HDU 1018.Big Number-Stirling(斯特林)公式 取N阶乘近似值

最近一堆题目要补，一直咸鱼，补了一堆水题都没必要写题解。备忘一下这个公式。

Stirling公式的意义在于：当n足够大时，n!计算起来十分困难，虽然有很多关于n!的等式，但并不能很好地对阶乘结果进行估计，尤其是n很大之后，误差将会非常大。但利用Stirling公式可以将阶乘转化成幂函数，使得阶乘的结果得以更好的估计。而且n越大，估计得越准确。

传送门:_(:з」∠)_

再来一个详细一点的，传送门:( ･´ω`･ )

Big Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 40645    Accepted Submission(s): 19863

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

 

Source

Asia 2002, Dhaka (Bengal)

 

 

 

代码:

 #include<iostream>
 #include<cstring>
 #include<cstdio>
 #include<algorithm>
 #include<cmath>
 using namespace std;
 const double e=exp();
 const double pi=acos(-1.0);
 int main(){
     int t;
     scanf("%d",&t);
     while(t--){
         int n;
         scanf("%d",&n);
         if(n==){printf("1\n");continue;}
         double s=log10(2.0*pi)/2.0+(n+0.5)*log10(n)-n*log10(e);
         int ans=ceil(s);
         printf("%d\n",ans);
     }
     return ;
 }