (int i=0; i<a.length; i++) a[i]=l.get(i);
}
// boolean Value
static void value(boolean val)
{
System.out.println(val ? "YES" : "NO");
System.out.flush();
}
// GCD
public static long GCD(long a , long b) {
if(b == 0) return a ;
return GCD(b, a%b) ;
}
// sieve
public static boolean [] sieveofEratosthenes(int n) {
boolean isPrime[] = new boolean[n+1] ;
Arrays.fill(isPrime, true);
isPrime[0] = false ;
isPrime[1] = false ;
for(int i = 2 ; i * i <= n ; i++) {
for(int j = 2 * i ; j<= n ; j+= i) {
isPrime[j] = false ;
}
}
return isPrime ;
}
// fastPower
public static long fastPowerModulo(long a, long b, long n) {
long res = 1 ;
while(b > 0) {
if((b&1) != 0) {
res = (res * a % n) % n ;
}
a = (a % n * a % n) % n ;
b = b >> 1 ;
}
return res ;
}
// check if sorted or not
public static boolean isSorted(int[] a)
{
for (int i = 0; i < a.length - 1; i++)
{
if (a[i] > a[i + 1]) {
return false;
}
}
return true;
}
// Palindrom or Not
static boolean isPalindrome(String str, int low, int high) {
while (low < high) {
if (str.charAt(low) != str.charAt(high)) return false;
low++;
high--;
}
return true;
}
}
| 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 sol(n):
ans = []
i = 2
cnt = 0
while n > 0 and i*i <= n:
if not(n%i):
cnt += 1
n //= i
ans.append(i)
i += 1
if cnt >= 2:
break
tr = cnt >= 2 and n not in ans
print("YES" if tr else "NO")
if tr:
print(*ans[:2], n)
t = int(input())
for _ in range(t):
n = int(input())
res = sol(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 | def helper(n):
upper = int(n**0.5) + 1
for i in range(2, upper):
if n % i == 0:
curr = n //i
for j in range(i + 1, upper):
if curr % j == 0:
new_curr = curr //j
if i < j < new_curr: return [i, j, new_curr]
return []
count = int(input())
for _ in range(count):
ans = helper(int(input()))
if len(ans) != 3:
print('NO')
else:
print('YES')
for val in ans:
print(val, 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 | import math
T=int(input())
for _ in range(T):
N=int(input())
n=N
L=[]
while(n%2==0):
L.append(2)
n=n//2
for i in range(3,int(math.sqrt(n))+1,2):
while n % i== 0:
L.append(i)
n = n // i
if(n>2):
L.append(n)
#print(L)
a=L[0]
b=1
c=1
ptr=0
for j in range(1,len(L)):
b=b*L[j]
if(b!=a):
ptr=j+1
break
for j in range(ptr,len(L)):
c=c*L[j]
if(a==b or b==c or a==c or a*b*c!=N or a<2 or b<2 or c<2):
print('NO')
else:
"""a=L[0]
b=1
c=1
ptr=0
for j in range(1,len(L)):
b=b*L[j]
if(b!=a):
ptr=j+1
break
for j in range(ptr,len(L)):
c=c*L[j]
G=list(set(L))
a=G[0]**(L.count(G[0]))
b=G[1]**(L.count(G[1]))
c=1
for j in range(0,len(L)):
if(L[j]!=G[0] and L[j]!=G[1]):
c=c*L[j]"""
print('YES')
print(a,b,c)
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | from math import ceil
t = int(input())
for i in range(t):
n = int(input())
ans = []
if n < 24:
print('NO')
else:
for i in range(2, ceil(n**(1/3))+1):
if n % i == 0:
ans.append(i)
break
if len(ans)==0:
print('NO')
else:
n1 = n//ans[0]
for j in range(ans[0]+1, ceil(n**(1/2))+1):
if n1 % j == 0:
ans.append(j)
break
if len(ans) != 2:
print('NO')
else:
n2 = n1 //ans[1]
if n2 <= ans[1]:
print('NO')
else:
print('YES')
print(f'{ans[0]} {ans[1]} {n2}') | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
template <typename T, typename TT>
ostream& operator<<(ostream& os, const pair<T, TT>& t) {
return os << t.first << " " << t.second;
}
template <typename T>
ostream& operator<<(ostream& os, const vector<T>& t) {
for (auto& i : t) os << i << " ";
return os;
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
;
long long t;
scanf("%lld", &t);
while (t--) {
long long n;
scanf("%lld", &n);
vector<long long> v;
long long tmp = n;
for (int i = 2; i * i <= tmp; i++) {
while (tmp % i == 0) {
v.push_back(i);
tmp /= i;
}
}
if (tmp > 1) v.push_back(tmp);
if (v.size() < 3) {
cout << "NO" << '\n';
continue;
}
long long a = -1, b = -1, c = -1;
a = v[0];
b = v[1];
if (b == a) b *= v[2];
c = n / (a * b);
if (c == b or c == a or c <= 1)
cout << "NO" << '\n';
else {
cout << "YES" << '\n' << a << " " << b << " " << c << '\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 | def getFacts(n):
facts = []
for i in range(2,int(n**0.5)+1):
if n%i == 0:
facts.append(i)
if i != n//i:
facts.append(n//i)
return facts
def solve(n):
facts = getFacts(n)
for i in range(len(facts)):
a = facts[i]
bc = n//a
f = getFacts(bc)
for k in f:
b = k
c = bc//k
if a != b and b != c and a != c:
print('YES')
print(a,b,c)
return
print('NO')
def main():
t = int(input())
for i in range(t):
n = int(input())
solve(n)
main()
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 |
import math
def printDivisors(n) :
i = 2
x = []
while i <= math.sqrt(n):
if (n % i == 0) :
if (n / i == i) :
x.append(i)
else :
x.append(i)
x.append(n//i)
i = i + 1
return x
t = int(input())
for test in range(t):
n = int(input())
x = printDivisors(n)
y = {}
for aa in x:
y[aa] = y.get(aa,0)+1
nn = len(x)
# print(y)
f=0
a,b,c=0,0,0
for i in range(nn-1):
for j in range(i+1,nn):
if (x[i]*x[j])<=n and (x[i]*x[j])!=0 and (n%(x[i]*x[j])==0):
check = n//(x[i]*x[j])
#print(check,x[i],x[j])
if check==x[i] or check==x[j]:
continue
elif check in y:
#print("SS")
f=1
a= x[i]
b = x[j]
c = check
break
if f==1:
break
if f==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=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 | for t in range(int(input())):
n=int(input())
p,q=0,0
"""
for i in range(2,int(n)):
if c==1:
break
elif n%i==0:
for j in range(2,int(pow(n,1/2))+1):
if j==i:
continue
if c==1:
break
if n%(i*j)==0:
r=n/(i*j)
if r==1:
continue
if i!=j and r!=j and i!=r:
if i*j*r==n:
#print (i," ",j," ",int(n/(i*j)))
p=i
q=j
r=int(n/(i*j))
c=1
break
"""
for i in range(2,int(pow(n,1/2)+1)):
if n%i==0:
p=i
n=n/i
break
for i in range(2,int(pow(n,1/2)+1)):
if n%i==0 and p!=i:
q=i
n=n/i
break
#print(p,q,n)
n=int(n)
if p!=0 and q!=0 and n!=1 and n!=p and n!=q:
print("YES")
print (p,q,int(n))
#print (p," ",q," ",r)
else:
print("NO")
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import java.io.DataInputStream;
import java.io.FileInputStream;
import java.io.IOException;
import java.util.ArrayList;
import java.util.List;
public class ProductofThreeNumbers {
static int mod = 1000000007;
static class Reader {
final private int BUFFER_SIZE = 1 << 16;
private DataInputStream din;
private byte[] buffer;
private int bufferPointer, bytesRead;
public Reader() {
din = new DataInputStream(System.in);
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public Reader(String file_name) throws IOException {
din = new DataInputStream(new FileInputStream(file_name));
