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
题意：求f(n)=1/1+1/2+1/3+1/4…1/n (1 ≤ n ≤ 108).，精确到10^-8。
题解：当n很小时，可直接求出结果，当n很大时，利用公式f(n)=ln(n)+C+1/(2*n)，在C++ math库中，log即为ln;
代码：
#include <iostream>
#include <cstdio>
#include <cmath>
using namespace std;
const double r=0.57721566490153286060651209;//欧拉常数
double a[];
int main()
{
a[]=;
for (int i=;i<;i++){//预先把小于10000的f(n)求出来
a[i]=a[i-]+1.0/i;
}
int n; cin>>n;
for (int kase=;kase<=n;kase++)
{
cin>>n;
if (n<){//n<10000时，可直接得出结果
printf("Case %d: %.10lf\n",kase,a[n]);
}
else{//否则利用欧拉公式
double a=log(n)+r+1.0/(*n);
printf("Case %d: %.10lf\n",kase,a);
}
}
return ;
}