问题:
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:
就是给你一个n,然后让你输出上面式子所得到的结果。
思路:
上面这个式子的结果时存在一个公式的,调和级数f(n) 约等于1/(2*n)+ln(n)+C,C是欧拉常数约等于0.57721566490153286060651209(发现欧拉这个人还真是无所不能),这个公式就是求上面式子的,但这是一个近似的结果,所以我们要对前好多项去直接用上面的式子进行打表,然后在对之后的项套公式(当n很大时,我们就认为f(n)的误差可以忽略)。
代码:
#define N 200001
#include<math.h>
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
double k[N],ol=0.57721566490153286060651209;
int n;
int main()
{
k[1]=1;
for(int i=2; i<N; i++)
k[i]=k[i-1]+1.0/i;
int Case=1,T;
scanf("%d",&T);
while(T--)
{
scanf("%d",&n);
if(n<N)
printf("Case %d: %.10lf\n",Case++,k[n]);
else
printf("Case %d: %.10lf\n",Case++,log(n)+1.0/(2*n)+ol);//C语言种log(n)就是ln(n)
}
return 0;
}