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)... | #include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
long long int n;
cin >> n;
for (long long int i = 0; i < n; i++) {
long long int a, x;
cin >> a;
x = a;
vector<int> v;
for (long long int i = 2; i <= sqrt(a); i++) {
if... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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=b=c=1
for i in range(2,int((n**0.5)+1)):
if n%i==0:
a=i
break
if a==1:
print('NO')
continue
x=n//a
for i in range(2,int(((x)**0.5)+1)):
if x%i==0 and i!=a:
b=i
break
... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | """
This template is made by Satwik_Tiwari.
python programmers can use this template :)) .
"""
#===============================================================================================
#importing some useful libraries.
import sys
import bisect
import heapq
from math import *
from collections import Cou... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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())
abc = set()
for i in range(2, int(n**0.5)+1):
if n % i == 0:
if len(abc) == 0:
abc.add(i)
n = n//i
else:
if i != n//i and n//i != 1:
abc.add(i)
... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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())
import math
temp = int(math.sqrt(n))
done = 0
for i in range(2,temp+1):
if n%i == 0:
a = i
for j in range(2,int(math.sqrt(a))+1):
if a%j == 0 and a/j != a:
x = n//i
... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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 libraries for input/ output handling
# on generic level
import atexit, io, sys
# A stream implementation using an in-memory bytes
# buffer. It inherits BufferedIOBase.
buffer = io.BytesIO()
sys.stdout = buffer
# print via here
@atexit.register
def write():
sys.__stdout__.write(buffer.get... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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 n, m, ni, mi;
long long cn;
string s;
long long num[100005];
long long ok = false;
deque<long long> v;
map<string, bool> M;
map<string, bool> MC;
long long sch(long long a, long long b, long long c, long long i) {
string s = to_string(a) + " " + to_string(b) + "... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | ////////////////////---------------------------SHUBHAM CHAUDHARI-------------------------------///////////////////////
import java.util.*;
public class Main {
public static void main(String[] args) {
Scanner in=new Scanner(System.in);
int t=in.nextInt();
StringBuilder res=new StringBuild... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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 collections
import itertools
import sys
from collections import defaultdict, Counter
from math import sqrt, ceil
input = sys.stdin.readline
############ ---- Input Functions ---- ############
def inp():
return (int(input()))
def inlt():
return (list(map(int, input().split())))
def insr():
s = ... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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
n = int(input())
for i in range(n):
f = int(input())
a = []
for j in range(2, int(sqrt(f))):
if f % j == 0:
a.append(j)
f = f // j
if len(a) == 2:
break
if len(a)==2 and a.count(f)==0:
print("YES")
print(a[0]... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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
3
1 2
3 4
5 6
'''
import math
# Method to print the divisors
def printDivisors(n) :
list = []
# List to store half of the divisors
a = []
x = n
for i in range(2, int(math.sqrt(n) + 1)) :
if (x % i == 0) :
a.append(i)
# i -= 1
x/=i
if len(a)==2:
p = 1
# for i in range(2,le... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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, floor
t = int(input())
for _ in range(t):
n = int(input())
if n < 24:
print("NO")
else:
m = floor(sqrt(n)) + 1
d = 0
divs = []
for i in range(2, m):
if n % i == 0:
d += 1
divs.append(i)
... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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());a=['YES'];i=j=2
while j<4and i*i<n:
if n%i<1:a+=i,;n//=i;j+=1
i+=1
print(*(a+[n],['NO'])[j<4][:4]) |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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 j in range(t):
n = int(input())
a = 1
b = 1
c = 1
for i in range(2, int(math.sqrt(n)) + 1):
if n % i == 0:
a = i
break
m = n // a
for i in range(a + 1, int(math.sqrt(m)) + 1):
if m % i == 0:
b = i
... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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;
double squrt(int64_t num) {
double high = num, low = 0, mid;
while (high - low > 0.00001) {
mid = low + (high - low) / 2;
if (mid * mid > num) {
high = mid;
} else {
low = mid;
}
}
return mid;
}
void solve() {
int64_t num;
cin >> num;... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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>
#pragma GCC optimize("O3")
using namespace std;
int add(int a, int b) {
long long x = a + b;
if (x >= 1000000007) x -= 1000000007;
if (x < 0) x += 1000000007;
return x;
}
long long mul(long long a, long long b) { return (a * b) % 1000000007; }
long long pw(long long a, long long b) {
... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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 MAX_N = -1;
const int inf = 2e9 + 19;
vector<long long> vec;
void primeDiv(long long n) {
for (int i = 2; i * i <= n; i++) {
if (n % i == 0) {
while (n % i == 0) vec.push_back(i), n /= i;
}
}
if (n) vec.push_back(n);
}
int main() {
ios::sync_... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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())
dvdrs = []
passs = True
if n < 24:
print("NO")
continue
for k in range(2,n):
if k*k*k > n:
passs = False
break
if n%k == 0:
dvdrs += [k]
n = int(n/k)
break
if passs:
for k in range(dvdrs[0]+1, n):
if k*k > n:
passs = Fal... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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 div(n):
l = []
i = 2
while i <= math.sqrt(n):
if (n % i == 0):
if (n / i == i):
l.append(int(i))
else:
l.append(int(i))
l.append(int(n/i))
i = i + 1
l.sort()
return l
for _ in range(int(input(... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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.lang.Math;
import java.io.* ;
public class Account {
class Pair {
int key ;
int val ;
Pair(int key ,int val) {
this.key = key ;
this.val = val ;
}
int getKey() {
return this.key ;
}
... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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):
li =[]
while n % 2 == 0:
li.append(2)
#print (2)
n = n//2
for i in range(3,int(math.sqrt(n))+1,2):
while n % i== 0:
##print (i)
li.append(i)
n = n//i
if n > 2:
li.append(... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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 lli = long long int;
using ld = long double;
const int N = 2e5 + 5, inf = 2e9;
bool valid(int a, int b, int c) {
if (a == b || b == c || a == c) return false;
return true;
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.ti... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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.*;
public class geek {
public static void main(String[] args) {
Scanner s = new Scanner(System.in);
int t = s.nextInt();
while (t-- > 0) {
int n = s.nextInt();
int a = 0;
int b = 0;
int c = 0;
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
from collections import defaultdict as dc
import math
from bisect import bisect, bisect_left,bisect_right
input=sys.stdin.readline
for _ in range(int(input())):
n=int(input())
i=2
f=0
#p=math.sqrt(n)
while(i<=math.sqrt(n)):
if n%i==0:
f=1
break
i+=1... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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(n):
if n == 1:
return []
if n == 2:
return [2]
i = 2
while i*i <= n:
if n%i == 0:
return prime(n//i)+[i]
i += 1
return [n]
for _ in range(int(input())):
n = int(input())
a = sorted(prime(n))
if len(set(a)) >= 3:
print("YE... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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 prod(li):
p = 1
for i in li: p*=i
return p
def getAns(n):
d = dict()
while n % 2 == 0:
if 2 in d.keys():
d[2] += 1
else:
d[2] = 1
n//=2
for i in range(3, int(math.sqrt(n)) + 3, 2):
while n % i == 0:
if i in d.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 string
import math
import collections
def LeReseauLMD(n):
orin = n
a,b,c = 0,0,0
for a in range(2,math.floor(n**0.5)+1):
if n%a==0:
n //=a
for b in range(2,math.floor((n)**0.5)+1):
if n%b == 0 and b != a:
"""for c in range(2,math.... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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 max6 = 1e6 + 6;
map<int, int> rem;
int e[max6], p[max6];
int main() {
int tests;
scanf("%d", &tests);
for (int i = 1; i <= 1000000; ++i) e[i] = i;
int sl = 0;
for (int i = 2; i <= 1000000; ++i)
if (e[i] == i) {
p[++sl] = i;
for (int j = 1... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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())):
