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 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).

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

12

1

2

3

4

5

6

7

8

9

90000000

99999999

100000000

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

题目大意：

求1 + 1/2 + 1/3 + 1/4 + 1/ 5 +...+ 1/ n(1 ≤ n ≤ 10⁸)

调和级数部分和，可以利用公式，（唉，然而我并不记得公式高数没学好-_-||）

如果直接循环的肯定会超时，那么我们开一个10^8/40 = 250万的数组用来分别存

1到1/40的和、1到1/80的和、1到1/120的和、1到1/160的和、... 、1到1/2500000的和

这样对于每一个n最多循环39次，节省了时间

#include<stdio.h>

#include<math.h>

#include<string.h>

#include<stdlib.h>

#include<algorithm>

using namespace std;

const int N = ;

const int M = 1e8 + ;

typedef long long ll;

double a[N];

int main()

{

    int t, n, p = ;

    double s = ;

    for(int i =  ; i < M ; i++)

    {

        s += (1.0 / i);

        if(i %  == )

            a[i / ] = s;

    }

    scanf("%d", &t);

    while(t--)

    {

        p++;

        scanf("%d", &n);

        int x = n / ;

        s = a[x];

        for(int i = x *  +  ; i <= n ; i++)

            s += (1.0 / i);

        printf("Case %d: %.10f\n", p, s);

    }

    return ;

}