Big Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 34084    Accepted Submission(s): 16111

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

看到这个题就像用java写，结果超时。其实这是一道规律题：

N！的长度=log10（i）【i：2~n】+1;

然后就简单了。

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