调和级数求和（分块）

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

分块：

#include<iostream>
#include<algorithm>
#include<cmath>
using namespace std;
const double p=0.577215664;
const int maxn=1e8+100;
/*
1/i求和为ln(n)+p 
*/
//数论分块 
double a[maxn/100+1];
void inint(){
    a[0]=0;
    a[1]=1.0;
    double t=1.0;
    for(int i=2;i<=maxn;i++){
        t=t+1.0/i;
        if(i%100==0){
            a[i/100]=t; 
        }
    }
}
int main(){
   inint();
   int t;
   cin>>t;
   int n;
   int kase=0;
   while(t--){
           cin>>n;
        int t=n/100;
        double ans=a[t];
        for(int i=t*100+1;i<=n;i++){
            ans+=1.0/i;
        } 
        printf("Case %d: %.10lf
",++kase,ans);
   }
   return 0;
}