Harmonic Number（打表法）

Harmonic Number Time Limit:3000MS     Memory Limit:32768KB     64bit IO Format:%lld & %llu

Submit Status Practice LightOJ 1234

Description

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

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

这题是单调级数部分求和，网上有公式，不过不知道公式也是没关系的，毕竟这个知识点也偏门了点。。。

我用的方法是打表记录1/i (1<=i<=n)，根据题意，n最大为一亿，将一亿个结果记录下来肯定是不可行的，但是可以记录百万级个结果。下面的代码中，我开了一个250万的数组，0到一亿范围内，每40个数记录一个结果，即是分别记录1/40,1/80,1/120,...,1/一亿，这样对于输入的每个n，最多只需执行39次求倒数运算，大大节省了时间。

注意的是，a[0] = 0,只是为了使得当n==1时不用单独判断。

AC Code：

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <cmath>

using namespace std;

const int maxn = 2500001;
double a[maxn] = {0.0, 1.0};

int main()
{
    int t, n, ca = 1;
    double s = 1.0;
    for(int i = 2; i < 100000001; i++)
    {
        s += (1.0 / i);
        if(i % 40 == 0) a[i/40] = s;
    }
    scanf("%d", &t);
    while(t--)
    {
        scanf("%d", &n);
        int x = n / 40;
       // int y = n % 40;
        s = a[x];
        for(int i = 40 * x + 1; i <= n; i++) s += (1.0 / i);
        printf("Case %d: %.10lf\n", ca++, s);
    }
    return 0;
}