LightOJ - 1234（欧拉常数 模板记）

链接：https://cn.vjudge.net/contest/276155#problem/I

1+1/2+1/3... = log(1+n) + r(欧拉常数),需要大精度时减去1.0/(2*n);

r = 0.57721566490153286060651209

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
const double r = 0.57721566490153286060651209;

int t,n;
double a[10005];

int main(){
	a[1] = 1.0;
		for(int i = 2;i<=10000;++i){
			a[i] = a[i-1] + 1.0/i;
			//cout<<a[i]<<endl;
		}
	scanf("%d",&t);
	int id = 0;
	while(t--){
		scanf("%d",&n);
		printf("Case %d: ",++id);
		if(n <= 10000)
	printf("%.10lf\n",a[n]);
		else{
			printf("%.10lf\n",log(1+n) + r - 1.0/(2*n));
		}
	}
	return 0;
}