hdu 1018

                     Big Number
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 34971 Accepted Submission(s): 16621


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

题目大意就是求N!的位数，我们先看看一个例子：

100000的位数等于log10(100000)+1=5+1=6

同理：n!的位数等于log10(n!)=log10(n)+log10(n).........

<span style="font-size:18px;">#include <iostream>#include<cstdio>#include<cmath>using namespace std;int main(){    int t,n;    scanf("%d",&t);    while(t--)    {        scanf("%d",&n);        double temp=0;        for(int i=2;i<=n;i++)            temp+=log10(i*1.0);        int ans=(int)temp+1;        printf("%d\n",ans);    }    return 0;}</span>