LightOJ 1234 Harmonic Number(打表 + 技巧)

http://lightoj.com/volume_showproblem.php?problem=1234

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

题目大意：

求1 + 1/2 + 1/3 + 1/4 + 1/ 5 +...+ 1/ n(1 ≤ n ≤ 10⁸)

调和级数部分和，可以利用公式，（唉，然而我并不记得公式高数没学好-_-||）

如果直接循环的肯定会超时，那么我们开一个10^8/40 = 250万的数组用来分别存

1到1/40的和、1到1/80的和、1到1/120的和、1到1/160的和、... 、1到1/2500000的和

这样对于每一个n最多循环39次，节省了时间

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<stdlib.h>
#include<algorithm>

using namespace std;
const int N = 2500010;
const int M = 1e8 + 10;
typedef long long ll;

double a[N];

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