I - Harmonic Number (调和级数+欧拉常数)

I - Harmonic Number

234 - Harmonic Number

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Time Limit: 3 second(s)

Memory Limit: 32 MB

In mathematics, the n^th 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 H_n.

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 n^th harmonic number. Errors less than 10^-8 will be ignored.

Sample Input	Output for Sample Input
12 1 2 3 4 5 6 7 8 9 90000000 99999999 100000000	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

 

           题目所给为调和级数，欧拉给出过一个近似公式：(n很大时)

  f(n)≈ln(n)+C+1/2*n    

  欧拉常数值：C≈0.57721566490153286060651209

           代码：

           

#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int maxn = 1e6+7;
double C = 0.57721566490153286060651209;//欧拉常数
double m[maxn];
void init()
{
    m[0] = 0;
    m[1] = 1.0;
    for(int i = 2; i < maxn; i++)
        m[i] = m[i - 1]+ 1/(1.0*i);
}

int main()
{
    int t, n, cas = 0;
    scanf("%d",&t);
    double a;
    init();
    while(t--)
    {
        scanf("%d",&n);
        if(n<maxn) a = m[n];
        else  a = log(n) + C + 1.0/(2 * n);//公式
        printf("Case %d: %.10lf\n", ++cas, a);
    }
    return 0;
}