题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1018

Big Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 27548    Accepted Submission(s): 12526

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：求一个数阶乘的位数。

分析：原本以为是一道大数题，其实是一道简单的数学题，需要用到“斯特灵”公式，log(n!)=log1 + log2 + ... + logn。

斯特灵公式链接：http://zh.wikipedia.org/wiki/%E6%96%AF%E7%89%B9%E9%9D%88%E5%85%AC%E5%BC%8F

 1 #include <cstdio>
 2 #include <cmath>
 3 using namespace std;
 4
 5 int main ()
 6 {
 7     int n,m,i;
 8     double sum;
 9     scanf ("%d",&n);
10     while (n--)
11     {
12         scanf ("%d",&m);
13         sum = 0;
14         for (i=1; i<=m; i++)
15         {
16             sum += log10((double)i);
17         }
18         printf ("%d\n",(int)sum+1);
19     }
20     return 0;
21 }