description stringlengths 35 9.39k | solution stringlengths 7 465k |
|---|---|
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | def prime_factors(n):
i = 2
factors = []
while i * i <= n:
if n % i:
i += 1
else:
n //= i
factors.append(i)
if n > 1:
factors.append(n)
return factors
m = int(input())
for i in range(m):
n = int(input())
if len(prime_factors(n)) <... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
for (int i = 0; i < t; i++) {
int n = sc.nextInt();
int cnt = 0;
int[] res = new int[3];
for (int j = 2; j * j <= n; j++) {
if(cnt == 2) break;
if... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | import math
t=int(input())
while(t):
t-=1
n=int(input())
ans=-1
if n<24:
print("NO")
continue
else:
root=int(math.sqrt(n))
factors=set()
for i in range(2,root+1):
if n%i==0:
factors.add(i)
factors.add(n//i)
# ... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | def min_dev(n,temp):
while temp<n**0.5+1:
if n%temp==0:
return temp
temp+=1
return 0
for _ in range(int(input())):
n=int(input())
a=min_dev(n,2)
if a==0:
print('NO')
continue
b=min_dev(int(n/a),a+1)
if b==0:
print('NO')
continue
... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;
public class _13ProductOfThreeNumbers {
public static void main(String[] args) {
FastInput input = new FastInput();
PrintWriter output = new Pri... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | import math
def prime(n):
flag=0
for i in range(2,int(math.sqrt(n))+1):
if(n%i==0):
flag=1
break
if(flag!=1):
return 1
else:
return 0
t=int(input())
while(t!=0):
n=int(input())
n1=n
flag=0
if((prime(n)==0) and n>=24):
for i in range... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | n = int(input())
for _ in range(n):
ans = []
x = int(input())
f = 2
while len(ans) < 2 and f ** 2 < x:
if x % f == 0:
ans.append(f)
x //= f
f += 1
if len(ans) == 2 and x > ans[1]:
print('YES')
print(ans[0], ans[1], x)
else:
print('... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | #include <bits/stdc++.h>
using namespace std;
void Doaa() {
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
}
int main() {
Doaa();
int n, t, num1 = 0, count = 0;
cin >> t;
while (t--) {
count = 0;
cin >> n;
int x = n;
bool f = true;
for (int i = 2; i <= sqrt(n); i++) {
if (n % ... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | from __future__ import division, print_function
def main():
for _ in range(int(input())):
n = int(input())
x = int(n**0.5)
i = 2
factors = []
while i <= x and n > 1:
if n%i==0:
factors.append(i)
n=n//i
else :
... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | def factorize(n):
l=[]
for i in range(2,int(n**0.5)+1):
if len(l)==2 and n>=i:
l.append(n)
break
if n<=1:
break
if n%i==0:
n=n//i
l.append(i)
return l
for _ in range(int(input())):
n=int(input())
l=factorize(n)
i... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | import java.io.*;
import java.math.BigInteger;
import java.util.*;
public class Main {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
PrintWriter pw = new PrintWriter(System.out);
static int MOD = 1000000007;
public static void main(String[] args) throws IOException {
... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... |
# Python program to print prime factors
import math
# A function to print all prime factors of
# a given number n
def primeFactors(n):
a=[]
# Print the number of two's that divide n
while n % 2 == 0:
a.append(2),
n = n // 2
# n must be odd at this point
... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | import math
t=int(input())
for _ in range(t):
n=int(input())
if(n<24):
print("NO")
else:
l=[]
temp=n
for i in range(2,math.ceil(math.sqrt(n))+1):
if(temp%i==0):
l.append(i)
temp=temp//i
if(len(l)==2):
bre... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | def req(n):
k = int(n ** (1 / 2))
for a in range(2, k):
if n % a == 0:
for b in range(2, k):
if b != a:
if n % (a * b) == 0:
c = n // (a * b)
if b != c and a != c:
return [a, b, c]... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | import math
def factor(n,a):
l=[]
r=n
k=int(math.sqrt(n))+1
d=1
while n%(2**d)==0:
if 2**d!=a and r//(2**d)!=a and r//(2**d)!=2**d and 2**d!=r:
l.append(2**d)
l.append(r//2**d)
n//=2**d
d+=1
break
... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | import math
from collections import defaultdict
