hud2156

Problem Description
We define the following matrix:
1/1 1/2 1/3
1/2 1/1 1/2
1/3 1/2 1/1
The elements on the diagonal of the matrix are always 1/1, the diagonal The denominators of the fractions on both sides increase one by one.
Request the sum of this matrix.

Input is
given an integer N (N<50000) for each row, which means that the matrix is ​​N*N. When N is 0, the input ends.

Output
outputs the answer with 2 decimal places.

#include <stdio.h>
int a[50001];
/*没什么好说的，
 就是找规律；
 列如：5*5的数组
 数字：     1 2 3 4 5
 出现次数： 5 8 6 4 2
 所以就很好计算了，每次乘以出现次数累加就行；`在这里插入代码片`
*/
int main()
{
    
    
    int n;
    while(scanf("%d",&n)!=EOF)
    {
    
    
        if(n==0)
            break;
        double sum=0;
        int i,k=n,m,num=2;//用数组存放出现次数；
        m=n*n;
        for(i=n;i>2;i--)
        {
    
    
            a[i]=num;
            k+=num;
            num+=2;
        }
        a[1]=n;
        a[2]=m-k;
        for(i=1;i<=n;i++)//累加
           sum+=a[i]*(1.0/i);
       printf("%.2lf\n",sum);
    }
    return 0;
}