LightOJ 1234 Harmonic Number

D - Harmonic Number
Time Limit:3000MS Memory Limit:32768KB 64bit IO Format:%lld & %llu
Submit Status Practice LightOJ 1234

Description

In mathematics, the nth 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 Hn.

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 nth 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

思路：就是求Hn那个公式的和 直接暴力会超内存 只存n/50个数就可以了 需要哪个数再算就好了

#include <iostream>
#include <math.h>
#include <stdio.h>
const int N=1e8+10;
using namespace std;
double a[N/50+10];
int main()
{
int i,n,t,k=1;
double sum=1.0;
a[0]=0.0;
a[1]=1.0;
for(i=2;i<=N;i++)
{
sum+=1.0/double(i);
if(i%50==0)
a[i/50]=sum;
}
cin>>t;
while(t--)
{
cin>>n;
int b=n/50;
double ans=a[b];
for(i=b*50+1;i<=n;i++)
ans+=1.0/double(i);
printf("Case %d: %.10lf\n",k++,ans);
}

return 0;
}