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

想事先打表、然后输出的，发现空间爆了，坑

思路：每50个单位长度的区间统计一下和，对每次输入的n/50 ，多出的部分就直接一个一个算

#include<cstdio>
#include<cmath>
double num[100000009/50];
int main()
{
num[0]=0;
int ans=1;
for(int i=1; i<=(1e8+50) / 50;++i){
double a=0;
for(int j=(i-1)*50+1;j<=i*50;j++)		//每单位长度为50的区间统计一次总和
a+=(1.0/j);
num[i]=num[i-1]+a;
}
int n,t=0;
scanf("%d",&n);
while(n--){
int a;
scanf("%d",&a);
int b=a/50;
double fin=num[b];
for(int i=b*50+1;i<=a;++i)	//多出来的的再来单个单个算、
fin+=1.0/i;
printf("Case %d: %.10lf\n",++t,fin);
}
}