Big Number（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): 31292 Accepted Submission(s): 14537



[align=left]Problem Description[/align]
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.

[align=left]Input[/align]
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.

[align=left]Output[/align]
The output contains the number of digits in the factorial of the integers appearing in the input.

[align=left]Sample Input[/align]

2
10
20

[align=left]Sample Output[/align]

7
19

[align=left]Source[/align]
Asia 2002, Dhaka (Bengal)

<pre name="code" class="html">题意为：求n的阶乘的位数，n为整数。
只需求出(log1+...+logn)+1即可。

#include<stdio.h>#include<string.h>#include<math.h>int main(){ int n,i,a; double sum; scanf("%d",&n); while(n--) { sum=0.0; scanf("%d",&a); for(i=1;i<=a;++i) { sum+=(log10(double(i))); } printf("%d\n",int(sum)+1); } return 0;}//取对数+1；