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public String readLine() throws IOException {
byte[] buf = new byte[10005]; // line length
int cnt = 0, c;
while ((c = read()) != -1) {
if (c == '\n')
break;
buf[cnt++] = (byte) c;
}
return new String(buf, 0, cnt);
}
public int nextInt() throws IOException {
int ret = 0;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public long nextLong() throws IOException {
long ret = 0;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public double nextDouble() throws IOException {
double ret = 0, div = 1;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (c == '.') {
while ((c = read()) >= '0' && c <= '9') {
ret += (c - '0') / (div *= 10);
}
}
if (neg)
return -ret;
return ret;
}
private void fillBuffer() throws IOException {
bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
if (bytesRead == -1)
buffer[0] = -1;
}
private byte read() throws IOException {
if (bufferPointer == bytesRead)
fillBuffer();
return buffer[bufferPointer++];
}
public void close() throws IOException {
if (din == null)
return;
din.close();
}
}
static class IntIntPair implements Comparable<IntIntPair> {
public final int first;
public final int second;
public static IntIntPair makePair(int first, int second) {
return new IntIntPair(first, second);
}
public IntIntPair(int first, int second) {
this.first = first;
this.second = second;
}
public boolean equals(Object o) {
if (this == o) {
return true;
}
if (o == null || getClass() != o.getClass()) {
return false;
}
IntIntPair pair = (IntIntPair) o;
return first == pair.first && second == pair.second;
}
public int hashCode() {
int result = first;
result = 31 * result + second;
return result;
}
public String toString() {
return "(" + first + "," + second + ")";
}
public int compareTo(IntIntPair o) {
int value = Integer.compare(first, o.first);
if (value != 0) {
return value;
}
return Integer.compare(second, o.second);
}
}
public static void main(String[] args) throws IOException {
Reader in = new Reader();
int i, t, n, x;
t = in.nextInt();
StringBuilder ans = new StringBuilder();
while (t-- > 0) {
n = in.nextInt();
List<IntIntPair> num = new ArrayList<>();
for (i = 2; i * i <= n; i++) {
// System.out.println(i + " " + (Math.log(n * 1.0) / Math.log(i * 1.0)));
if (n % i == 0) {
x = (int) (Math.log10(n * 1.0) / Math.log10(i * 1.0));
int z = 1;
for (int j = 1; j <= x + 1; j++) {
if (n % (int) (Math.pow(i, j)) == 0) {
z = j;
}
}
num.add(IntIntPair.makePair(i, z));
n = n / (int) (Math.pow(i, z));
}
}
// System.out.println(n);
if (n != 1) {
num.add(IntIntPair.makePair(n, 1));
}
// System.out.println(num);
if (num.size() == 0) {
ans.append("NO\n");
continue;
}
if (num.size() == 1) {
if (num.get(0).second >= 6) {
ans.append("YES\n");
x = num.get(0).first;
int a = x, b = x * x, c = (int) Math.pow(x, num.get(0).second - 3);
ans.append(a + " " + b + " " + c + "\n");
} else {
ans.append("NO\n");
}
continue;
}
if (num.size() == 2) {
if (num.get(0).second >= 3) {
ans.append("YES\n");
x = num.get(0).first;
int a = x, b = (int) Math.pow(x, num.get(0).second - 1),
c = (int) Math.pow(num.get(1).first, num.get(1).second);
ans.append(a + " " + b + " " + c + "\n");
} else if (num.get(1).second >= 3) {
ans.append("YES\n");
x = num.get(1).first;
int a = x, b = (int) Math.pow(x, num.get(1).second - 1),
c = (int) Math.pow(num.get(0).first, num.get(0).second);
ans.append(a + " " + b + " " + c + "\n");
} else if (num.get(0).second == 2 && num.get(1).second == 2) {
ans.append("YES\n");
x = num.get(1).first;
int a = x, b = num.get(0).first, c = a * b;
ans.append(a + " " + b + " " + c + "\n");
} else
ans.append("NO\n");
continue;
}
int a, b, c;
x = 1;
for (i = 0; i < num.size() - 2; i++) {
x = x * (int) Math.pow(num.get(i).first, num.get(i).second);
}
int y = num.size();
b = (int) Math.pow(num.get(y - 2).first, num.get(y - 2).second);
c = (int) Math.pow(num.get(y - 1).first, num.get(y - 1).second);
ans.append("YES\n" + x + " " + b + " " + c + "\n");
}
System.out.println(ans);
}
}
| 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 compute_problem(n):
if n < 24:
print("NO")
elif n % 10 == 0:
a = 2
n_div_10 = int(n/10)
b = min(5, n_div_10)
c = max(5, n_div_10)
if b != c:
print("YES")
print(a, b, c, sep=" ")
else:
print("NO")
else:
result_found = False
max_a = int(n ** (1. / 3.)) + 1
for a in range(2, max_a):
#print("a: ", a)
if not n % a == 0:
continue
max_b = int((int(n/a) + 1)**(1./2.)) + 1
for b in range(a + 1, max_b):
#print("b: ", b)
prod_ab = a * b
c = int(n/prod_ab)
#print("c: ", c)
#print(( (n % prod_ab == 0) , (prod_ab != n) , (c != a) , (c != b)))
result_found = (n % prod_ab == 0) and (prod_ab != n) and (c != a) and (c != b)
#print(result_found)
if result_found:
print("YES")
print(a, b, c, sep=" ")
break
if result_found:
break
if not result_found:
print("NO")
num_tests = int(input())
for test_num in range(0, num_tests):
test_case = input()
if test_case:
aim_num = int(test_case)
compute_problem(aim_num) | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible 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 = []
d = 2
while len(a) < 2 and d * d <= n:
if 0 == n % d:
a += d,
n //= d
d += 1
if a and n > a[-1]:
a += n,
if len(a) < 3:
print('NO')
else:
print('YES')
print(*a)
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int main() {
int t;
scanf("%d", &t);
while (t--) {
int n;
scanf("%d", &n);
int arr[3];
int k = 0;
for (int i = 2; i * i <= n && k < 2; i++) {
if (n % i == 0) {
n /= i;
arr[k++] = i;
}
}
if (k == 2 && arr[1] < n) {
printf("YES\n%d %d %d\n", arr[0], arr[1], n);
} else {
printf("NO\n");
}
}
return 0;
}
| CPP |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | def Find(n):
for i in range(2,int(n**0.5)+1):
if (not n%i):
p = int(n/i)
if (p >= 2):
for j in range(2,int(p**0.5)+1):
if (i != j) and (not p%j):
k = int(p/j)
if (k>=2) and (i != k) and (j != k):
return "YES\n"+str(i)+" "+str(j)+" "+str(k)
return "NO"
def main():
t = int(input())
outcome = [Find(int(input())) for _ in range(t)]
print("\n".join(outcome))
main()
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | t = int(input())
while t:
n = int(input())
i = 2
a = 0
while i*i<=n:
if n%i==0:
a=i
n=n/i
break
i+=1
i =2
b =0
while i*i<=n:
if n%i==0 and a!=i:
b= i
n=n/i
c = n
break
i+=1
if(a>=2 and b>=2 and c>=2 and c!=b and c!=a):
print("YES")
print(int(a),int(b),int(c))
else:
print("NO")
t-=1 | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import math
from collections import Counter
def primeFactors(n):
a = []
c = 0
while n % 2 == 0:
a.append(2)