def divisors(n, ans):
i = 2
while i <= math.sqrt(n):
if (n % i == 0):
if (n / i == i):
ans.append(i)
else:
ans.append(i)
ans.append(n // i)
i = i + 1
for _ in range(int(input... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | #include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;
using pll = pair<ll, ll>;
using pii = pair<int, int>;
const int MOD = 1e9 + 7;
const int MAX = 1e5 + 5;
const ll INF = 1e18 + 1;
void err(...) {}
template <typename T, typename... Args>
void err(istream_iterator<string> it, T a,... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | import math
def splitPrime(n):
arr = {}
for i in range(2, math.ceil((math.sqrt(n))) + 1):
while n != 0 and n % i == 0:
f = arr.get(i, 0)
arr[i] = int(f + 1)
n = n // i
if n > 1:
arr[n] = 1
return arr
t = int(input())
for i in range(t):
n = int(inp... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | import math
t=int(input())
while(t):
n=int(input())
a=[]
for i in range(2,int(math.sqrt(n))+1):
if n%i==0:
a.append(i)
n=n//i
if len(a)==2:
break
if len(a)==2 and n!=1:
if n not in a:
print("YES")
print(a[0],a[1],n)
... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | # -*- coding: utf-8 -*-
import sys
def input(): return sys.stdin.readline().strip()
def list2d(a, b, c): return [[c] * b for i in range(a)]
def list3d(a, b, c, d): return [[[d] * c for j in range(b)] for i in range(a)]
def list4d(a, b, c, d, e): return [[[[e] * d for j in range(c)] for j in range(b)] for i in range(a... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | #include <bits/stdc++.h>
using namespace std;
long long q, n;
long long ans1, ans2, ans3;
long long cnt, a[1000005];
signed main() {
cin >> q;
while (q--) {
cnt = 0;
cin >> n;
for (long long i = 2; i * i <= n; i++)
if (n % i == 0) {
a[++cnt] = i;
a[++cnt] = n / i;
}
if (c... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | # cook your dish here
import math
def primeFactors(n):
arr=[]
while n % 2 == 0:
arr.append(2)
n = n / 2
for i in range(3,int(math.sqrt(n))+1,2):
while n % i== 0:
arr.append(i)
n = n / i
if n > 2:
arr.append(n)
return arr
t=in... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | import java.io.*;
import java.math.*;
import java.util.*;
public class A {
public static void main(String[] agrs) throws IOException {
// PrintWriter pw = new PrintWriter(System.out);
FastScanner sc = new FastScanner();
int yo = sc.nextInt();
while(yo-->0) {
int n = sc.nextInt();
if(n < 8) ... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | def factor(x):
p = []
d = 2
while d * d <= x:
if x % d == 0:
p.append(d)
x //= d
else:
d += 1
if x > 1:
p.append(x)
return p
t = int(input())
for i in range(t):
n = int(input())
p = factor(n)
a, b, c = 1, 1, 1
ans = 'NO'
... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | import sys, itertools, math
T = int(sys.stdin.readline())
def solve(n):
for a in range(2, int(math.sqrt(n))+1):
if n % a != 0:
continue
for b in range(a+1, int(math.sqrt(n))+1):
r = a * b
c = n // r
if n % r == 0 and c != a and c != b:
print("YES")
print(a, b, c)
... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | #-----------------
# cook your dish here
#############-----------------
######-----
import math
###########
#####
try:
#######
def func1(t2,io):#############
######
##########
if len(io) >= 3 and len(t2) >= 3:######
##########
y = t2[1]######
#######
... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | #C
for _ in range(int(input())):
n = int(input())
cnt=0
ans = []
for i in range(2,int(n**0.5)+1):
if n % i == 0:
cnt+=1
ans.append(i)
if cnt == 2 and n % (ans[0]*ans[1]) == 0 and n//(ans[0]*ans[1]) != ans[0] and n//(ans[0]*ans[1]) != ans[1]:
break
... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | t=int(input())
for q in range(t):
n=int(input())
c=[]
i=2
while len(c)<2 and i*i<n:
if n%i==0:
c.append(i)
n=n/i
i=i+1
if len(c)==2:
print("YES")
print(*c,int(n))
else:
print("NO") |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | for t in range(int(input())):
n = int(input())
try:
a = next(i for i in range(2, int(n**0.5)+1) if n%i == 0)
n//=a
b = next(i for i in range(a+1, int(n**0.5)+1) if n % i == 0)
if n//b <= b:
raise
else:
print("YES")
print(a, b, n//b)
... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... |
#create a string in which no two same char comes together.
