HDU 1018 (斯特林公式)

Big Number

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

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

问题分析

这里lz就介绍两种解法，都是公式。像lz这种蒟蒻就打算背住QAQ。
①即n的位数是(int)log10(n)+1；
代码为

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

②斯特林公式

log(n!) = (1/2)log(2*π*n)+n*log(n/e)

#include<iostream>
#include<cstdio>
#include<math.h>
#include<algorithm>
#define PI 3.141592654
#define e 2.71828182
using namespace std;

int main()
{
    int t,n;
    double x;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        x = (1.0/2*log10(2*PI*n)+n*log10(n/e));
        printf("%d\n",(int)x+1);
    }
    return 0;
}