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 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
... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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 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
... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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())
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")
... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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.*;
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));
}
Stri... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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 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... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | # _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 productOfThreeNumber... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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 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"... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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):
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())
w... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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() {
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;
co... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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())
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=... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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
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... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | # 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:
... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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
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)
... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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 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;
... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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 = 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... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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--) {
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);
}
}
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)... | 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... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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 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[... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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 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 "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 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')
... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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 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++) ... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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 : 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()
... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | #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
... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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
# 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... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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 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 = {};
... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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 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;
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)... | # -*- 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
w... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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, 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;
... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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
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):
... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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 = 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 = []... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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>
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 !=... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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 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, ... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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;
#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... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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 class FastReader
{
final private int BUFFER_SIZE = 1 << 16;
private DataInputStream din;
private byte[] buffer;
private int bufferPointer, bytesRead;
public Fast... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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() {
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;
... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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())
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)
... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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())
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 * ... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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 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())
... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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)... | 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]... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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 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 []
... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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())
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)
... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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
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')
el... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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, 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(... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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 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 = g... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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 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 ... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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())
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") |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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())
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
... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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.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 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)... |
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:
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)... | 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)
|
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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;
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) ... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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 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):
... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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())
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("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 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 , ... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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 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) {
... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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=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 di... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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
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:... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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 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
... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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 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)
... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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() {
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<... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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())
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 ... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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 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 ... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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
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... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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 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()... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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 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... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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())
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
... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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 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:
an... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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 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... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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 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:
... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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 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;
}
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 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... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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 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 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 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 == differe... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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())
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 gen... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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
############ ---- 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... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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
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
... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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;
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) {
... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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
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: ... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... |
/*
_oo0oo_
o8888888o
88" . "88
(| -_- |)
0\ = /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 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
... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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>
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 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)... | #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 ==... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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() {
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;
... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | # ░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░
# ░░░░░░░▄▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▄░░░░░░
# ░░░░░░█░░▄▀▀▀▀▀▀▀▀▀▀▀▀▄░░█░░░░░
# ░░░░░░█░█░░▀░░░░░░░░▀░░█░█░░░░░
# ░░░░░░█░█░░░▀░░░░░░▀░░░█░█░░░░░
# ░░░░░░█░█░░░░▀░░░░▀░░░░█░█░░░░░
# ░░░░░░█░█▄░░░░▀░░▀░░░░▄█░█░░░░░
# ░░░░░░█░█░░░░░░██░░░░░░█░█░░░░░
# ░░░░░░█░▀░░░░░░░░░░░░░░▀░█░░░░░
# ░░░░░░█░░░░... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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
# 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 ... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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 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... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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 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 Non... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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
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
... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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
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 ra... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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
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
... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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 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):
f... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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())
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)), ... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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())
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
... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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 #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:
pr... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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_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... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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())
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],... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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())
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... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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 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){... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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.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() {
... |
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