Harmonic Number （LightOJ-1234）

Problem Description

In mathematics, the nth harmonic number is the sum of the reciprocals of the first n natural numbers:

 

In this problem, you are given n, you have to find Hn.

Input

Input starts with an integer T (≤ 10000), denoting the number of test cases.

Each case starts with a line containing an integer n (1 ≤ n ≤ 108).

Output

For each case, print the case number and the nth harmonic number. Errors less than 10-8 will be ignored.

Sample Input

12
1
2
3
4
5
6
7
8
9
90000000
99999999
100000000

Sample Output

Case 1: 1
Case 2: 1.5
Case 3: 1.8333333333
Case 4: 2.0833333333
Case 5: 2.2833333333
Case 6: 2.450
Case 7: 2.5928571429
Case 8: 2.7178571429
Case 9: 2.8289682540
Case 10: 18.8925358988
Case 11: 18.9978964039
Case 12: 18.9978964139

题意：给出 t 组数，每组输入一个数 n，求这个数的调和级数 H(n)

思路：

当 n 很小时，直接打表求即可

当 n 很大时，需要使用欧拉给出的近似公式：，其中 C 是欧拉常数，约等于 0.5772

Source Program

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<string>
#include<cstring>
#include<cmath>
#include<ctime>
#include<algorithm>
#include<utility>
#include<stack>
#include<queue>
#include<vector>
#include<set>
#include<map>
#define PI acos(-1.0)
#define E 1e-9
#define INF 0x3f3f3f3f
#define LL long long
const int MOD=10007;
const int N=10000+5;
const int dx[]= {-1,1,0,0};
const int dy[]= {0,0,-1,1};
using namespace std;
double a[N];
int main(){

    a[1]=1;
    for(int i=2;i<=10000;i++)
        a[i]=a[i-1]+1.0/i;

    int t;
    scanf("%d",&t);

    int Case=1;
    while(t--){
        int n;
        scanf("%d",&n);


        if(n<=10000){
            printf("Case %d: %.10lf\n",Case++,a[n]);
        }
        else{
            double res=log(n)+1.0/(2*n)+0.57721566490153286060651209;
            printf("Case %d: %.10lf\n",Case++,res);
        }
    }
    return 0;
}