HDU1018求大数N！的位数【取对数】

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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     19

题目分析：

第一种做法：

N！=1*2*3....*n

求位数我们一般用对一个数取对数就可以了，

log10(n!)=log10(1)+ log10(2) +log10(3)...+log10(n);

所以循环求和就可以了！

但是这里注意一点结果要加1！因为这里计算出来的 log10(1)=0 ！

所以结果要加上这个误差 ‘1’

第二种做法：

这就是我最近研究的斯特林数，第一类斯特林数就可以做这个！

补充一点，斯特林数能够做一切关于阶乘有关的大数运算要深入学习！

这里给出递归公式：

log10(n!)=1.0/2*log10(2*pi*n)+n*log10(n/e)

#include <iostream>#include <stdio.h>#include <algorithm>#include <iomanip>#include <math.h>#include <string.h>#include <stdlib.h>using namespace std;#define e 2.718281828459045235#define pi acos(-1)int main()//取对数{    int t,ans,i,n;    double sum;    cin>>t;    while(t--)    {        scanf("%d",&n);        sum=1;        for(i=1; i<=n; i++)            sum+=log10(i);        printf("%d\n",((int)sum));    }    return  0;}int main()//Striling公式{    int t,ans,n;    cin>>t;    while(t--)    {        scanf("%d",&n);        ans=(int)(1.0/2.0*log10(2.0*pi*n)+1.0*n*log10(n/e)+1);        printf("%d\n",ans);    }    return 0;}

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