n = n // 2
c+=1
# 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(i)
n = n // i
c+=1
# Condition if n is a prime
# number greater than 2
if n > 2:
a.append(n)
c+=1
return Counter(a),c
cases = int(input())
for t in range(cases):
n = int(input())
d,c = primeFactors(n)
v = len(d)
if v>2:
v = list(d.keys())
print("YES")
print(v[0],v[1],n//(v[0]*v[1]))
elif v==2:
if c>3:
v = list(d.keys())
print("YES")
print(v[0],v[1],n//(v[0]*v[1]))
else:
print("NO")
else:
if c>5:
v = list(d.keys())[0]
print("YES")
print(v,v**2,n//(v**3))
else:
print("NO")
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
template <typename T>
void print(set<T> v) {
for (auto x : v) cout << x << " ";
cout << endl;
}
set<long long int> count_div(long long int m, long long int j) {
set<long long int> s;
for (long long int i = 1; i <= sqrtl(m); i++) {
if (m % i == 0 && i > j) {
if (i != m / i && (m / i != j)) {
s.insert(i);
s.insert(m / i);
return s;
} else
continue;
}
}
return s;
}
set<long long int> divisors(long long int n) {
set<long long int> s;
for (long long int i = 1; i <= sqrtl(n); i++) {
if (n % i == 0 && i >= 2) {
if (i != n / i) {
s = count_div(n / i, i);
if (s.size() == 2) {
s.insert(i);
return s;
}
}
}
}
return s;
}
signed main() {
long long int t;
cin >> t;
while (t--) {
long long int n;
cin >> n;
set<long long int> s = divisors(n);
if (s.size() == 3) {
cout << "YES" << endl;
print(s);
} else
cout << "NO" << endl;
}
}
| CPP |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | for _ in range(int(input())):
n=int(input())
ans=False
for i in range(2,int(n**(0.5))+1):
a=i
if n%i==0:
div=n//i
for j in range(2,int(div**0.5)+1):
if div%j==0 and j!=a:
div2=div//j
if div2!=1 and div2!=j and div2!=a:
ans=True
break
if ans:
break
if ans:
print("YES")
print(a,div//div2,div2)
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
res=[1,1,1]
def del1(a,k,d):
sqrtlam=lambda x : math.sqrt(x)
for i in range(d,math.ceil(sqrtlam(a))):
if a%i==0:
if(k)==1:
if(int(a/i)!=res[0]):
res[1]=int(i)
res[2]=int(a/i)
return 1
else:
res[0]=i
if(del1(a/i,k+1,i+1)):
return 1
return 0
n=int(input())
for i in range(n):
res=[1,1,1]
b=int(input())
if(del1(b,0,2)==0 ):
print("NO")
else:
print("YES")
print (*res)
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible 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 wa(a):
for i in range(2,int(math.sqrt(a)),1):
if(a%i==0):
return False
return True
def ans(a):
count=0
f=0
l=[]
s=''
d=int(math.sqrt(a))
for i in range(2,d+1,1):
if(a%i==0):
l.append(i)
count+=1
a=a//i
if(count==2 and l[0]!=a and l[1]!=a):
l.append(a)
f=1
break
if(f):
print("YES")
for i in l:
print(i,end=' ')
print()
else:
print("NO")
t=int(input())
while(t>0):
a=int(input())
if(wa(a)==True):
print("NO")
elif(a%10==0):
x=a//10
y=2
z=5
if(x*z*y==a and x!=y and x!=z and z!=y and x>1 and y>1 and z>1):
print("YES")
if(z>x):
print(y,x,z)
else:
print(y,z,x)
else:
print("NO")
else:
ans(a)
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
def function(num):
var = int(math.sqrt(num)) + 1
for i in range(2,var+1):
if (num%i == 0) and (num//i != i):
return list((i,num//i))
t = int(input())
for _ in range(t):
count = 0
n = int(input())
num = int(math.sqrt(n)) + 1
for i in range(2,num):
if n%i == 0:
func = function(n//i)
if func == None:pass
else:
if len(set([i,func[0],func[1]])) == 3:
count = 1
print('YES')
print('{} {} {}'.format(i,func[0],func[1]))
break
else:pass
if count == 0:
print("NO")
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
void solve() {
set<int> s;
int n, i, a, l, p = 1, pr = 1;
cin >> n;
l = n;
for (i = 2; i * i <= l; i++) {
if (n % i == 0 && p < 3) {
s.insert(i);
n = n / i;
if (p == 2 && n != 1) {
s.insert(n);
}
p++;
}
}
for (set<int>::iterator i = s.begin(); i != s.end(); i++) {
pr = pr * (*i);
}
if (s.size() > 2 && pr == l) {
cout << "YES\n";
for (set<int>::iterator i = s.begin(); i != s.end(); i++) cout << *i << " ";
cout << "\n";
} else
cout << "NO\n";
}
int main() {
int t;
cin >> t;
while (t--) {
solve();
}
}
| CPP |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | t = int(input())
for i in range(t):
n = int(input())
ans = 'NO'
p = q = 1
for j in range(2, int(n ** (1 / 3)) + 2):
if n % j == 0:
p = j
n = n // p
ans = 'YES'
break
if ans == 'YES':
ans = 'NO'
for j in range(p + 1, int(n ** (1 / 2)) + 2):
if n % j == 0:
q = j
n = n // q
ans = 'YES'
break
if n <= q:
ans = 'NO'
print(ans)
if ans == 'YES':
print(p, q, n)
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import math
def divisorGenerator(n):
large_divisors = []
for i in range(1, int(math.sqrt(n) + 1)):
if n % i == 0:
yield i
if i*i != n:
large_divisors.append(int(n / i))
for divisor in reversed(large_divisors):
yield divisor
n = int(input())
for x in range(n):
a = int(input())
t = (list(divisorGenerator(a)))
t.pop(-1)
t.pop(0)
b1 = False
b2 = False
if len(t)<=2:
print('NO')
continue
for z in range(0, len(t)-2):
for zz in range(z+1, len(t)-1):
for zzz in range(zz+1, len(t)):
if t[z]*t[zz]*t[zzz] == a:
print('YES')
print(t[z], t[zz], t[zzz])
b1 = True
break
if b1 == True:
b2 = True
break
if b2 == True:
break
if b1 == False and b2 == False:
print('NO') | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | from sys import stdin
input=stdin.readline
#a=[]
import math
def com(p,q,r):
if p==q or q==r or p==r:
return -1
else:
return 1
def pd(n) :
t=[]
i = 1
while i <= math.sqrt(n):
if (n % i == 0 and i!=1) :
if (n // i == i) :
t.append([i,i])
else :
t.append([i,n//i])
i+=1
return t
for _ in range (int(input())):
n=int(input())
a=pd(n)
ans=-1
for i in range(len(a)):
if ans==1:
break
x=a[i][0]
y=a[i][1]
b=pd(x)
c=pd(y)
for i in range(len(b)):
ans = com(b[i][0],b[i][1],y)
if ans==1:
ans1=[b[i][0],b[i][1],y]
break
if ans==1:
break
for i in range(len(c)):
ans = com(c[i][0],c[i][1],x)
if ans==1:
ans1=[c[i][0],c[i][1],x]
break
if ans==1:
print('YES')
print(*ans1)
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 f(n):
d={}
for i in range(2,int(pow(n,0.5))+1):
if n%i==0:
d[i]=0
while n%i==0:
d[i]=d[i]+1
n=n//i
if n!=1:
d[n]=1
return(d)
final=[]
t=int(input())
for i in range(0,t):
n=int(input())
d=f(n)
values = list(d.values())
keys=list(d.keys())
if len(keys)>=3:
ans='YES'
x=n//(keys[0]*keys[1])
final=[x,keys[0],keys[1]]
elif len(keys)==2:
if values[0] ==2 and values[1]==2:
ans='YES'
final=[keys[0],keys[1],keys[0]*keys[1]]
elif values[0]>=3 or values[1]>=3:
ans='YES'
if d[keys[0]]>=3:
final=[keys[0],keys[0]**(d[keys[0]]-1),keys[1]**(d[keys[1]])]
else:
final=[keys[1],keys[1]**(d[keys[1]]-1),keys[0]**(d[keys[0]])]
else:
ans='NO'
else:
