Problem Description：

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：

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.

Output：

The output contains the number of digits in the factorial of the integers appearing in the input.

Sample Input：

2

10

20

Sample Output：

7

19

解题思路：

这道题很好理解，就一句话：一个数的位数等于它以10为底取对数再加一。

程序代码：

#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;
int main()
{
  int t,n,i;
  double sum;
  scanf("%d",&t);
  while(t--)
  {
    sum=0.0;
    scanf("%d",&n);
    for(i=2;i<=n;i++)
      sum+=log10(i);
    printf("%d\n",(int)sum+1);
  }
  return 0;
}