OJ之求N的阶乘的位数（大整数）

题目来源 http://poj.org/

编号1423

题目

Big Number

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 <= m <= 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

 

分析

求输入指定大数阶乘的位数。

一般公式为：

log10(n!)=log10(1*2*3…*n)=log10(1)+log10(2)+…+log10(n)

求位数：

log10(n!)=log10(1*2*3…*n)=log10(1)+log10(2)+…+log10(n)，对log10(n!)的值取整加1就是n!的位数。

《计算机程序设计艺术》中给出了另一个公式

n! = sqrt(2*π*n) * ((n/e)^n) * (1 + 1/(12*n) + 1/(288*n*n) + O(1/n^3))

π = acos(-1)

e = exp(1)

两边对10取对数

忽略log10(1 + 1/(12*n) + 1/(288*n*n) + O(1/n^3)) ≈ log10(1) = 0

得到公式

log10(n!) = log10(sqrt(2 * pi * n)) + n * log10(n / e)

 

源码

#include <stdio.h>
#include <math.h>

#define NUM 65535

int main(int argc, char* argv[])
{
    long iloop = 0;
    long counter = 0;
    long target = 0;
    int result[NUM] = {0};

    double Pi,E;  
    Pi = acos(-1.0);  
    E = exp(1.0);

    scanf("%d", &counter);

    iloop = counter;
    while (0 < iloop)
    {
        scanf("%d", &target);

        if(target == 1)
        {

            result[iloop] = 1;
        }
        else
        {
            result[iloop] = (int)(log10l(sqrtl(2*Pi*target)) + target*log10l(target/E)+1);
        }

        iloop--;
    }

    iloop = counter;
    while (0 < iloop)
    {
        printf("%d\r\n", result[iloop]);

        iloop--;
    }

    return 0;
}