HDU-1018 Big Number 斯特林公式

最新推荐文章于 2025-09-12 18:08:49 发布

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最新推荐文章于 2025-09-12 18:08:49 发布

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文章标签： java

本文介绍了一种计算大数阶乘位数的方法，利用斯特林公式避免直接计算大数阶乘，提供了C语言实现代码。

Big Number

Time Limit: 20000/10000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10040 Accepted Submission(s): 4476

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 ≤ 10 ⁷ 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
  

Source

Asia 2002, Dhaka (Bengal)

Recommend

JGShining

刚看到题目，第一反应便想到大数，但是看看题目，给出的数据可能达到 10^7 的级别，这么大的数，求阶乘！ oh，my god！要知道，10000000 的阶乘有六千万多位。

还好，我们有公式， log10(n!)=(0.5*log(2*PI*n)+n*log(n)-n)/log(10) ，这里利用了 log10 (N)=ln(N)/ln(10);

公式的名字是斯特林公式维基百科的链接 http://zh.wikipedia.org/wiki/%E6%96%AF%E7%89%B9%E6%9E%97%E5%85%AC%E5%BC%8F

#include<stdio.h>
#include<stdlib.h>
#include<math.h>
#include<string.h>
#include<time.h>
#define PI 3.1415926
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        int digits,n;
        scanf("%d",&n);
        if(n==0)  //没有也过了..
        {
            printf("1\n");
            continue;
        }
        digits=(int)((0.5*log(2*PI*n)+n*log(n)-n)/log(10));
        printf("%d\n",digits+1);   //加1是必须的。
    }
    return 0;
}