# assumption only small albhabetical letters used
#input format: just enter the string without any space
import sys
import math
input = sys.stdin.readline
############ ---- Input Functions ---- #######Start#####
def inp():
return(int(input()))
def inlt():
... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | #include <bits/stdc++.h>
using namespace std;
int ans1, ans2, n, T;
inline int read() {
int x = 0;
int flag = 0;
char ch;
ch = getchar();
while (!isdigit(ch)) {
if (ch == '-') flag = 1;
ch = getchar();
}
while (isdigit(ch)) {
x = (x << 3) + (x << 1) + ch - 48;
ch = getchar();
}
if (fla... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | import java.util.*;
// scanner.nextInt();
public class Solution {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
int test_case = scanner.nextInt();
for(int z = 0; z < test_case; ++z)
{
int n = scanner.nextInt();
... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | #include <bits/stdc++.h>
using namespace std;
int main() {
int t;
cin >> t;
while (t--) {
set<int> s;
int count = 0;
int a;
cin >> a;
for (int i = 2; i * i <= a; i++) {
if (a % i == 0) {
int u = a / i;
s.insert(i);
for (int j = i + 1; j * j <= u; j++) {
... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | import sys
input = sys.stdin.readline
def solve():
n = int(input())
ans, ind = [], 2
while len(ans) < 2 and ind < n ** (1 / 2):
if not n % ind:
n = n // ind
ans.append(ind)
ind += 1
return ans + [n]
for _ in range(int(input())):
sol = solve()
if len(so... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | #!/usr/bin/env python3
R = lambda: list(map(int, input().split()))
def solve():
n = R()[0]
a, cnt, i = [0, 0], 0, 2
while i * i < n and cnt < 2:
if n % i == 0:
a[cnt] = i
cnt += 1
n //= i
i += 1
if cnt < 2:
print("NO")
else:
prin... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | #include <bits/stdc++.h>
using namespace std;
const int INF = 1e8 + 1;
const long double PI = 3.141592653;
bool isp(int n) {
for (int i = 2; i * i <= n; i++) {
if (n % i == 0) return 0;
}
return 1;
}
int main() {
int tc;
cin >> tc;
while (tc--) {
int n;
cin >> n;
if (isp(n)) {
cout << ... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | def process(n,i,sol):
for j in range(i+1,int(n**(0.5))+1):
if n%j==0:
sol.append(j)
if len(sol)==1:
process(n//j,j,sol)
break
if len(sol)==2:
if j!=n//j:
sol.append(n//j)
break
return ... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | #include <bits/stdc++.h>
using namespace std;
int32_t main() {
ios::sync_with_stdio(false);
cin.tie(NULL);
;
long long t;
cin >> t;
while (t--) {
long long n;
cin >> n;
if (n < 24) {
cout << "NO" << endl;
continue;
}
long long p = 0;
for (long long i = 2; i <= sqrt(n); i+... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | def main():
T = int(input())
for t in range(T):
m = int(input())
n = m
dic = {}
i = 2
while i * i <= n:
while n % i == 0:
dic[i] = dic.get(i, 0) + 1
n //= i
i += 1
if m % n == 0 and n != 1:
dic[n]... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | def nonprime(number,start=2):
from math import sqrt
for i in range(start,int(sqrt(number))+1):
if number%i==0:
return [True,i]
return [False]
testcases=int(input())
for k in range(testcases):
numbertobetested=int(input())
initial=nonprime(numbertobetested)
if initial[0]:
... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | import java.util.Scanner;
public class productod3{
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
int t = scan.nextInt();
for(int k = 0; k < t; k++){
int n = scan.nextInt();
if(!func(n)){
System.out.println("NO");
}
}
}
