Harmonic Number 数论

Harmonic Number

In mathematics, the nth 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 Hn.

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 nth 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

刚开始看见  ：

是不是很熟悉，反正对于我这种刚上完大一的高数一眼瞄上去这不就是调和级数吗。

个人赶脚 这种求一个有规律式子的没公式基本白瞎。

所以就立马上网搜了一波调和级数的公式，

H(n) = C+log(n)+/(*n);

然后 C=

#include
    
     
#include
     
      
#include
      
       
using namespace std;


double C=0.57721566490153286060651209;
double a[10000+10];
void init()
{
    a[0]=0;
    for(int i=1;i<10000;i++)
    {
        a[i]=a[i-1]+1.0/i;
    }
}
typedef long long  ll;
int  main()
{
     int T;
     ll n;
     init();
     scanf("%d",&T);
     for(int ii=1;ii<=T;ii++)
     {
         scanf("%lld",&n);
         double ans=0;
         if(n<10000)
         {
             ans=a[n];
         }
         else
         {
             ans=C+log(n)+1.0/(2*n);
         }
         printf("Case %d: %.10lf\n",ii,ans);
     }
     return 0;
}


// f(n)=ln(n)+C+1/(2*n);