Light OJ 1234 Harmonic Number

1234 - Harmonic Number

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Time Limit: 3 second(s)

Memory Limit: 32 MB

In mathematics, the n^th harmonic number isthe 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^thharmonic 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

 

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Problem Setter: Jane Alam Jan

题意：计算1+1/2+1/3+……+1/n ;

思路：分段打表，将每1000的和存一下，需要计算时找出对应的1000的权值，再计算后面的部分相加即可

下面附上我的代码：

#include <cstdio>
#include <cmath>
double a[100005];

void init(){
	int tm = 1;
	a[0] = 0;
	for (int i = 1; i <= 100002; ++i){
		a[i] = a[i-1];//a[i]计算的从(i-1)*1000+1到i*1000的和 
		for (int j = tm; j <= tm+999; ++j){
			a[i] += 1.0/j;
		}
		tm = tm+1000;
	}
}
int main(){
	int t;
	scanf("%d",&t);
	init();
	int Case = 1;
	while (t--){
		int n;
		scanf("%d",&n);
		int r = n/1000*1000;//找出他的对应1000的位置 
		for (int i = r+1; i <= n; ++i){//计算一下剩余的值的和 
			tmp += 1.0/i;
		}
		double tt;
		 tt = a[n/1000];
		//printf("%lf %lf\n",tt,tmp);
		double ans = tt+tmp;
		printf("Case %d: %.9lf\n",Case++,ans);
	}
	return 0;
}