public static Boolean func(int n)... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... |
n = int(input())
for k in range(n):
z = int(input())
l = []
for i in range(2 , z):
if z % i == 0:
l.append(i)
z = z // i
break
if (i * i) > z:
break
for j in range(2 , z):
if z % j == 0 and j not in l:
l.append(j)
... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | import java.util.*;
import java.io.*;
import java.math.*;
import java.security.*;
public class Main
{
public static void main(String[] args)throws Exception
{
Reader reader = new Reader();
reader.readLine();
int t = reader.readInt(0);
StringBuilder res = new StringBuilder();
while (t-- > 0)
{
r... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | def solve():
n = int(input())
primes = {}
i = 2
while i * i <= n:
while n % i == 0:
primes[i] = primes.get(i, 0) + 1
n //= i
i += 1
if n > 1:
primes[n] = primes.get(n, 1)
k = list(primes.keys())
v = list(primes.values())
if len(primes) >= 3... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | from math import sqrt as s
def factorization(n):
a=[]
while n%2==0:
a.append(2)
n = n//2
for i in range(3,int(s(n))+1,2):
while n%i==0:
a.append(i)
n = n//i
if n>2:
a.append(n)
return a
for i in range(int(input())):
... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | t = int(input())
for i in range(t):
l = []
n = int(input())
for i in range(2,n):
if n%i == 0:
l.append(i)
n /= i
if len(l) == 2 or i*i>n:
break
if len(l) == 2 and n!=l[0] and n!= l[1] and n>1:
print("YES")
print(l[0],l[1],int(n))
else:
print("NO") |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | from sys import stdin, stdout
from collections import defaultdict
import math
rl = lambda: stdin.readline()
rll = lambda: stdin.readline().split()
def main():
cases = rll()
for line in stdin:
n = int(line)
ans = []
for f1 in range(2, math.ceil(math.sqrt(n))):
if n % f1 == 0:
ans.append(f1)
break
i... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | #include <bits/stdc++.h>
using namespace std;
long long myCeil(long long num, long long divider) {
long long result = num / divider;
if ((result * divider) == num) return result;
return result + 1;
}
long long gcd(long long a, long long b) {
if (!b) return a;
return gcd(b, a % b);
}
bool isPrime(long long n) ... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | t = int(input())
for i in range(t):
n = int(input())
p = []
j = 2
while j * j <= n:
while (n % j == 0):
p.append(j)
n //= j
j += 1
if n > 1:
p.append(n)
if len(p) < 3:
print("NO")
elif len(p) == 3:
if p[0] != p[1] and p[1] != p[2] and p[0] != p[2]:
print("YES")
print(p[0], p[1], p[2])
els... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | for _ in range(int(input())):
n = int(input())
c = []
i=2
while len(c)<2 and i**2<n:
if n % i == 0:
c.append(i)
n=n/i
i=i+1
if len(c)==2 and n not in c:
print("YES")
print(*c,int(n))
else:
print("NO")
|
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | #include <bits/stdc++.h>
using namespace std;
long long myCeil(long long num, long long divider) {
long long result = num / divider;
if ((result * divider) == num) return result;
return result + 1;
}
long long gcd(long long a, long long b) {
if (!b) return a;
return gcd(b, a % b);
}
bool isPrime(long long n) ... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | #include <bits/stdc++.h>
using namespace std;
int tab[50000];
int main() {
int t;
int long long n, a, b, c, maxi;
scanf("%lld", &t);
while (t--) {
int s = 0;
int long long i = 2, j = 2, sq;
scanf("%lld", &n);
sq = sqrt(n);
while (n != 0 && i <= sq) {
if (n % i == 0) {
tab[s] = ... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | from math import sqrt
