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
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
double sum[1000005];
int main()
{
double s=0;
sum[0]=0;
for(int i=1;i<=1e8;i++)
{
s+=1.0/i;
if(i%100==0)
{
sum[i/100]=s;
}
}
int n,cas;
scanf("%d",&cas);
for(int tt=1;tt<=cas;tt++)
{
scanf("%d",&n);
double ans=sum[n/100];
for(int i=(n/100)*100+1;i<=n;i++)
{
ans+=1.0/i;
}
printf("Case %d: %.10lf\n",tt,ans);
}
return 0;
}