| Time Limit: 3000MS | Memory Limit: 32768KB | 64bit IO Format: %lld & %llu |
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
Source
Problem Setter: Jane Alam Jan
/**
题意：如题
做法：因为n为1e8比较大，然后可以进行打表，每100个存储，然后在进行剩余的
**/
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <cmath>
#include <algorithm>
#include <queue>
const int maxn = 1e8 + ;
using namespace std;
double mmap[maxn/+];
void init()
{
mmap[] = ;
mmap[] = ;
double sum = 1.0;
for(int i=;i<=maxn;i++)
{
sum += 1.0 / (i*1.0);
if(i % == ) mmap[i/] = sum;
}
}
int main()
{
int Case = ;
int T;
scanf("%d",&T);
init();
while(T--)
{
long long n;
scanf("%lld",&n);
int res = n/;
double tot = mmap[res];
for(int i=res*+;i<=n;i++)
{
tot += (1.0)/(i*1.0);
}
printf("Case %d: %.10f\n",Case++,tot);
}
return ;
}