hdu 1018 -Big Number(求n的阶乘的位数)

最新推荐文章于 2019-10-18 20:20:43 发布

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最新推荐文章于 2019-10-18 20:20:43 发布

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分类专栏：数学 OJ题目

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本文介绍了一种计算任意整数阶乘位数的方法，通过使用对数特性简化问题，实现了快速求解。适用于安全传输、加密等应用场景中对大整数的需求。

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

思路：

n的位数可以用log 10(n)+1表示；

故n!的位数可以用log10(n!)+1表示；

log10(n!)=log10(1)+......+log10(n);

代码如下：

#include<stdio.h>
#include<math.h>
int main ()
{
   double sum;
   int d;
    int i,n,m;
    scanf ("%d",&n);
    while (n--)
    {
        sum=0;
        scanf ("%d",&m);
        for (i=2;i<=m;i++)
            sum+=log10(i*1.0);
            d=sum+1;
            printf ("%d\n",d);
    }
    return 0;
}