def fatores(n):
fat = {}
for p in range(2, int(sqrt(n)+1)):
while n % p == 0:
n/=p
if not fat.get(p):
fat[p] = 0
fat[p]+=1
p+=1
if n > 1:
fat[int(n)] = 1
return fat
for _ in range(int(input())):
... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | t = int(input())
for _ in range(t):
n = int(input())
d = []
a = n
i = 2
while i * i <= a:
while a % i == 0:
d.append(i)
a //= i
i += 1
if a != 1:
d.append(a)
a = d[0]
b = 1
c = 1
for i in range(1, len(d)):
if a == b or b == ... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | import sys
import math
from collections import defaultdict
from collections import deque
from itertools import combinations
from itertools import permutations
input = lambda : sys.stdin.readline().rstrip()
read = lambda : list(map(int, input().split()))
go = lambda : 1/0
def write(*args, sep="\n"):
for i in args:
... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | t=int(input())
for tt in range(t):
n=int(input())
d={}
for i in range(2,int(n**0.5)+1):
if n%i==0 and i not in d:
d[i]=1
n=n//i
break
for i in range(2,int(n**0.5)+1):
if n%i==0 and i not in d:
d[i]=1
n=n//i
break
if len(d)<2 or (n in d) or n==1:
print("NO")
else:
print("YES")
d[n]=1... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | import java.io.*;
import java.math.BigInteger;
import java.util.*;
public class Main {
static InputReader in = new InputReader(System.in);
static PrintWriter out = new PrintWriter(System.out);
static int oo = (int)1e9;
// static long oo = (long)1e15;
static int mod = 1_000_000_007;
// static int mod = 998... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | from math import ceil
def find(n,p,ans):
for i in range(2,ceil(n**(1/p))+1,1):
if n%i==0:
if p==3:
ans+=find(n//i,2,[i])
if len(ans)==3:
return ans
ans=[]
elif p==2 and n//i>1:
new=ans+[(n//i),i]
... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | # -*- coding: utf-8 -*-
from math import sqrt
def main():
for _ in range(int(input())):
n = int(input())
d = n
ret = []
for i in range(2, int(sqrt(n))):
if d % i == 0:
d /= i
ret.append(i)
if len(ret) == 2:
bre... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | t = int(input())
def find_one_factor(num, a):
for i in range(2, int(num ** 0.5) + 1):
if num % i == 0:
if a == i:
continue
return i, num // i
return -1, -1
while t:
t -= 1
n = int(input())
flag = True
factors = []
a, num = find_one_factor... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | from math import sqrt
T = int(input())
for case in range(T):
n = int(input())
ans = False
move = True
for i in range(2,int(sqrt(n)) + 1):
if move is False:break
if n%i == 0:
a = i
b = int(n/a)
for j in range(a + 1 , int(sqrt(n)) + 1):
if (b%j == 0):
b = b//j
c = j
if (c==b or c==a or... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | //Contribted By - Rahul Roy
import java.util.*;
import java.lang.*;
import java.io.*;
public class Main {
public static PrintWriter out = new PrintWriter (new BufferedOutputStream(System.out));
public static void main(String sp[])throws IOException{
FastReader sc = new FastReader();
int t = sc.nextInt();
whil... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | for _ in range(int(input())):
n = int(input())
ans = []
i = 2
x = n
while i * i <= x:
if n % i == 0:
if len(ans) < 2:
ans.append(i)
n //= i
i += 1
if n not in ans and len(ans) < 3:
ans.append(n)
if len(ans) > 2:
prin... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | from math import *
for ct in range(int(input())):
n = int(input())
flag = False
for i in range(2, int(sqrt(n))):
if flag:
break
if n % i == 0:
par = n//i
for j in range(2, int(sqrt(par))+1):