if len(keys)==0:
ans='NO'
else:
if values[0]>=6:
ans='YES'
final=[keys[0],keys[0]**2,n//(keys[0]**3)]
else:
ans='NO'
if ans=='NO':
print(ans)
else:
print(ans)
for i in range(0,len(final)):
print(final[i],end=' ')
print() | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import math
def sf(s,n):
for i in range(s,int(math.sqrt(n)+1)):
if n%i==0:
n=n//i
return i,n
return 1,n
t=int(input())
#t=1
for _ in range(t):
a,b,c=1,1,1
n=int(input())
a,n=sf(2,n)
b,n=sf(a+1,n)
c=n
#print(a,b,c)
if a!=b and b!=c and c!=1 and a!=c and a!=1 and b!=1:
print("YES")
print(a,b,c)
else:
print("NO")
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | t = int(input())
while t:
t-=1
n = int(input())
factors = []
for i in range(2,int(n**0.5)+1):
if n%i==0:
factors.append(i)
if n//i!=i:
factors.append(n//i)
if len(factors)<2:
print("NO")
continue
factors.sort()
done = False
for i in range(len(factors)):
for j in range(i+1,len(factors)):
a = factors[i]
b = factors[j]
c = n//(a)
if c%b!=0:
continue
c = c//b
if c>=2 and a!=c and b!=c :
done = True
print("YES")
print(a,b,c)
break
if done:
break
if not done:
print("NO") | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | t = int(input())
for T in range(t):
n = int(input())
ans = []
for i in range(2, int(n ** 0.5) + 1):
if n % i == 0:
ans.append(i)
val = n // i
for j in range(2, int(val ** 0.5) + 1):
ans = [i]
if val % j == 0:
ans.append(j)
ans.append(val // j)
if len(set(ans)) == 3:
break
## print(ans)
if len(set(ans)) == 3:
break
ans.append(n // i)
val = i
for j in range(2, int(val ** 0.5) + 1):
if val % j == 0:
ans.append(j)
ans.append(val // j)
if len(set(ans)) == 3:
break
ans = [n // i]
## print(ans)
if len(set(ans)) == 3:
break
ans = []
if len(ans) == 3:
print("YES")
print(*ans)
else:
print("NO")
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | def factors(n):
i = 2
f = []
s = n
while i*i <= n:
while s % i == 0:
s //= i
f.append(i)
i += 1
else:
if s > 1:
f.append(s)
return f
def solve():
f = factors(n)
s = set()
i = 0
x = 1
for i in f:
if i not in s:
s.add(i)
if len(s) == 2: break
else:
x *= i
if x != 1 and x not in s:
s.add(x)
if len(s) == 2: break
x = 1
ss = list(s)
r = n // (ss[0]*ss[1]) if len(ss) == 2 else 1
if len(s) == 2 and r != 1 and r != ss[0] and r != ss[1]:
s.add(r)
if len(s) >= 3:
print("YES")
print(' '.join(map(str, list(s)[:3])))
else:
print("NO")
def main():
global n
for _ in range(int(input())):
n = int(input())
solve()
main()
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | t = int(input())
def factorization(n):
ind = []
arr = []
temp = n
for i in range(2, int(-(-n**0.5//1))+1):
if temp%i==0:
cnt=0
while temp%i==0:
cnt+=1
temp //= i
ind.append(cnt)
arr.append(i)
if temp!=1:
ind.append(1)
arr.append(temp)
if arr==[]:
ind.append(1)
arr.append(n)
return arr, ind
for i in range(t):
n = int(input())
arr, ind = factorization(n)
if len(arr) == 1:
if ind[0] >= 6:
print('YES')
print(arr[0], arr[0]**2, arr[0]**(ind[0]-3))
else:
print('NO')
elif len(arr) == 2:
if ind[0] >= 3:
print('YES')
print(arr[0], arr[0]**2, (arr[0]**(ind[0]-3))*(arr[1]**ind[1]))
elif ind[1] >= 3:
print('YES')
print(arr[1], arr[1]**2, (arr[1]**(ind[1]-3))*(arr[0]**ind[0]))
elif ind[0] >= 2 and ind[1] >= 2:
print('YES')
print(arr[0], arr[1], (arr[0]**(ind[0]-1))*(arr[1]**(ind[1]-1)))
else:
print('NO')
else:
print('YES')
print(arr[0], arr[1], (n//arr[0])//arr[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() {
long long t, n;
cin >> t;
while (t--) {
cin >> n;
int a[3] = {}, count = 0;
for (long long i = 2; i * i < n; i++) {
if (n % i == 0) {
a[count] = i;
count++;
n /= i;
}
if (count >= 2) break;
}
if (count != 2)
cout << "NO\n";
else {
cout << "YES\n";
cout << a[0] << " " << a[1] << " " << n << endl;
}
}
}
| CPP |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | # import sys
# file = open('test1')
# sys.stdin = file
def ii():
a = int(input())
return a
def ai():
a = list(map(int, input().split()))
return a
def mi():
a = map(int, input().split())
return a
from math import sqrt
def fact(n):
lst = []
for i in range(2, int(sqrt(n))+1):
if n%i==0:
a = i
# print(a,"a")
b = n//i
for j in range(2, int(sqrt(b))+1):
# print(j,"j")
if b%j == 0 and j!=a and (b//j)!=a and j!=(b//j):
lst.append(a)
lst.append(j)
lst.append(b//j)
return lst
for _ in range(ii()):
n = ii()
lst = fact(n)
if lst:
print("YES")
print(*lst)
else:
print("NO")
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | def isPrime(n : int):
if n == 1:
return False
if n == 2:
return True
else:
for i in range(2, n // 2):
if n % i == 0:
return False
return True
t = int(input())
for i in range(t):
n = int(input())
y = n
res = []
k = 2
for k in range(2, int(y ** 0.5) + 1):
if n % k == 0:
res.append(k)
n = n // k
k += 1
if len(res) == 2:
break
if len(res) == 2 and res[-1] < n:
print("YES")
print(*res, 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 main():
n = int(input())
different = [0, 0, 0]
k = 0
for i in range(2, int(math.sqrt(n)) + 1):
if k == 2:
break
if n % i == 0 and k < 2:
n = n // i
different[k] = i
k += 1
if n == 1 or n == different[0] or n == different[1] or different[1] == 0:
print("NO")
return
different[2] = n
print("YES")
print(different[0], different[1], different[2])
if __name__ == "__main__":
t = int(input())
for i in range(t):
main()
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import math
t = int(input())
def primefactorize(n):
prime_factors_list = []
while n%2 == 0:
prime_factors_list.append(2)
n = n/2
for i in range(3,int(math.sqrt(n))+1,2):
while n%i == 0:
prime_factors_list.append(i)
n = n/i
if n > 2:
prime_factors_list.append(n)
return prime_factors_list
def generate_factors(prime_factors_list,n):
a = 1
b = 1
c = 1
for i in range(len(prime_factors_list)):
if a == 1 :
a *= prime_factors_list[i]
elif b == 1 or a == b:
b *= prime_factors_list[i]
else:
c *= prime_factors_list[i]
return a,b,c
for tt in range(t):
n = int(input())
prime_factors = primefactorize(n)
# print(prime_factors)
a,b,c = generate_factors(prime_factors,n)
if (a!=1 and b!=1 and c!=1) and (a!=b and b!=c and a!=c):
print("YES")
print(str(int(a))+ " " + str(int(b)) + " " + str(int(c)))
else:
print("NO") | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import sys
import math
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()))
t = inp()
for i in range(t):
n = inp()
i = 2
check = False
while(i*i <= n):
if(n % i == 0):
check = True
nn = n//i
j = i + 1
check2 = False
while(j *j <= nn):
if(nn % j == 0):
c = nn // j
if(c != j):
if(c != i):
print("YES")
print(str(i) + " " + str(j) + " " + str(c))