i = 2
l = []
n = int(input())
size = 0
while(i*i<=n and size < 2):
if(n % i == 0):
l.append(i)
n //= i
size += 1
i += 1
if(size == 2 and (l[0]!=n and l[1]!=n)):
print('YES')
print(l[0],l[1],n)
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)... | def read_ints(): return map(int, input().split())
def read_int(): return int(input())
def factors(n):
factors_p = dict()
twos = 0
while n % 2 == 0:
n //= 2
twos += 1
if twos > 0:
factors_p[2] = twos
d = 3
while d * d <= n:
p = 0
while n % d == 0:
... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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 floor, sqrt, ceil
t = int(input())
for _ in range(t):
a = -1
b = -1
n = int(input())
i = 2
while i*i < n:
if n % i == 0:
a = i
n //= a
break
i += 1
i = 2
while i*i < n:
if n % i == 0 and i != a:
b = i
... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | m=int(input())
for l in range(m):
n=int(input())
q,t=int(n**.5)+1,0
for i in range(2,q):
if n%i==0:
a=n//i
w=int(a**.5)+1
for j in range(2,w):
if a%j==0:
r=a//j
if r!=j and j!=i and r!=i:
... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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
m=1000000007
def fact(n):
ans=1
for i in range(1,n+1):
ans=((ans%m)*(i%m))%m
return ans
def power_2(n):
ans=1
for i in range(n):
ans=((ans%m)*(2))%m
return ans
for z in range(int(input())):
n=int(input())
fin=int(math.ceil(n**(1/3)))
cnt,i=0,2
#... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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)
int1 = lambda x: int(x) - 1
p2D = lambda x: print(*x, sep=" ")
def II(): return int(sys.stdin.readline())
def MI(): return map(int, sys.stdin.readline().split())
def LI(): return list(map(int, sys.stdin.readline().split()))
def LLI(rows_number): return [LI() for _ in range(row... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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
t=int(input())
for i in range(t):
x=int(input())
ar=[]
l=0
a=1
b=1
c=1
ct=0
bl=False
for j in range(2,int(x**(0.5))):
if(x%j==0 and ct==0):
a=j
ct+=1
elif(x%j==0 and ct==1):
temp=j
if... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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
import sys
INF=1e9
import bisect
def get_int(): return int(stdin.readline().strip())
def get_ints(): return map(int,stdin.readline().strip().split())
def get_array(): return list(map(int,stdin.readline().strip().split()))
def get_string(): return stdin.readline().strip()
def op(c): return... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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())):
num = int(input())
list_of_nums = [1]
for i in range(2, int(pow(num, .5)+1)):
if num % i == 0:
num = num // i
list_of_nums.append(i)
break
# print(list_of_nums)
for i in range(2, int(pow(num, .5))+1):
if num % i == 0 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)... | 'use strict'
const problem = (n) => {
for (let i = 2; i < Math.cbrt(n); i++) {
if (n % i === 0) {
n /= i;
for (let j = i + 1; j < Math.sqrt(n); j++) {
if (n % j === 0) return `YES\n${i} ${j} ${n/j}`;
}
}
}
return 'NO';
}
let t = +readline... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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 my_def(x):
a, b, c = 1, 1, 1
for j in range(2, int(x ** 0.5) + 2):
while x % j == 0:
if a == 1:
a = j
elif b == 1 or b == a:
b *= j
else:
return (a, b, x)
x //= j
return (a, b, x)
n = int(input())
... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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 calcul(x,message):
for j in range (2,int(sqrt(x))):
if x%j==0:
m=x/j
s=int(sqrt(m))
if j+1>s:
s=int(m/2)
for k in range (j+1,s+1):
if (m%k==0) and (int(m/k)!=k) and (int(m/k)!=j):
... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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 w in range(t):
n=int(input())
c=0
ans=[]
for i in range(2,int(sqrt(n))+1):
if n==1 or c==2:
break
if n%i==0:
ans.append(i)
n=n//i
c+=1
if c<2 or n==ans[1] or n==ans[0]:
print("NO")
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)... | import math
from collections import Counter,defaultdict
I =lambda:int(input())
M =lambda:map(int,input().split())
LI=lambda:list(map(int,input().split()))
for _ in range(I()):
n=I();i=2;a,b,c=-1,-1,-1
while i*i<=n:
while n%i==0:
n//=i
if a==-1:
a=i
eli... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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 t;
scanf("%d", &t);
while (t--) {
int n;
scanf("%d", &n);