if par % j == 0:
k = par //... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | import sys
from os import path
if(path.exists('input.txt')):
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
get_int = lambda: int(input())
get_ints = lambda: map(int, input().split())
get_ints_list = lambda: list(get_ints())
from math import sqrt
def factorize(n):
i = 2
facto... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | for _ in [0]*int(input()):
n = int(input())
i = 2
tt = False
delim = []
while i*i <= n:
if not n % i:
delim.append(i)
i += 1
for i in range(len(delim)):
if tt == True:
break
for j in range(i+1, len(delim)):
f = n // (delim[i]*de... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | #include <bits/stdc++.h>
using namespace std;
int main() {
int t;
cin >> t;
for (int i = 0; i < t; i++) {
long long int a, b, c, n;
cin >> n;
int e = 0;
long long int d;
for (int j = 2; j <= (int)sqrt(n); j++) {
if (n % j == 0) {
e = 1;
d = n / j;
a = j;
b... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | #include <bits/stdc++.h>
using namespace std;
template <typename T>
void inpos(T &x) {
x = 0;
register T c = getchar();
while (((c < 48) || (c > 57)) && (c != '-')) c = getchar();
bool neg = 0;
if (c == '-') neg = 1;
for (; c < 48 || c > 57; c = getchar())
;
for (; c > 47 && c < 58; c = getchar()) x =... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | t = int(input())
for _ in range(t):
n = num = int(input())
arr = []
div = 2
while n != 1:
if div > int(n**0.5):
arr.append(n)
n //= div
break
if n % div == 0:
n //= div
arr.append(div)
continue
div += 1
... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | from math import sqrt
t = int(input())
for _ in range(t):
n = int(input())
i = 2
fact = []
while n%i == 0:
n = n//i
fact.append(i)
for i in range(3, int(sqrt(n))+1, 2):
while n%i == 0:
n = n//i
fact.append(i)
if n != 1:
fact.append(n)
if len(fact) < 3:
print('NO')
continue
ans = 1
i = 0
fac... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | for _ in range(int(input())):
n = int(input())
prime = []
for i in range(2, int(n ** 0.5) + 1):
if n % i == 0:
prime.append(i)
n //= i
if n != 1:
prime.append(n)
if len(prime) < 3:
print("NO")
else:
if len(set(prime)) < 3:
print... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | from math import sqrt
t= int(input())
while t:
n= int(input())
r=n
a=[]
for i in range(2,int(sqrt(n))):
if n%i==0:
a.append(i)
n=int(n/i)
if len(a)==2:
break
if len(a)==2 and a[0]!=a[1] and a[1]!=n and a[0]!=n and n>=2:
print("YES",a[0],a[1],n)
e... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | #include <bits/stdc++.h>
using namespace std;
vector<int> primefactors(int n) {
vector<int> ans;
while (n % 2 == 0) {
ans.push_back(2);
n = n / 2;
}
for (int i = 3; i * i <= n; i += 2) {
while (n % i == 0) {
ans.push_back(i);
n = n / i;
}
}
if (n > 2) {
ans.push_back(n);
}
... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.StringTokenizer;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.InputStream;
/**
* Built using CHelper plug-in
* Actual soluti... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | t = int(input())
for _ in range(t):
n = int(input())
#n,k = map(int,input().split())
for a in range(2,int(n**0.5)+1):
if n%a == 0:break
else:
print ("NO")
continue
ne = n//a
for b in range(2,int(ne**0.5)+1):
if b == a:
continue
if ne%b == 0:b... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | for i in range(int(input())):
n=int(input())
a=[]
d=2
while (d*d<n) and len(a)<2:
if n%d==0:
a.append(d)
n//=d
d+=1
if len(a)<2:
print("NO")
else:
print("YES")
print(n,*a)
|
