light oj 1234 Harmonic Number

Harmonic Number

Time Limit:3000MS    Memory Limit:32768KB     64bit IO Format:%lld & %llu

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Description

In mathematics, the n^th harmonic number is the sum of the reciprocals of the firstn natural numbers:

 

Hn=1/1+1/2+1/3+1/4…1/n

 

In this problem, you are given n, you have to findH_n.

Input

Input starts with an integer T (≤ 10000), denoting the number of test cases.

Each case starts with a line containing an integern (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

题意：计算调和级数的前n项和(n<=1e8)。

这题看了好多人的写法都是用公式，看到一个比较好的解法记录一下，^ ^,博客地址为：http://www.cnblogs.com/shentr/p/5296462.html，数据范围有点大，时间限制是3s，如果直接打表的话时间是够的但是1e8的数组肯定MLE了，这里可以考虑每40组记录一个值，输入的数直接除以40就可以得到一个数，最多再进行39次运算就能得到结果了

#include<iostream>#include<algorithm>#include<cmath>#include<cstdio>#include<queue>#include<cstring>#include<string>#include<map>#include<set>using namespace std;typedef long long ll;#define for1(i,a) for(int (i)=1;(i)<=(a);(i)++)#define for0(i,a) for(int (i)=0;(i)<(a);(i)++)#define sf scanf#define pf printf#define mem(i,a) memset(i,a,sizeof i)#define pi acos(-1.0)#define eps 1e-10const int Max=1e8;const int M=Max/40;double s[M+1]={0.0,0.0};int m,n;int main(){    double d=1;    for(int i=2;i<=Max;i++)//打表求出1e8的所有值每40个分一个组，时间大概为1s多一点    {        d+=(1.0/i);        if(!(i%40))            s[i/40]=d;    }    sf("%d",&m);    for1(i,m)    {        sf("%d",&n);        int j=n/40;        double ans=s[j];//求出n所对应的数组，再继续算剩余的部分即可        for(int k=j*40+1;k<=n;k++)            ans+=1.0/k;        pf("Case %d: %.10f\n",i,ans);    }    return 0;}