Harmonic Number

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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

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

题意：求f(n)=1/1+1/2+1/3+1/4…1/n (1 ≤ n ≤ 10⁸).，精确到10^-8

^知识点：

^{调和级数(即f(n))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}

^{c++ math库中，log即为ln。}

^题解:

^{公式：f(n)=ln(n)+C+1}^/(2*n);

#include<stdio.h>#include <iostream>#include<cstdio>#include<iostream>#include<algorithm>#include<math.h>#include<string.h>#include<map>#include<queue>#include<vector>#include<deque>#define ll long long#define inf 0x3f3f3f3f#define mem(a,b) memset(a,b,sizeof(a))using namespace std;const double r=0.57721566490153286060651209;double s[100005];int main(){    int t;    scanf("%d",&t);    s[1]=1;    for(int i=2;i<=10000;i++)    {        s[i]=s[i-1]+1.0/i;    }    int g=0;    while(t--)    {        ++g;       int n;       scanf("%d",&n);       if(n<=10000)        printf("Case %d: %.10lf\n",g,s[n]);       else       {       double sum;       sum=log(n)+r+1.0/(2*n);       printf("Case %d: %.10lf\n",g,sum);       }    }}

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