hdu 2086 A1=?

hdu 2086 A1=?

思路（来自discuss）

纯粹数学题，找规律:
An = (1/2)An-1 + (1/2)An+1 - Cn
An-1 = (2/3)An-2 + (1/3)An+1 - (2/3)Cn - (4/3)Cn-1
An-2 = (3/4)An-3 + (1/4)An+1 - (1/2)Cn - Cn-1 - (3/2)Cn-2
An-3 = (4/5)An-4 + (1/5)An+1 - (2/5)Cn - (4/5)Cn-1 - (6/5)Cn-2 - (8/5)Cn-3
......
接着:
A1 = (n/(n+1))A0 + (1/(n+1))An+1 - (2/(n+1))Cn - (4/(n+1))Cn-1 - ... -(2n/(n+1))C1
      = [ nA0 + An+1 - 2(Cn + 2Cn-1 + 3Cn-2 + ... + nC1) ]/(n+1)***/

#include <iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int main()
{
    int n,i,k;
    double a0,an1,b[3100],sum,num;
    while(scanf("%d",&n)!=EOF)
    {
        sum=0.0;k=1;
        scanf("%lf%lf",&a0,&an1);
        for(i=1;i<=n;i++)
        {
            scanf("%lf",&b[i]);
        }
        for(i=n;i>=1;i--)
        {
            sum+=b[i]*k;
            k++;
        }
        num=n*a0+an1;
        sum=(num-2*sum)/(n+1);
        printf("%.2lf\n",sum);
    }
    return 0;
}