int flag, a, b, c;
flag = 1;
if (n % 2 == 0) {
a = 2;
flag = 0;
} else
for (a = 3; a * a * a <= n; a += 2)
if (n % a == 0) {
flag = 0;
break;
... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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.*;
public class code {
static int sqrt(int x) {
for(int i=2; true; i++) {
if(i*i==x) {
return i;
}
if(i*i>x) {
return i-1;
}
}
}
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
for(int i=0; i<n; i++) {
int a= sc.... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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_multipliers(n):
if n < 8:
return None
a = b = c = None
for i in range(2, int(n**0.5 + 1)):
if n % i == 0:
a = i
break
if a is None:
return None
n /= a
for i in range(2, int(n**0.5 + 1)):
if (i != a) and (n % i == 0):
... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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 Sieve(n):
ret = []
divlis = [-1] * (n+1)
flag = [True] * (n+1)
flag[0] = False
flag[1] = False
ind = 2
while ind <= n:
if flag[ind]:
ret.append(ind)
ind2 = ind ** 2
while ind2 <= n:
flag[ind2] = False
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)... | #include <bits/stdc++.h>
using namespace std;
int main() {
long long int T;
cin >> T;
while (T--) {
long long int n, k = 0, i, j, b, c, flag = 0, p = 0;
cin >> n;
vector<int> a;
for (int i = 1; i <= sqrt(n); i++) {
if (n % i == 0) {
if (n / i == i)
a.push_back(i);
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)... | for _ in range(int(input())):
n=int(input())
l=[]
a=2
m=n
while a*a<=n:
if(n%a==0):
l.append(a)
n=n//a
break
a+=1
b=2
while b*b<=n:
if(n%b==0):
if b not in l:
l.append(b)
n=n//b
... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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())
answer = []
F = 0
for i in range(2,int(N**0.5)+1):
if N%i == 0:
X = i
Z = N//i
for j in range(2,int(Z**0.5)+1):
if Z%j==0:
ANS1=Z//j
if ANS1!=j and 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)... | from bisect import bisect_left as bl
from bisect import bisect_right as br
import heapq
import math
from collections import *
from functools import reduce,cmp_to_key
import sys
input = sys.stdin.readline
# M = mod = 998244353
def factors(n):return sorted(set(reduce(list.__add__, ([i, n//i] for i in range(1, int(n**0.... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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 static java.lang.Math.max;
public class Main {
public static void main(String[] args) throws IOException {
run();
end();
}
static void run() throws IOException {
int t = nextInt();
while (t-- > 0) {
solve(nextInt(... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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 factors(x):
l=[]
while(x>1):
a=0
for i in range(2,int(math.sqrt(x)+1)):
if(x%i==0):
l.append(i)
a=1
break
if(a==0):
l.append(x)
break
x=x//l[-1]
return l
t=int(input())
while(t... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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())
l = []
i = 2
c = 0
while(c < 2 and i*i < n):
if(n % i == 0):
l.append(i)
n //= i
c += 1
i += 1
#print(*l)
if(c == 2 and n not in l):
print('YES')
print(*l, n)
else:
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)... | # -*- coding: utf-8 -*-
"""
Created on Tue Feb 4 23:28:21 2020
@author: User
"""
import math
n = int(input())
num = []
for i in range(n):
num.append(int(input()))
for x in num:
flag = 0
counter = 0
y = int(math.sqrt(x))
for j in range(2,y+1) :
if flag == 1 : break
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)... | /*package whatever //do not write package name here */
import java.util.*;
import java.lang.*;
import java.io.*;
public class GFG {
public static void main (String[] args) {
Scanner sc=new Scanner(System.in);
int t=sc.nextInt();
for(int i=0;i<t;i++)
{int n=sc.nextInt();mm(n);}
... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | #-------------Program--------------
#----Kuzlyaev-Nikita-Codeforces----
#-------------Round615-------------
#----------------------------------
t=int(input())
for i in range(t):
n=int(input())
a=[]
for i in range(2,int(n**0.5)+2):
if len(a)==2:
a.append(n)
break
... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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 minp():
return sys.stdin.readline().strip()
def mint():
return int(minp())
def mints():
return map(int, minp().split())
def solve():
n = mint()
a = []
nn = n
for i in range(2,n+1):
if i*i > n:
break
if n % i == 0:
c = 0
while n % i == 0:
n //= i
if len(a) == 2:
break
... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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;
inline void setPrecision(int n) { cout.precision(n); }
long long int INF = 1e10;
int MOD = 1e9 + 7;
inline bool same2(int x, int y, int z) {
return (x == y) || (y == z) || (z == x);
}
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
int tt;
cin... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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())
raj=set()
for i in range(2, math.ceil(math.sqrt(n))):
if(n%i==0):
raj.add(i)
raj.add(n//i)
flg=0
for i in raj:
a=i
dup=n//a
andro=set()
for j in range(2, math.ceil(math.sqrt(... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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 div(n, r):
sq = math.sqrt(n)
d = 1
for x in range(2, int(sq) + 1):
if n % x == 0 and x != r:
d = x
break
return d
t = int(input())
for i in range(0, t):
y = int(input())
a = div(y, 1)
if a == 1:
print("NO")
continue
else... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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 deliteli_chisla(cur):
mas = []
ot = "NO"
n = cur
for i in range(2, math.ceil(cur ** 0.5)):
if n % i == 0:
mas.append(i)
for i in range(len(mas)):
del1 = mas[i]
nn = n // del1
s = nn
for j in range(i, len(mas)):
nn = s
... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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;
cin >> n;
int a, b, c, c1 = 0;
for (int i = 2; i < sqrt(n) + 1; i++) {
if (n % i == 0) {
if (c1 == 0) {
a = i;
c1++;
n = n / i;
} else if (c1 == 1) {... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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 = []
i=2
while(len(a)<2 and i*i <n):
if n%i ==0:
n=n//i
a.append(i)
i+=1
if len(a)==2 and n not in a:print("YES");print(n,*a)
else:print("NO") |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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())
a=[]
for j in range(2,int(n**0.5)+1):
if(n%j==0):
a.append(j)
n=n/j
break
if(len(a)==1):
for k in range(j+1,int(n**0.5)+1):
if(n%k==0):
a.append(k)
n=int(n/k)... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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=list()
for i in range(2,int(n**(2/3))+1):
if len(ans) == 2:
break
if n%i==0:
ans.append(i)
n//=i
if len(ans) == 2 and ans[1] < n:
print('YES')
print(*ans, n)
else:
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)... | import math
T=int(input())
for i in range((T)):
x=input()
x=int(x)
d=0
a,b,c=-1,-1,-1
p=x
for j in range(2,int(math.sqrt(x))+1):
if x%j==0:
a=j
x=x//j
break
if a!=-1:
for j in range(a+1,int(math.sqrt(x))+1):
if x%j==0:
... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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())
f = 0
for x in range(2, n):
if x * x * x >= n: break
for y in range(x + 1, n):
#print(x, y)
if x * y * y >= n: break
if n % (x * y) == 0:
f = 1
print("YES")
print(x, y, n // (x * y))
break
if f > 0: ... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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 t;
std::cin >> t;
for (; t > 0; t--) {
long n;
std::cin >> n;
int i = 2;
std::vector<int> primes;
while (i * i <= n) {
if (n % i == 0) {
while (n % i == 0) {
primes.push_back(i);
n /= i;
}
}
i++;
... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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.Arrays;
import java.util.Comparator;
import java.util.HashSet;
import java.util.LinkedList;
import java.util.Queue;
import java.util.Scanner;
import java.util.Set;
public class Main {
private static long gcd(long a, long b) {
return (a == 0 ? b : gcd(b % a, a));
}
public static void main(Strin... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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 check(n):
k=0
for i in range(2,int(sqrt(n+1))):
if k==2 and n>=i:
return a,b,n
if k==3:
return a,b,c
if k<3 and n<i:
return 'NO'
if n%i==0:
k+=1
n //= i
... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100)... | # Python3 implementation of the approach
from math import sqrt
# Function to find the required triplets
def findTriplets(x):
# To store the factors
fact = [];
factors = set();
# Find factors in sqrt(x) time
for i in range(2, int(sqrt(x))):
if (x % i == 0):
fact.append(i);
... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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
# -*- coding: utf-8 -*-
"""
Created on Wed Jan 22 08:37:17 2020
@author: huynguyen
"""
def findTripple(n):
res = []
for i in range(2,int(n**.5)+1):
if n%i == 0:
res.append(i)