check2 = True
break
j = j + 1
if(check2 == False):
check = False
break
i = i + 1
if(check == False):
print("NO") | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | from sys import stdin
input=lambda:stdin.readline().strip()
from collections import defaultdict
def fun(n,dict1):
if n%2==0 and (n//2)!=2 and dict1[2]==0:
#print(dict1[2])
return [2,n//2]
elif n%3==0 and (n//3)!=3 and dict1[n//3]==0 and dict1[3]==0:
return [3,n//3]
else:
i=2
while i*i<n:
if n%i==0 and (n//i)!=i and dict1[n//i]==0 and dict1[i]==0:
#print(dict1[i],'i',i)
return [i,n//i]
i+=1
return -1
for _ in range(int(input())):
n=int(input())
if n<8:
print('NO')
else:
dict1=defaultdict(int)
a=fun(n,dict1)
if a==(-1):
print('NO')
else:
a,b=a
dict1[a]+=1
#print(dict1)
c=fun(b,dict1)
if c==(-1):
print('NO')
else:
print('YES')
print(a,c[0],c[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;
void solve() {
int n;
cin >> n;
vector<int> factors;
for (int i = 2; 1LL * i * i <= (long long int)n; ++i) {
if (n % i == 0) {
factors.push_back(n / i);
if (n / i != i) {
factors.push_back(i);
}
}
}
if (factors.size() < 2) {
cout << "NO\n";
} else {
sort(factors.begin(), factors.end());
int poss = 0;
for (int i = 0; i < factors.size() and !poss; ++i) {
for (int j = i + 1; j < factors.size() and !poss; ++j) {
int a = factors[i];
int b = factors[j];
int c = n / a;
if (c % b != 0) {
continue;
}
c /= b;
if (c >= 2 and c != a and c != b) {
poss = 1;
cout << "YES\n";
cout << a << " " << b << " " << c << '\n';
break;
}
}
}
if (!poss) cout << "NO\n";
}
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(nullptr);
cout.tie(nullptr);
int t;
cin >> t;
while (t--) solve();
return 0;
}
| CPP |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import sys
import math
import bisect
from sys import stdin, stdout
from math import gcd, floor, sqrt, log2, ceil
from collections import defaultdict
from bisect import bisect_left as bl, bisect_right as br
from collections import Counter
from collections import deque
ip = lambda : int(stdin.readline())
inp = lambda: map(int,stdin.readline().split())
ips = lambda: stdin.readline().rstrip()
t = ip()
for _ in range(t):
n = ip()
ch = []
for i in range(2,int(n**0.5)+1):
if n%i == 0:
ch.append(i)
ch.append(n//i)
break
if len(ch) == 2:
a = ch[0]
b = ch[1]
ch = []
for i in range(2,int(a**0.5)+1):
if a%i == 0:
if i != b and a//i != b and i != a//i:
ch.append(i)
ch.append(a//i)
break
if len(ch) == 2:
print("YES")
print(b,ch[0],ch[1])
else:
ch = []
for i in range(2,int(b**0.5)+1):
if b%i == 0:
if i != a and b//i != a and i!= b//i:
ch.append(i)
ch.append(b//i)
break
if len(ch) == 2:
print("YES")
print(a,ch[0],ch[1])
else:
print("NO")
else:
print("NO")
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 |
/*
_oo0oo_
o8888888o
88" . "88
(| -_- |)
0\ = /0
___/`---'\___
.' \\| |// '.
/ \\||| : |||// \
/ _||||| -:- |||||- \
| | \\\ - /// | |
| \_| ''\---/'' |_/ |
\ .-\__ '-' ___/-. /
___'. .' /--.--\ `. .'___
."" '< `.___\_<|>_/___.' >' "".
| | : `- \`.;`\ _ /`;.`/ - ` : | |
\ \ `_. \_ __\ /__ _/ .-` / /
=====`-.____`.___ \_____/___.-`___.-'=====
`=---='
*/
import java.util.function.Consumer;
import java.util.*;
import java.math.BigDecimal;
import java.math.BigInteger;
import java.math.RoundingMode;
import java.text.DecimalFormat;
import java.io.*;
import java.lang.Math.*;
public class KickStart2020{
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());}
float nextFloat() {return Float.parseFloat(next());}
String nextLine() {
String str = "";
try {str = br.readLine();}
catch (IOException e) { e.printStackTrace();}
return str; }}
static boolean isBracketSequence(String s, int a, int b) {
Stack<Character> ss = new Stack<>();
boolean hachu = true;
for(int i = a; i <= b; i++) {
if(s.charAt(i) == ')' && ss.isEmpty()) {hachu = false; break;}
if(s.charAt(i) == '(') ss.add('(');
else ss.pop();
}
return ss.empty() && hachu;
}
static String reverseOfString(String a) {
StringBuilder ssd = new StringBuilder();
for(int i = a.length() - 1; i >= 0; i--) {
ssd.append(a.charAt(i));
}
return ssd.toString();
}
static char[] reverseOfChar(char a[]) {
char b[] = new char[a.length];
int j = 0;
for(int i = a.length - 1; i >= 0; i--) {
b[i] = a[j];
j++;
}
return b;
}
static boolean isPalindrome(char a[]) {
boolean hachu = true;
for(int i = 0; i <= a.length / 2; i++) {
if(a[i] != a[a.length - 1 - i]) {
hachu = false;
break;
}
}
return hachu;
}
static long gcd(long a, long b)
{
if (b == 0)
return a;
return gcd(b, a % b);
}
static long power(long x, long y, long mod){
long ans = 1;
x = x % mod;
if (x == 0)
return 0;
int i = 1;
while (y > 0){
if ((y & 1) != 0)
ans = (ans * x) % mod;
y = y >> 1;
x = (x * x) % mod;
}
return ans;
}
static boolean check(String a) {
boolean hachu = true;
for(int i = 0; i < a.length(); i++) {
if(a.charAt(0) != a.charAt(i)) {hachu = false; break;}
}
return hachu;
}
public static class Pair implements Comparable<Pair> {
public final int index;
public final int value;
public Pair(int index, int value) {
this.index = index;
this.value = value;
}
@Override
public int compareTo(Pair other) {
//multiplied to -1 as the author need descending sort order
return -1 * Integer.valueOf(this.value).compareTo(other.value);
}
}
static boolean equalString(int i, int j, int arr[], String b) {
int brr[] = new int[26];
for(int k = i; k <= j; k++) brr[b.charAt(k) - 'a']++;
for(int k = 0; k < 26; k++) {
if(arr[k] != brr[k]) return false;
}
return true;
}
static boolean cequalArray(int arr[], int brr[], int a, int b) {
int count[] = new int[101];
int count1[] = new int[101];
for(int i = a; i <= b; i++) count[arr[i]]++;
for(int i = a; i <= b; i++) count1[brr[i]]++;
for(int i = 0; i < 101; i++) if(count[i] != count1[i]) return false;
return true;
}
public static void main(String[] args) throws Exception{
FastReader sc = new FastReader();
PrintWriter out = new PrintWriter(System.out);
int t =sc.nextInt();
while(t-- > 0) {
int n = sc.nextInt();
ArrayList<Integer> ss = new ArrayList<>();
int i = 2;
int z = (int)Math.sqrt(n);
while(true) {
if(n % i == 0) {
ss.add(i);
n /= i;
}
i++;
if(i > z) break;
if(ss.size() == 2) break;
}
if(ss.size() == 2 && n != 1 && n != (int)ss.get(0) && n != (int)ss.get(1)) {
out.println("YES");
out.println(ss.get(0) + " " + ss.get(1) + " " + n);
}
else out.println("NO");
}
out.close();
}
} | JAVA |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import sys
sys.setrecursionlimit(10**9)
IA =lambda: map(int,input().split())
ans=[0,0,0,0]
def solve(n):
num=int(0)
i=int(2)
tmp=1
m=n
while i*i<=n:
if n%i==0:
n=n//i
tmp*=i
if tmp not in ans:
num+=1
ans[num]=tmp
tmp=1
if num>=3:
ans[num]*=n
return 1
i+=1
if n>1:
if tmp*n not in ans:
num+=1
ans[num]=n*tmp
if num>=3: return 1
else:return -1
T=int(input())
for t in range(0,T):
ans=[0,0,0,0]
n=int(input())
if solve(n)==1:
print("YES")
for i in range(1,4):
print(ans[i],end=" ")
print()
else:print("NO")
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | #include <bits/stdc++.h>
const long long int MOD = 1e9 + 7;
const int cfnum = 1e5 + 5;
using namespace std;
long long int gcd(long long int a, long long int b) {
if (a == 0) return b;
return gcd(b % a, a);
}
int main() {
int q;
cin >> q;
while (q--) {
long long int n;
cin >> n;
vector<long long int> ans;
for (int i = 2; i < sqrt(n); i++) {
if (n % i == 0) {
if (ans.size() == 0) {
ans.push_back(i);
n /= i;
} else {
ans.push_back(i);
ans.push_back(n / i);
break;
}
}
}
if (ans.size() == 3) {
cout << "YES"
<< "\n";
cout << ans[0] << " " << ans[1] << " " << ans[2] << "\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 | #include <bits/stdc++.h>
using namespace std;
int main() {
long long n, t, a, b, c, var;
cin >> t;
vector<long long> fact1, fact2, fact3;
while (t--) {
cin >> n;
var = n;
a = b = c = 1;
fact1.clear();
fact2.clear();
fact3.clear();
for (int i = 2; i * i <= n; i++) {
if (n % i == 0) {
fact1.push_back(i);
fact1.push_back(n / i);
}
}
sort(fact1.begin(), fact1.end());
if (fact1.size() == 0) {
cout << "NO\n";
continue;
}
n = n / fact1[0];
a = fact1[0];
for (int i = 2; i * i <= n; i++) {
if (n % i == 0) {
fact2.push_back(i);
fact2.push_back(n / i);
}
}
fact2.push_back(n);
sort(fact2.begin(), fact2.end());
for (int i = 0; i < fact2.size(); i++) {
if (fact2[i] != a) {
b = fact2[i];
break;
}
}
c = var / (a * b);
if (a != b && b != c && c != a && c != 1) {
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 | #include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
int q, n;
cin >> q;
while (q--) {
cin >> n;
set<int> A;
for (int i = 2; i * i < n; i++) {
if (n % i == 0) {
A.insert(i);
n /= i;
break;
}
}
for (int i = 2; i * i < n; i++) {
if (n % i == 0) {
if (A.count(i)) continue;
A.insert(i);
n /= i;
break;
}
}
if (A.count(n) || n == 1 || A.size() < 2) {
cout << "NO\n";
} else {
cout << "YES\n";
A.insert(n);
for (int a : A) cout << a << " ";
cout << "\n";
}
}
}
| CPP |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | # ░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░
# ░░░░░░░▄▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▄░░░░░░
# ░░░░░░█░░▄▀▀▀▀▀▀▀▀▀▀▀▀▄░░█░░░░░
# ░░░░░░█░█░░▀░░░░░░░░▀░░█░█░░░░░
# ░░░░░░█░█░░░▀░░░░░░▀░░░█░█░░░░░
# ░░░░░░█░█░░░░▀░░░░▀░░░░█░█░░░░░
# ░░░░░░█░█▄░░░░▀░░▀░░░░▄█░█░░░░░
# ░░░░░░█░█░░░░░░██░░░░░░█░█░░░░░
# ░░░░░░█░▀░░░░░░░░░░░░░░▀░█░░░░░
# ░░░░░░█░░░░░░░░░░░░░░░░░░█░░░░░
# ░░░░░░█░░░░░░░░░░░░░░░░░░█░░░░░
# ░░░░░░▀░░░░░░░░░░░░░░░░░░▀░░░░░
# ░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░
import math
for i in range(int(input())):
n = int(input())
l = []
p =math.sqrt(n)
p = int(p)
i=2
while len(l)<2 and i<p:
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,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
input = sys.stdin.readline
# sys.setrecursionlimit(400000)
def I(): return input().strip()
def II(): return int(input().strip())
def LI(): return [*map(int, input().strip().split())]
import copy, string, math, time, functools, random, fractions
from heapq import heappush, heappop, heapify
from bisect import bisect_left, bisect_right
from collections import deque, defaultdict, Counter, OrderedDict
# from itertools import permutations, combinations, groupby
from operator import itemgetter
import itertools
for _ in range(II()):
n = II()
l = []
while(n%2==0):
l.append(2)
n = n//2
for i in range(3,int(math.sqrt(n))+1,2):
while n%i == 0:
l.append(i)
n = n//i
if n>2:
l.append(n)
p = len(set(l))
z = len(l)
if p == 1:
if z >= 6:
print("YES")
print(l[0],end = ' ')
print(l[1]*l[2],end = ' ')
print(l[0]**(z-3))
else:
print("NO")
elif p == 2:
if z>=4:
l.sort()
y = 1
m = list(set(l))
v = m[0]
k = m[1]
l.remove(v)
l.remove(k)
print("YES")
print(v,end = ' ')
print(k,end = ' ')
for i in range(len(l)):
y = y*l[i]
print(y)
else:
print("NO")
elif p>2:
print("YES")
x=1
if z == 3:
print(*l)
else:
l.sort()
for i in range(3, z):
x = x * l[i]
print(x, l[0], l[1] * l[2])
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | t = int(input())
for x in range(t):
n = int(input())
i = 2
rt = int(n**0.5)
l = []
dp = [True]*(int(n**0.5)+1)
while i <= rt and len(l) < 3:
if n%i!=0:
i+=1
else:
if dp[i]:
l.append(i)
dp[i] = False
if len(l) == 2:
if l[0] != n//i and i != n//i:
l.append(n//i)
n = n//i
rt = int(n**0.5)
i = 2
else:
i += 1
if len(l) == 3:
print("YES")
print(l[0], l[1], l[2])
else:
print("NO") | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import math
def mul3(n):
original_n = n
divisors = set()
divisor = 2
sqrt_n = int(math.sqrt(n))
while divisor <= sqrt_n and len(divisors) < 2:
if n % divisor == 0:
divisors.add(divisor)
n //= divisor
divisor += 1
if len(divisors) < 2:
return None
list_divisors = list(divisors)
multiplication = list_divisors[0] * list_divisors[1]
third_number = original_n // multiplication
if third_number in divisors:
return None
return divisors | set([third_number])
# 24 -> 2, 3, 4
# 8 -> None
# 7 -> None
# 6 -> None
# 13 * 17 * 19 -> 13, 17, 19
number_of_tests = int(input())
for _ in range(number_of_tests):
n = int(input())
result = mul3(n)
if not result:
print('NO')
continue
print('YES')
print(' '.join(map(str, result)))
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import math
import collections
import sys
import math
def primeFactors(n):
l = []
count = 0
while n % 2 == 0:
count+=1
n = n // 2
if count > 0:
l.append([2, count])
count = 0
for i in range(3, int(math.sqrt(n))+1, 2):
while n % i == 0:
count+=1
n = n // i
if count > 0:
l.append([i, count])
count = 0
if n > 2:
l.append([n, 1])
return l
input = sys.stdin.readline
for _ in range(int(input())):
n = int(input())
l = primeFactors(n)
if len(l) >= 3:
print("YES")
m1, m2, m3 = 1, 1, 1
m1 = l[0][0]**l[0][1]
m2 = l[1][0]**l[1][1]
for i in range(2, len(l)):
m3*=l[i][0]**l[i][1]
print(m1, m2, m3)
else:
if len(l) == 1:
if l[0][1] >= 6:
print("YES")
print(l[0][0]**1, l[0][0]**2, l[0][0]**(l[0][1]-3))
else:
print("NO")
else:
if l[0][1] + l[1][1] >= 4:
print("YES")
print(l[0][0], l[1][0], (l[0][0]**(l[0][1]-1))*(l[1][0]**(l[1][1]-1)))
else:
print("NO") | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import math
for _ in range(int(input())) :
