算法与数据结构实验题1.1 单身狗进化

1、题目:

2、代码:

#include<stdio.h>
#define PI 3.141592654
#define E 2.71828182846
#include<math.h>
int main()
{
	int n;
	scanf("%d",&n);
	int result=1;
	result=log10(2*PI*n)/2+n*log10(n/E)+1;
	printf("%d
",result);
	return 0;
}
原文地址:https://www.cnblogs.com/laixiaolian/p/5937145.html