N! (N factorial) can be quite irritating and difficult to compute for large values of N. So instead of calculating N!, I want to know how many digits are in it. (Remember that N! = N * (N - 1) * (N - 2) * ... * 2 * 1)
Each line of the input will have a single integer N on it 0 < N < 1000000 (1 million). Input is terminated by end of file.
For each value of N, print out how many digits are in N!.
1 3 32000Sample Output
1 1
130271
#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<string.h>
#include<cmath>
using namespace std;
int main()
{
int n;
while(cin>>n)
{
double a=1;
for(double i=1; i<=n; i++)
{
a+=log(i)/log(10);
}
int ans=a;
cout<<ans<<endl;
}
return 0;
}