n=int(input())
s=[]
try:
for i in range(2,int(math.sqrt(n))+1):
if n%i==0:
g=(n//i)
a=i+0
break
for i in range(a+1,int(math.sqrt(g))+1):
if g%i==0:
b=i
break
for i in range(2,int(math.sqrt(n))+1):
if n%i==0:
s.append(i)
except:
print("NO")
continue
if len(s)>=3:
print("YES")
print(a,b,n//(a*b))
else:
print("NO") | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import sys
for _ in range(input()):
n=input()
flag=0
for i in range(2,int(n**(0.5))+1):
if n%i==0:
a=i
s=n/i
b=0
c=0
for j in range(2,int(s**(0.5)+1)):
if s%j==0 and a!=j:
b=j
c=s/j
if a!=b and b!=c and a!=c and a>=2 and b>=2 and c>=2:
print "YES"
arr= [a,b,c]
arr.sort()
for i in arr:
print i,
print
flag=1
break
if flag:
break
else:
print "NO"
| PYTHON |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import math
from bisect import bisect_left, bisect_right
from sys import stdin, stdout
input = lambda: stdin.readline().strip()
print = lambda s: stdout.write(s)
primes = []
def SieveOfEratosthenes(n):
prime = [True for i in range(n+1)]
p = 2
while (p * p <= n):
if (prime[p] == True):
for i in range(p * p, n+1, p):
prime[i] = False
p += 1
for p in range(2, n):
if prime[p]:
primes.append(p)
SieveOfEratosthenes(10**6)
t = int(input())
for _ in range(t):
n = int(input())
ls = []
for i in primes:
if n%i==0:
ls.append(i)
if len(ls)==2:
break
if len(ls)==0:
print('NO\n')
elif len(ls)>=2:
if n//(ls[0]*ls[1]) in ls or n//(ls[0]*ls[1])==1:
print('NO\n')
else:
print('YES\n')
print(str(ls[0])+' '+str(ls[1])+' '+str(n//(ls[0]*ls[1]))+'\n')
elif n==ls[0] or n==ls[0]**2 or n==ls[0]**3 or n==ls[0]**4 or n==ls[0]**5:
print('NO\n')
else:
copy = n
cnt = 0
while copy%ls[0]==0:
copy//=ls[0]
cnt+=1
if cnt>=3:
print('YES\n')
print(str(ls[0])+' '+str(ls[0]**2)+' '+str(n//(ls[0]**3))+'\n')
else:
print('NO\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 | N = int(input())
from math import *
for _ in range(N):
n = int(input())
sn = int(sqrt(n))+1
cnt = 0
ans = []
for i in range(2, sn):
if n % i == 0:
ans.append(i)
n //= i
cnt += 1
if cnt == 2:
cnt = i
break
if n not in ans and len(ans) == 2:
print("YES")
print(' '.join(map(str, ans)), n)
else:
print("NO")
"""
TEST 1:
5
64
32
97
2
12345
===
YES
2 4 8
NO
NO
NO
YES
3 5 823
///
TEST 2:
2
98
196
===
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())
d=dict()
i=2
while(i*i<=n):
if i not in d and n%i==0:
d[i]=i
n//=i
break
i+=1
j=2
while(j*j<=n):
if j not in d and n%j==0:
d[j]=j
n//=j
break
j+=1
if len(d)<2 or n in d or 1 in d:
print('NO')
else:
print('YES')
d[n]=1
print(" ".join(str(a) for a in d))
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible 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 #2 2 2 3| 2 2 3
def fac(x):
i=2
a = []
while i*i<=x:
if x%i==0:
a.append(i)
x//=i
else: i+=1
if x>1:
a.append(x)
return a
def res(x):
f = fac(x)
if len(f)<=2 or len(set(f))==1 and len(f)<6 or len(set(f))==2 and len(f)<=3:
print("NO")
return 0
a = f[0]
l = 1
r = len(f)-1
s = 1
for i in f[l:r+1]:
s*=i
b = s
h = 31
c = 1
while h>0:
if c!=1 and c!=a and c!=b: break
else:
c=int(c*(f[r]))
b= int(b/(f[r]))
r-=1
if l == r: break
h-=1
print("YES")
b, c = min(b,c),max(b,c)
print(a,b,c)
n = int(input())
while n>0:
x = int(input())
res(x)
n-=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 | def prime_divisors(n):
res = []
d = 2
while d * d <= n:
if n % d == 0:
if len(res) > 0 and res[-1][0] == d:
res[-1][1] = res[-1][1] + 1
else:
res.append([d, 1])
n = n // d
else:
d = d + 1
if n > 1:
if len(res) > 0 and res[-1][0] == n:
res[-1][1] = res[-1][1] + 1
else:
res.append([n, 1])
return res
t = int(input())
for i in range(t):
a = int(input())
pds = prime_divisors(a)
first_divisor = pds[0][0]
if len(pds) >= 3:
second_divisor = pds[1][0]
ans = [first_divisor, second_divisor, a // (first_divisor * second_divisor)]
print("YES")
print(*ans)
elif len(pds) == 1:
if pds[0][1] <= 5:
print("NO")
else:
print("YES")
print(first_divisor, first_divisor * first_divisor, a // (first_divisor * first_divisor * first_divisor))
else:
if pds[0][1] + pds[1][1] >= 4:
second_divisor = pds[1][0]
print("YES")
print(first_divisor, second_divisor, a // (first_divisor * second_divisor))
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())
a = 0
i, arr= 2, []
for i in range(2, int(n**0.5)+1):
if len(arr) == 2:
break
if n%i == 0:
n //= i
arr.append(i)
if len(arr) == 2 and n not in arr and n != 1:
print("YES")
print(arr[0], arr[1], n)
else:
print("NO")
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import math
t = int(input())
for _ in range(t):
n = int(input())
d = []
temp = n
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 = int(temp / (a * b))
else:
b = d[1]
c = int(temp / (a * b))
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 java.io.*;
import java.util.*;
public class MyClass {
public static void main(String args[]) {
FastReader sc = new FastReader(); //For Fast IO
//func f = new func(); //Call func for swap , permute, upper bound and for sort.
int t = sc.nextInt();
while(t-->0){
long n = sc.nextLong();
TreeSet<Long> ts = new TreeSet<>();
long num=n;
long curr=1;
for(int i=2;i*i<=num;i++){
if(n%i==0){
while(n%i==0){
curr *= i;
n /= i;
if(!ts.contains(curr)){
ts.add(curr);
curr=1;
}
if(ts.size()==2){
break;
}
}
if(ts.size()==2){
break;
}
}
}
if(n>1){
ts.add(n);
}
if(ts.size()==3){
System.out.println("YES");
for(Long n1 : ts){System.out.print(n1+" ");}
System.out.println();
}
else{
System.out.println("NO");
}
}
}
}
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;
}
}
class func{
static ArrayList<String> al = new ArrayList<>();
public static void sort(long[] arr) {
int n = arr.length, mid, h, s, l, i, j, k;
long[] res = new long[n];
for (s = 1; s < n; s <<= 1) {
for (l = 0; l < n - 1; l += (s << 1)) {
h = Math.min(l + (s << 1) - 1, n - 1);
mid = Math.min(l + s - 1, n - 1);
i = l;
j = mid + 1;
k = l;
while (i <= mid && j <= h) res[k++] = (arr[i] <= arr[j] ? arr[i++] : arr[j++]);
while (i <= mid) res[k++] = arr[i++];
while (j <= h) res[k++] = arr[j++];
for (k = l; k <= h; k++) arr[k] = res[k];
}
}
}
public static void permute(char a[] , int i , int n){
if(i==n-1){
String s = new String(a);
al.add(s); // al stores all permutations of string.
return;
}
for(int j=i;j<n;j++){
swap(a,i,j);
permute(a,i+1,n);
swap(a,i,j);
}
}
public static void swap(char a[],int i, int j){
char temp = a[i];
a[i] = a[j];
a[j] = temp;
}
public static void swap(int a[],int i, int j){
int temp = a[i];
a[i] = a[j];
a[j] = temp;
}
//-It returns index of first element which is strictly greater than searched value.