n//=i
if len(res) ==2:
break
if n!=1:
if res:
... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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
q = int(input())
for _ in range(q):
n = int(input())
cnt = 0
save = [0] * 6
for i in range(2, int(sqrt(n)) + 2):
if n % i == 0:
cnt += 1
save[cnt] = i
n = int(n/i)
if cnt == 2:
break
if cnt < 2:
print("NO")... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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, i;
long n, d, arr[2];
cin >> t;
while (t--) {
cin >> n;
i = 0;
for (d = 2; d * d < n && i < 2; d++) {
if (n % d == 0) {
arr[i++] = d;
n /= d;
}
}
if (i == 2)
cout << "YES\n" << arr[0] << ' ' <... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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())
l=[]
i=2
while len(l)<2 and n>i**2:
if n%i==0:
l.append(i)
n=n/i
i+=1
if len(l)==2 and n not in l:
print('YES')
print(*l,int(n))
else:print('NO') |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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 os
import time
import collections
from collections import Counter, deque
import itertools
import math
import timeit
import random
#########################
# imgur.com/Pkt7iIf.png #
#########################
def sieve(n):
if n < 2: return list()
prime = [True for _ in range(n + 1)]
p = 3... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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
import math
primes = [2]
for n in range(3,math.ceil(math.sqrt(1e9)),2):
is_prime = True
sqrt_n = math.sqrt(n)
for p in primes:
if p > sqrt_n:
break
if n%p == 0:
is_prime = False
break
if is_prime:
primes.append(n)
t =... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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 Jan 22 19:40:08 2020
@author: DELL
"""
import math
# A function to print all prime factors of
# a given number n
def primeFactors(n):
li=[]
# Print the number of two's that divide n
while n % 2 == 0:
li.append(2)
n = n / 2
... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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())
t=pow(n,0.5)+3
l=[]
co=0
i=2
while i<t:
if n%i==0:
n=n//i
l.append(i)
else:
i+=1
# print(l)
if n!=1:
l.append(n)
flag="YES"
t=len(l)
if t>=3:
a=l[-1]
l.pop(-1)
b=l[-1]
l.pop(-1)
if a==b:
b=b*l[-1]
l.pop(-1)
c=1
for i in l:
c=c... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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.*;
public class Product_of_Three_Numbers
{
public static void main(String args[])
{
Scanner ob=new Scanner(System.in);
int t=ob.nextInt();
while(t-->0)
{
int n=ob.nextInt();
int a=0,b=0;
for(int i=2;i*i<n;i++)
{
if(n%i==0)
{
n/=i;
if(a==0)
a=i;
else{
b=i;
break;
}
}
}
if(a!=0 && b!=0 && n>b) {
System.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)... | t = int(input())
for _ in range(t):
n = int(input())
li = []
i = 2
while i*i<n:
if n%i==0:
li.append(i)
n/=i
if len(li)==2:
break
i+=1
if len(li)!=2 or n in li:
print('NO')
else:
print("YES")
print(li[0... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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 factorial
from collections import Counter
from sys import stdin
def RL(): return map(int, stdin.readline().rstrip().split() )
def comb(n, m): return factorial(n)/(factorial(m)*factorial(n-m)) if n>=m else 0
def perm(n, m): return factorial(n)//(factorial(n-m)) if n>=m else 0
def mdis(x1, y1, x2, y2): 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)... | # Python program to print prime factors
import math
# a given number n
def prime(n):
a= []# Print the number of two's that divide n
while n % 2 == 0:
a.append(2)
n = n / 2
# n must be odd at this point
# so a skip of 2 ( i = i + 2) can be used
for i in range(3,int(math.sqrt(n))+1,2):
# wh... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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 = []
while n % 2 == 0:
d.append(2)
n //= 2
for i in range(3, (int(math.sqrt(n)) + 1), 2):
while n % i == 0:
d.append(i)
n //= i
if n > 2:
d.append(n)
if len(d) < 3:
... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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 as mt
n = int(input())
i = 0
while i < n:
case = int(input())
n1 = case
a, b, c = 1, 1, 1
divisor = []
for p in range(2, mt.ceil(mt.sqrt(case + 1))):
if len(divisor) == 2:
divisor.append(int(n1 / (divisor[0] * divisor[1])))
break
elif case % p == 0... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int n, q;
cin >> q;
while (q--) {
cin >> n;
vector<int> res;
for (int i = 2; res.size() < 2 && i * i < n; ++i) {
