LightOJ1234——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

差不多就是求调和级数的前n项和
n比较小用暴力求 n比较大用近似值公式有可能超时或精度wa 调整一下这个边界

代码：

#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
using namespace std;
const double C=0.5772156649015328606;
int main(void){
    int t;
    int n;
    scanf("%d",&t);
    int c=1;
    while(t--){
        scanf("%d",&n);
        double ans=0;
        if(n<1000000){
            for(int i=1;i<=n;i++){
                ans+=1.0/i;
            }
        }
        else{
            ans=log((double)n)+C+1.0/(2*n);
        }
        printf("Case %d: %.10lf\n",c++,ans);
    }
    return 0;
}