public static int upperBound(long[] array, int length, long value) {
int low = 0;
int high = length;
while (low < high) {
final int mid = (low + high) / 2;
if (value >= array[mid]) {
low = mid + 1;
} else {
high = mid;
}
}
return low;
}
/*If searched element doesn't exist function returns index of first element which is bigger than searched value.<br>
* -If searched element is bigger than any array element function returns first index after last element.<br>
* -If searched element is lower than any array element function returns index of first element.<br>
* -If there are many values equals searched value function returns first occurrence.<br>*/
public static int lowerBound(long[] array, int length, long value) {
int low = 0;
int high = length;
while (low < high) {
final int mid = (low + high) / 2;
if (value <= array[mid]) {
high = mid;
} else {
low = mid + 1;
}
}
return low;
}
}
| 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.math.*;
import java.io.*;
import java.util.*;
import java.awt.*;
public class CP {
public static void main(String[] args) throws Exception {
new Solver().solve();
}
}
class Solver {
final Helper hp;
final int MAXN = 1000_006;
final long MOD = (long) 1e9 + 7;
Solver() {
hp = new Helper(MOD, MAXN);
hp.initIO(System.in, System.out);
}
void solve() throws Exception {
int i, j, k;
for (int tc = hp.nextInt(); tc > 0; --tc) {
long N = hp.nextLong();
hp.println(factorise(N));
}
hp.flush();
}
String factorise(final long N) {
Random rnd = new Random();
for (int t = 47; t > 0; --t) {
long[] prod = new long[] {1, 1, 1};
long i, n = N;
for (i = 2; i * i <= n; ++i) if (n % i == 0) {
while (n % i == 0) {
prod[rnd.nextInt(3)] *= i;
n /= i;
}
}
if (n > 1) prod[rnd.nextInt(3)] *= n;
Arrays.sort(prod);
if (hp.min(prod) > 1 && prod[0] < prod[1] && prod[1] < prod[2]) {
return "YES\n" + hp.joinElements(prod);
}
}
return "NO";
}
}
class Helper {
final long MOD;
final int MAXN;
final Random rnd;
public Helper(long mod, int maxn) {
MOD = mod;
MAXN = maxn;
rnd = new Random();
}
public static int[] sieve;
public static ArrayList<Integer> primes;
public void setSieve() {
primes = new ArrayList<>();
sieve = new int[MAXN];
int i, j;
for (i = 2; i < MAXN; ++i)
if (sieve[i] == 0) {
primes.add(i);
for (j = i; j < MAXN; j += i) {
sieve[j] = i;
}
}
}
public static long[] factorial;
public void setFactorial() {
factorial = new long[MAXN];
factorial[0] = 1;
for (int i = 1; i < MAXN; ++i) factorial[i] = factorial[i - 1] * i % MOD;
}
public long getFactorial(int n) {
if (factorial == null) setFactorial();
return factorial[n];
}
public long ncr(int n, int r) {
if (r > n) return 0;
if (factorial == null) setFactorial();
long numerator = factorial[n];
long denominator = factorial[r] * factorial[n - r] % MOD;
return numerator * pow(denominator, MOD - 2, MOD) % MOD;
}
public long[] getLongArray(int size) throws Exception {
long[] ar = new long[size];
for (int i = 0; i < size; ++i) ar[i] = nextLong();
return ar;
}
public int[] getIntArray(int size) throws Exception {
int[] ar = new int[size];
for (int i = 0; i < size; ++i) ar[i] = nextInt();
return ar;
}
public String[] getStringArray(int size) throws Exception {
String[] ar = new String[size];
for (int i = 0; i < size; ++i) ar[i] = next();
return ar;
}
public String joinElements(long[] ar) {
StringBuilder sb = new StringBuilder();
for (long itr : ar) sb.append(itr).append(" ");
return sb.toString().trim();
}
public String joinElements(int[] ar) {
StringBuilder sb = new StringBuilder();
for (int itr : ar) sb.append(itr).append(" ");
return sb.toString().trim();
}
public String joinElements(String[] ar) {
StringBuilder sb = new StringBuilder();
for (String itr : ar) sb.append(itr).append(" ");
return sb.toString().trim();
}
public String joinElements(Object[] ar) {
StringBuilder sb = new StringBuilder();
for (Object itr : ar) sb.append(itr).append(" ");
return sb.toString().trim();
}
public long gcd(long a, long b) {
return b == 0 ? a : gcd(b, a % b);
}
public int gcd(int a, int b) {
return b == 0 ? a : gcd(b, a % b);
}
public long max(long[] ar) {
long ret = ar[0];
for (long itr : ar) ret = Math.max(ret, itr);
return ret;
}
public int max(int[] ar) {
int ret = ar[0];
for (int itr : ar) ret = Math.max(ret, itr);
return ret;
}
public long min(long[] ar) {
long ret = ar[0];
for (long itr : ar) ret = Math.min(ret, itr);
return ret;
}
public int min(int[] ar) {
int ret = ar[0];
for (int itr : ar) ret = Math.min(ret, itr);
return ret;
}
public long sum(long[] ar) {
long sum = 0;
for (long itr : ar) sum += itr;
return sum;
}
public long sum(int[] ar) {
long sum = 0;
for (int itr : ar) sum += itr;
return sum;
}
public void shuffle(int[] ar) {
int r;
for (int i = 0; i < ar.length; ++i) {
r = rnd.nextInt(ar.length);
if (r != i) {
ar[i] ^= ar[r];
ar[r] ^= ar[i];
ar[i] ^= ar[r];
}
}
}
public void shuffle(long[] ar) {
int r;
for (int i = 0; i < ar.length; ++i) {
r = rnd.nextInt(ar.length);
if (r != i) {
ar[i] ^= ar[r];
ar[r] ^= ar[i];
ar[i] ^= ar[r];
}
}
}
public long pow(long base, long exp, long MOD) {
base %= MOD;
long ret = 1;
while (exp > 0) {
if ((exp & 1) == 1) ret = ret * base % MOD;
base = base * base % MOD;
exp >>= 1;
}
return ret;
}
static byte[] buf = new byte[2048];
static int index, total;
static InputStream in;
static BufferedWriter bw;
public void initIO(InputStream is, OutputStream os) {
try {
in = is;
bw = new BufferedWriter(new OutputStreamWriter(os));
} catch (Exception e) {
}
}
public void initIO(String inputFile, String outputFile) {
try {
in = new FileInputStream(inputFile);
bw = new BufferedWriter(new OutputStreamWriter(
new FileOutputStream(outputFile)));
} catch (Exception e) {
}
}
private int scan() throws Exception {
if (index >= total) {
index = 0;
total = in.read(buf);
if (total <= 0)
return -1;
}
return buf[index++];
}
public String next() throws Exception {
int c;
for (c = scan(); c <= 32; c = scan()) ;
StringBuilder sb = new StringBuilder();
for (; c > 32; c = scan())
sb.append((char) c);
return sb.toString();
}
public int nextInt() throws Exception {
int c, val = 0;
for (c = scan(); c <= 32; c = scan()) ;
boolean neg = c == '-';
if (c == '-' || c == '+')
c = scan();
for (; c >= '0' && c <= '9'; c = scan())
val = (val << 3) + (val << 1) + (c & 15);
return neg ? -val : val;
}
public long nextLong() throws Exception {
int c;
long val = 0;
for (c = scan(); c <= 32; c = scan()) ;
boolean neg = c == '-';
if (c == '-' || c == '+')
c = scan();
for (; c >= '0' && c <= '9'; c = scan())
val = (val << 3) + (val << 1) + (c & 15);
return neg ? -val : val;
}
public void print(Object a) throws Exception {
bw.write(a.toString());
}
public void printsp(Object a) throws Exception {
print(a);
print(" ");
}
public void println() throws Exception {
bw.write("\n");
}
public void println(Object a) throws Exception {
print(a);
println();
}
public void flush() throws Exception {
bw.flush();
}
} | JAVA |
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