Harmonic Number 调和级数的欧拉公式

最新推荐文章于 2025-02-28 07:32:05 发布

Big_Rui

最新推荐文章于 2025-02-28 07:32:05 发布

阅读量1.6k

点赞数

分类专栏：数论

本文链接：https://blog.youkuaiyun.com/Big_Rui/article/details/76668386

版权

数论专栏收录该内容

18 篇文章

订阅专栏

数学中，第n个调和数是前n个自然数倒数的和。本文将探讨如何求解Hn，即调和级数的n项和。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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

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：

给定n,求调和级数Hn；

注意：

输出五花八门，关键是保留十位小数，直接输出%.10f就行

n<=1e8,直接开数组打表会ML,这时候要用欧拉公式,但是n很大时公式才成立，所以一部分打表，另一部分直接算，数组开到1e4就行。

Hn=ln(n)+c+1.0/(2*n)

c=0.57721566490153286060651209;

#include<cstdio>
#include<iostream>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<map>;
using namespace std;
const double c=0.57721566490153286060651209;
double ans[10000];

void mul()
{
    memset(ans,0,sizeof(ans));
    for(int i=1;i<10000;i++)
     ans[i]=ans[i-1]+1.0/i;
}

int main()
{
    int t,n;
    mul();
    scanf("%d",&t);
    for(int i=1;i<=t;i++)
    {
        scanf("%d",&n);
        if(n<10000)
          printf("Case %d: %.10f\n",i,ans[n]);
        else
          printf("Case %d: %.10f\n",i,log(n)+c+1.0/(2*n));
    }
   return 0;
}