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | import sys
def fastio():
from io import StringIO
from atexit import register
global input
sys.stdin = StringIO(sys.stdin.read())
input = lambda : sys.stdin.readline().rstrip('\r\n')
sys.stdout = StringIO()
register(lambda : sys.__stdout__.write(sys.stdout.getvalue()))
fastio()
for _ in rang... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | from math import *
for _ in range(int(input())):
n = int(input())
e = []
a = False
for i in range(2, int(sqrt(n)) + 1):
m = n / i
if m.is_integer() == True:
for j in range(2, int(sqrt(m)) + 1):
if m % j == 0:
if (n / (i * j)).is_integer() =... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | for _ in range(int(input())):
n=int(input())
k=[]
for i in range(2,10000):
if n%i==0:
k.append(i)
n=n//i
break
for j in range(2,10000):
if n%j==0:
if j not in k:
k.append(j)
n=n//j
break
... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... |
import java.util.Scanner;
import java.util.Stack;
public class Main {
public static void main(String[] args) {
Scanner s = new Scanner(System.in);
int t = s.nextInt();
while (t-- > 0) {
int n = s.nextInt();
int i = 2;
int a = -1;
int b = -... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | import math
a = int(input())
for x in range(a):
a = int(input())
prime = 2
count = list()
while prime <= math.ceil(math.sqrt(a)):
if a % prime == 0:
a /= prime
count.append(prime)
if len(count) == 2:
if not a in count:
prin... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | import sys
import math
input = sys.stdin.readline
ins = lambda: input().rstrip()
ini = lambda: int(input().rstrip())
inm = lambda: map(int, input().split())
inl = lambda: list(map(int, input().split()))
t = ini()
ans = []
andsindex = [0] * t
for _ in range(t):
n = ini()
i = 2
tmp = []
while n >= i * i:... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | import java.util.*;
import java.io.*;
public class Main
{
public static void main(String []args) throws IOException {
FastScanner in = new FastScanner(System.in);
PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)), false);
solve(in, out);
in.close();
o... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | #include <bits/stdc++.h>
using namespace std;
int n, m;
int main() {
int a, b, c, d, t;
scanf("%d", &t);
while (t--) {
int flag = 0;
scanf("%d", &n);
a = 0;
b = 0;
for (int i = 2; i < sqrt(n); i++) {
if (n % i == 0) {
if (a == 0) {
a = i;
n /= i;
} els... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | #include <bits/stdc++.h>
using namespace std;
void solve() {
int n;
cin >> n;
int a = 2;
while (a * a <= n) {
if (n % a == 0) {
n /= a;
break;
}
++a;
}
int b = a + 1, c = -1;
while (b * b < n) {
if (n % b == 0) {
if (n / b != a) {
c = n / b;
cout << "YES" ... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | import math
t = int(input())
for i in range(t):
n = int(input())
a=0
b=0
c=0
j=2
while(j<=math.sqrt(n)):
if(n%j==0):
if(a==0):
a=j
n=int(n/j)
j+=1
else:
b=j
c=int(n/j)
break
else:
j+=1
if(a!=0 and b!=0 and c!=0) and(a!=b and b!=c and a!=c) and (a>=2 and b>=2 and c>=2):
pri... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | 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) {
... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | '''input
5
64
32
97
2
12345
'''
from collections import defaultdict as dd
from collections import Counter as ccd
from itertools import permutations as pp
from itertools import combinations as cc
from random import randint as rd
from bisect import bisect_left as bl
from bisect import bisect_right as br
import heapq as ... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | #include <bits/stdc++.h>
