C

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

坑点：

1.输出第一个样例时可以是1.0000000000

2.对于double类型的数据进行除法时要采用1.0/i的形式，否则结果会有误差

题解:

调和级数公式：f(n) = ln(n) + C + 1.0/2*n;

其中C是欧拉常数其值等于C ≈ 0.57721566490153286060651209；

对于较小的数据公式的误差会比较大，所以对于前10000个数据采用打表的方法来求解

AC代码

#include<iostream>
#include<cstdio>
#include<cmath>
const double C = 0.57721566490153286060651209;  //欧拉常数

using namespace std;

int main()
{
    double a[10005];
    int t, n;
    int flag = 0;
    scanf("%d", &t);
    a[0] = 0;
    for(int i = 1; i <= 10000; i++)
    {
        a[i] = a[i-1] + 1.0/i;
    }

    while(t--)
    {
        scanf("%d", &n);
        if(n <= 10000)
            printf("Case %d: %.10lf
", ++flag, a[n]);
        else
        {
            double sum;
            sum = log(n) + C + 1.0/(2*n); //这里是1.0不然的话算的结果有偏差
            printf("Case %d: %.10lf
", ++flag, sum);
        }

    }

    return 0;
}