Thursday, 6 April 2017

Geeksforgeeks Solution For " No of Factors of n! "

GeeksforGeeks Solution For Hard Domain .Below You Can Find The Solution Of  School Basic ,Easy ,Medium . Or Hackerrank Solution You Can Also Direct Submit Your Solution to Geeksforgeeks Same Problem .You Need to login then you can submit you answers 

Problem :- No of Factors of n! 

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Solution :- 

#include <bits/stdc++.h>
using namespace std;
void sieve(int n, bool prime[])
{
for (int i=1; i<=n; i++)
prime[i] = 1;
   prime[1] = 0;
for (int i=2; i*i<=n; i++)
{
if (prime[i])
{
for (int j=i*i; j<=n; j += i)
prime[j] = 0;
}
}
}
int expFactor(int n, int p)
{
int x = p;
int exponent = 0;
while ((n/x) > 0)
{
exponent += n/x;
x *= p;
}
return exponent;
}

long long int  countFactors(int n)
{
long long int  ans = 1;
bool prime[n+1];
sieve(n, prime);
for (int p=1; p<=n; p++)
{
if (prime[p]==1)
ans *= (expFactor(n, p) + 1);
}

return ans;
}

int main()
{
int t;
cin>>t;
while(t--)
{
int n;
cin>>n;
cout<<countFactors(n);
cout<<endl;
}
return 0;
}

Output:-



Geeksforgeeks Solution For " No of Factors of n! "

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