Light OJ 1234 Harmonic Number

Description

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

思路：没思路啊 这个是高数的东西 发散 n足够大时它无穷大 直接公式解

#include <cstdio>
#include <cstring>
#include <cmath>
#include <string>
#include <algorithm>
#include <iostream>
using namespace std;
const int maxn = 1000010;
const double phi = 0.57721566490153286060651209;///这个数字就是后来称作的欧拉常数。不过遗憾的是，我们对这个常量还知之甚少，连这个数是有理数还是无理数都还是个谜。但是此题中所用是ln（n+0.5）+phi；double a[maxn];

int main()
{
    for(int i = 1; i <= 1000000; i++)
        a[i] = a[i-1] + 1.0/i;
    int cas = 1;
    int T;
    scanf("%d", &T);
    while(T--)
    {
        int n;
        scanf("%d", &n);
        double ans;
        if(n <= 1000000)
            ans = a[n];
        else
            ans = log(n+0.5) + phi;
        printf("Case %d: %.10lf
", cas++, ans);
    }
    return 0;
}