Problem 2: Digits in Factorial

Points:

Small Case: 10 Answer: 3223

Large Case: 20 Answer: 11565705518104

Factorial of a number 'n' is defined by

n!= 1 x 2 x 3 x 4 x ....... x n

The problem is to find the number of digits in n!.

Small Case: n=1215

Large Case: n=10^12

Solution:

n! can be expressed by stirling formula as

Total number of digit in a number x can be given by
floor(log(x))+1 where log is calculated in base 10.
Hence,the number if digits in n! is
floor(log(n!))+1

#include<iostream>
#include<cmath>
using namespace std;
#define PI 3.141592653589793
int main()
{
    long long n;
    long long digit;
    cin>>n;
    digit = floor(log10(pow(2*PI*n,.5))+(double)n*(log10((double)n)-((double)1/log(10))))+1;
    cout<<digit<<"\n";
}