LightOJ Harmonic Number 1234【技巧】

1234 - Harmonic Number

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Time Limit: 3 second(s)

Memory Limit: 32 MB

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

 

解题思路：

该题可以利用公式。不过我不知道。。。。

普通的直接求会超时的。在网上看见其他人是每隔几位记录了值这样就快很多。

此题完全忽视结果，记得保留10位小数即可。

AC代码：

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

using namespace std;

double a[2500010];
void init()
{
	double s=0;
	for(int i=1;i<100000010;i++){
		s+=(1.0/i);
		if(i%40==0) a[i/40]=s;
	}
}

int main()
{
	init();
	int t;
	scanf("%d",&t);
	int xp=1;
	while(t--){
		double n;
		scanf("%lf",&n);
		double ans=1.0;
		int x=n/40;
		ans=a[x];
		for(int i=40*x+1;i<=n;i++){
			ans+=(1.0/i);
		}
		printf("Case %d: %.10lf\n",xp++,ans);
	}
}