if (n % i == 0) {
res.push_back(i);
n /= i;
}
... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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):
d = 2
while d * d <= n and n % d != 0:
d += 1
return d * d > n
t = int(input())
ans = []
for j in range(t):
n = int(input())
divs = []
i = 2
while i*i <= n:
if n % i == 0:
x = n // i
u = i + 1
while u*u <= x:
if... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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.Math.*;
public class CodeForces615C {
public static void main (String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int t = Integer.valueOf(br.readLine());
for (int i = 0;i<t ;i++ ... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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.ArrayList;
import java.util.List;
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int t = Integer.parseInt(sc.next());
for(int i = 0; i < t; i++){
int n = 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)... | from sys import stdin, stdout
from collections import defaultdict
import math
rl = lambda: stdin.readline()
rll = lambda: stdin.readline().split()
def main():
cases = int(input())
for line in stdin:
n = int(line)
ans = []
f1 = 2
while f1 <= math.sqrt(n):
if n % f1 == 0:
ans.append(f1)
break
f1... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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
n = int(input())
for i in range(n):
num = int(input())
ans = []
for mult in range(2, ceil(sqrt(num))):
#print('Try ', mult,' for ', num)
if num < mult:
break
if num % mult == 0:
ans.append(mult)
num = num / mult
... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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.lang.*;
import java.io.*;
public class Codeforces {
static FastReader sc=new FastReader();
static PrintWriter out=new PrintWriter(System.out);
static long mod=1000000007;
public static void main (String[] args) throws java.lang.Exception
{
// your code goes here
// StringBuf... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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 python
# coding: utf-8
# In[7]:
import math
def divisor(n):
for i in range(2,int(math.sqrt(n))+1):
if n%i==0:
k=n//i
for j in range(2,int(math.sqrt(k))+1):
if j!=i and k%j==0:
t=k//j
if t!=i and t!=j:
... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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.math.*;
public class S{
static class Pair implements Comparable<Pair>{
int a;
int b;
public Pair(int x,int y){a=x;b=y;}
public Pair(){}
public int compareTo(Pair p1){
if(a == p1.a)
return b - p1.b;
return a - p1.a;
}
}
stati... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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 q;
cin >> q;
for (int i = 0; i < q; ++i) {
int n;
cin >> n;
set<int> used;
for (int i = 2; i * i <= n; ++i) {
if (n % i == 0 && !used.count(i)) {
used.insert(i);
n /= i;
break;
}
}
for (int i... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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 pm(n):
arr = []
while(n%2==0):
arr.append(2)
n//=2
for i in range(3,int(pow(n,0.5))+1,2):
while n%i==0:
arr.append(i)
n//=i
if n>2:
arr.append(n)
return arr
tc = int(input())
for ii in range(tc):
n = int(input())
arr = pm(n)
a... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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//=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)... | import math
from functools import reduce
t = int(input())
for _ in range(t):
n = int(input())
m = n
sqrt_n = int(math.sqrt(n))
res = []
for i in range(2, sqrt_n):
if n == 1:
break
while n > 1 and n % i == 0:
res.append(i)
n //= i
if n > 1:
res.append(n)
if len(res) < 3:
print('NO')
elif len(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)... | # from sys import stdin, stdout
# print = lambda x: stdout.write(x)
# input = stdin.readline
from math import sqrt
#Ref: Geeksforgeeks
def primeFactors(n):
arr = []
# Print the number of two's that divide n
while n % 2 == 0:
arr.append(2)
n = n // 2
# n must be odd at this point
# ... |
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can 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 factor(n):
s = int(sqrt(n))
rLst = [1, n]
for i in xrange(2, s+1):
if n%i==0:
rLst.append(i)
if n//i != i:
rLst.append(n//i)
rLst = list(set(rLst))
rLst.sort()
return rLst
from sys import stdin
raw_input = lambda: stdin.readline().rstrip()
input = lambda: int(raw_input())
I=... |
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