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
公式。。。记住就行
欧拉常数y=0.57721566490153286060651209
ans=log(n)+y+1.0/(2*n)
#include <iostream>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <stdio.h>
using namespace std;
typedef long long ll;
const int INF=0x3f3f3f3f;
const int maxn=1e4;
double f[maxn+5];
int main()
{
int T;
cin>>T;
f[1]=1;
for(int i=2;i<=maxn;i++)
f[i]=f[i-1]+1.0/i;
for(int cas=1;cas<=T;cas++)
{
int n;
cin>>n;
if(n<=maxn)
printf("Case %d: %.10lf
",cas,f[n]);
else
{
double ans=log(n)+0.57721566490153286060651209+1.0/(2*n);
printf("Case %d: %.10lf
",cas,ans);
}
}
return 0;
}