LightOJ - 1234 I - Harmonic Number-优快云博客

本文链接：https://blog.youkuaiyun.com/Vace___yun/article/details/82319588

本文介绍了一种高效计算第N个调和数的方法，通过预计算并存储每一百个自然数的倒数之和来减少计算量，避免了在大数据集上直接求和可能导致的时间超时问题。

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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：根据公式，求前n项的和

分析:因为n的范围太大，求和肯定会超时，所以可以把每一百个数存一下，进行打表，求前n项的和最多遍历100个数。

代码：
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
double sum[1000000+10];
int main()
{
    double ans=0;
    sum[0]=0;
    for(int i=1;i<=1e8;i++)  //打表
    {
        ans+=1.0/i;
        if(i%100==0)
            sum[i/100]=ans;
    }
    int t,k=1,n;
    scanf("%d",&t);
    while(t--)
    {
        double cout;
        scanf("%d",&n);
        cout=sum[n/100];
        for(int i=(n/100)*100+1;i<=n;i++) //不够一百个数的遍历
            cout+=1.0/i;
        printf("Case %d: %.10lf\n",k++,cout);
    }
}