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
题意：
求f（n）=n/1+n/2+n/3+......+n/n，
分析：
调和奇数求和，暴力超时，打表超内存，但一般这种题都打表，我们没100个记录一次ans就可以了。对于每一个输入最多遍历100次
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<set>
#include<queue>
#include<stack>
#include<map>
#define PI acos(-1.0)
#define in freopen("in.txt", "r", stdin)
#define out freopen("out.txt", "w", stdout)
#define kuaidian ios::sync_with_stdio(0);
using namespace std;
double ans[1000010];
int init()
{
double sum=0;
ans[0]=0;
for(int i=1;i<=1e8;i++)
{
sum+=(1.0)/i;
if(i%100==0)
ans[i/100]=sum;
}
}
int main()
{
int t;
init();
scanf("%d",&t);
int cas=0;
while(t--)
{
int n;
scanf("%d",&n);
double ans1 = 0;
for(int i=n/100*100+1;i<=n;i++)
{
ans1+=(1.0)/i;
}
ans1+=ans[n/100];
printf("Case %d: %.10lf\n",++cas,ans1);
}
return 0;
}
