Harmonic Number LightOJ - 1234 (暴力打表，区域保存)

In mathematics, the nth harmonic number is the sum of the reciprocals of the first n natural numbers。

Hn = 1 + 1 / 2 +１/3 +…… + 1/n

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

暴力打表，这道题学会了按区域来保存的技巧，（把结果分成很多段，保存每段的第一个结果，搜索时找到对应区间再暴力该区间就可以）

#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<queue>
#include<algorithm>

#define ll long long
#define inf 0x3f3f3f3f
#define maxn 1e8+7

using namespace std;

double a[2000100];

void Init()
{
a[0] = 0.0;
a[1] = 1.0;
double ans = 1;
for( int i = 2; i < maxn; i ++)
{
ans += 1.0 / i;
if( i % 50 == 0)
a[i/50] = ans;
}
}

int main()
{
int t,n;
Init();
//  for(int i = 0;i <= 10; i++)
//      printf("%lf\n",a[i]);
scanf("%d",&t);
for(int i = 1; i <= t;i ++)
{
scanf("%d",&n);
int tmp = n / 50;
double ans = a[tmp];
for(int j = tmp * 50 + 1; j <= n; j ++)
{
ans += 1.0 / j;
}
printf("Case %d: %.9lf\n",i,ans);
}
return 0;
}