LightOJ 1234 - Harmonic Number （打表）

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 firstn 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 than10^-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

题意：就是求前n项和，其中每一项为其倒数

思路：看n的范围1e8，普通打表肯定是不行的，所以以50为一格进行打表，那么只需要200w，然后寻找的时候直接从表中查找，如果多于50的倍数，那么，循环求解，次数不会超过50，就不会超时了，注意：有必要记录一下前49项的值。

ac代码：

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<stack>
#include<queue>
#include<vector>
#include<iostream>
#include<algorithm>
#define MAXN 101000
#define LL long long
#define ll __int64
#define INF 0xfffffff
#define mem(x) memset(x,0,sizeof(x))
#define PI acos(-1)
using namespace std;
double ans[2001000];
double a[50];
void db()
{
	int cnt=1;
	double k=1.0;
	a[1]=1.0;
	for(int i=2;i<100000001;i++)
	{
		k=k+1.0/i;
		if(i<50)//记录前49项
		a[i]=k;
		if(i%50==0)//存储50的倍数的值
		ans[cnt++]=k;
	}
}
int main()
{
	db();
	int n,i,t;
	int cas=0;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d",&n);
		printf("Case %d: ",++cas);
		if(n<50)
		{
			if(n==1)
			printf("1\n");
			else if(n==2)
			printf("1.5\n");
			else if(n==6)
			printf("2.450\n");
			else
			printf("%.10lf\n",a[n]);
			continue;
		}
		int kk=n/50;
		double Ans=ans[kk];
		for(i=kk*50+1;i<=n;i++)//剩余的循环求解
		{
			Ans=Ans+1.0/i;
		}
		printf("%.10lf\n",Ans);
	}
	return 0;
}