using namespace std;
int main() {
long long int t;
cin >> t;
while (t--) {
long long int n;
cin >> n;
long long int k = n;
vector<long long int> b;
for (int i = 2; i < sqrt(k) + 1; i++) {
if (n % i == 0) {
b.push_back(i);
n /= i;
}
if ... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | def getFactors(n):
fact = []
i =2
while i*i <=n:
if n%i==0:
fact.append(i)
if(i!=n//i):
fact.append(n//i)
i+=1
return fact
t = int(input())
while t>0:
n = int(input())
fact = getFactors(n)
fact.sort()
found = False
for i in ran... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | import math
def primeFactors(n,d):
# Print the number of two's that divide n
while n % 2 == 0:
try:
if d[2]>=0:
d[2]+=1
except:
d[2]=1
n = n / 2
# n must be odd at this point
# so a skip of 2 ( i = i + 2) can be used ... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | //package main;
import java.io.*;
import java.util.*;
public final class Main {
BufferedReader br;
StringTokenizer stk;
public static void main(String[] args) throws Exception {
new Main().run();
}
{
stk = null;
br = new BufferedReader(new InputStreamReader(System.in)... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | from math import *
t = int(input())
for z in range(t):
n = int(input())
na = True
sq = ceil(sqrt(n)) + 1
for a in range(2, sq):
if not na:
break
if n % a == 0:
q = n // a
sq2 = ceil(sqrt(q)) + 1
for b in range(a + 1, sq2):
i... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | import java.io.*;
import java.util.*;
public class Bacteria {
public static void main(String args[]) throws IOException{
Scanner sc = new Scanner(System.in);
PrintWriter pw = new PrintWriter(System.out);
int t = sc.nextInt();
while(t--!=0){
int n = sc.nextInt();
... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | import math
def primeFactors(n):
vals = []
# Print the number of two's that divide n
two = False
while n % 2 == 0:
vals.append(int(n))
if not two:
vals.append(2)
two = True
n = n / 2
# n must be odd at this point
# so a skip of 2 ( i = i + 2) can ... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | /*
Author: Anthony Ngene
Created: 09/09/2020 - 06:52
*/
import java.io.*;
import java.util.*;
public class C {
// CodeForces.Div3.CF1385_656.P1.A mind is like a parachute. It doesn't work if it isn't open. - Frank Zappa
void solver() throws IOException {
int cases = in.intNext();
for (int ... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | import java.util.*;
import java.io.*;
public class Main{
public static void main(String[] args)throws IOException{
Scanner sc = new Scanner(System.in);
long a,b,c,d,e,f,g,h,i,j,k;
a = sc.nextLong();
for(b = 0;b < a;b++){
c = sc.nextLong();
d = (long)Math.sqrt(c);
if(c % 2 == 0){
... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | t=int(input())
for _ in range(t):
n=int(input())
tf=False
for i in range(2,int(n**(1/2))+1):
if n%i==0:
a=i
b=n//i
tf=True
break
if tf:
tf=False
for i in range(2,int(b**(1/2))+1):
if b%i==0:
if i!=a and b... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | from math import ceil, sqrt
t = int(input())
for _ in range(t):
n = int(input())
if n < 24:
print('NO')
else:
try:
for i in range(4, ceil(sqrt(n)) + 1):
if n % i == 0:
a = n // i
b = i
for j in range(2,... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | import math
t = int(input())
for i in range(t):
n = int(input())
L, a, r = [], n, []
m = math.ceil(math.sqrt(n))
f = False
for j in range(2, m+1):
if a % j == 0:
c = 0
while a % j == 0:
a //= j
c += 1
L.append((j, c))
... |
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