求一个数阶乘后位数问题

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问题：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

回答：题意求一个数阶乘的位数.

方法一：

#include <stdio.h>
#include <math.h>
void main()
{
 int count,var,j;
 double sum;
 scanf("%d",&count);
 while(count--)
 {
  sum=0;
  scanf("%d",&var);
  for(j=1;j<=var;j++)
   sum+=log10(j);
  printf("%d/n",(int)sum+1);
 }
}
这是最简单的方法，但是耗时长，406ms。
好吧！我们换个方法：
这需要组合数学的知识。log10(n!) = log10(sqrt(2 * pi * n)) + n * log10(n / e)
 方法二：
#include <iostream>
#include <cmath>
#define PI 3.14159265
int n;
int num,ans;
void solve()
{
    double t;
    int i;
    t = (num*log(num) - num + 0.5*log(2*num*PI))/log(10);
    ans = t+1;
    printf("%d/n",ans);
}
int main()
{
    while(scanf("%d",&n)!=EOF)
    {
        while(n--)
        {
            scanf("%d",&num);
            solve();
        }
    }
    return 0;
}

求一个数阶乘后位数问题

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原文地址：http://www.cnblogs.com/benchao/p/4529635.html