Harmonic Number(调和级数+欧拉常数)

Harmonic Number

https://vjudge.net/contest/288520#problem/I

In mathematics, the n^th 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 H_n.

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 n^th 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

 

 

欧拉常数：

const double oula=0.57721566490153286060651209;

近似公式：ln(n)+C+1/(2*n)    C++中log即为ln

在n小的时候不精确，需要暴力求解

#include<bits/stdc++.h>
using namespace std;
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define sqr(x) ((x)*(x))
#define pb push_back
#define eb emplace_back
#define maxn 10000005
#define eps 1e-8
#define pi acos(-1.0)
#define rep(k,i,j) for(int k=i;k<j;k++)
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<long long,int>pli;
typedef pair<int,char> pic;
typedef pair<pair<int,string>,pii> ppp;
typedef unsigned long long ull;
const long long mod=1e9+7;
const double oula=0.57721566490153286060651209;
using namespace std;

int t;

int main(){
   // std::ios::sync_with_stdio(false);
    int t;
    cin>>t;
    ///log(n)+C+1/(2*n)  
    for(int _=1;_<=t;_++){
        int n;
        cin>>n;
        double ans=0;
        if(n<=100000)
            for(int i=1;i<=n;i++){
                ans=ans+1.0/i;
            }
        else{
            ans=log(n)+oula+1.0/2/n;
        }
        printf("Case %d: %.8f\n",_,ans);

    }
}