LightOJ1234(Harmonic Number)

Harmonic Number

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

分析：调和级数至今没有一个完全正确的公式，但欧拉给出过一个近似公式：(n很大时)

      f(n)≈ln(n)+C+1/2*n    

      ^{欧拉常数值：C≈0.57721566490153286060651209}

#include<cstdio>
#include<cmath>
double C=0.57721566490153286060651209;
double a[120000];
int main()
{
    int T,n,cas=1;
    a[1]=1;
    for(int i=2;i<120000;i++)
    a[i]=a[i-1]+1.0/i;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        printf("Case %d: ",cas++);
        if(n<120000) printf("%.10lf
",a[n]);
        else
        {
            printf("%.10lf
",log(n)+C+1.0/(2*n));
        }
    }
    
    return 0;
}