lightoj 1234 - Harmonic Number（技巧打表）

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

这题让求1/1 + 1/2 + 1/3 + .... + 1/n 的和，然而n有1e8这么大，所以直接打表是会MLE的，所以就用到了一点技巧，每隔50个存一个和，这样200W个数还是可以存的下的。两个之间的数直接加上就行了。

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

using namespace std;

const int N = 1e8;
double sum[N/50+10];
void init()
{
    sum[0] = 0;
    double t = 0;
    for(int i=1;i<=N;i++)
    {
        t += 1.0/i;
        if(i % 50 == 0)
            sum[i/50] = t;
    }
}
int main(void)
{
    int T,i,n;
    init();
    scanf("%d",&T);
    int cas = 1;
    while(T--)
    {
        scanf("%d",&n);
        int t = n/50;
        double ans = sum[t];
        for(i=t*50+1;i<=n;i++)
            ans += 1.0/i;
        printf("Case %d: %.8f\n",cas++,ans);
    }

    return 0;
}