Number lengths FZU

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)

Input
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.

Output
For each value of N, print out how many digits are in N!.

Sample Input
1
3
32000
Sample Output
1
1
130271

//N！位数应该是 log10(1)+log10(2)+···+log10(n) 取整加1,
//注意：log10（n）传入的值必须是double型，所以要进行强制转换！
#include<map>
#include<queue>
#include<stack>
#include<vector>
#include<math.h>
#include<cstdio>
#include<sstream>
#include<numeric>//STL数值算法头文件
#include<stdlib.h>
#include<string.h>
#include<iostream>
#include<algorithm>
#include<functional>//模板类头文件
using namespace std;

const int INF=1e9+7;
const int maxn=110;
typedef long long ll;

int n;

int main()
{
    while(~scanf("%d",&n))
    {
        int i,j=0;
        double sum=0;
        for(i=1; i<=n; i++)
            sum+=log10(double(i));
        j=(int)sum+1;
        printf("%d
",j);
    }
    return 0;
}