In many applications very large integers numbers are required. Some of these applications are using keys for secure transmission of data, encryption, etc. In this problem you are given a number, you have to determine the number of
digits in the factorial of the number.
Input consists of several lines of integer numbers. The first line contains an integer n, which is the number of cases to be tested, followed by n lines, one integer 1 ≤ n ≤ 107 on each line.
The output contains the number of digits in the factorial of the integers appearing in the input.
Asia 2002, Dhaka (Bengal)
意解：用对数来求n的阶乘的位数；
AC代码：
#include <iostream>
#include <cstring>
#include <cstdio>
#include <cmath>
using namespace std;
typedef long long ll;
int main()
{
int n,m;
cin>>m;
while(m--)
{
scanf("%d",&n);
double sum = 0;
for(int i = 1; i <= n; i++)
{
sum += log10((double)i);
}
printf("%d\n",(int)sum + 1);
}
return 0;
}
