HDU 2086 A1 = ?

/*
Ai=(Ai-1+Ai+1)/2 - Ci,
A1=(A0 + A2)/2 - C1;
A2=(A1 + A3)/2 - C2 , ...
=>    A1+A2 = (A0+A2+A1+A3)/2 - (C1+C2)
2[(A1+A2)+(C1+C2)] = A0+A2+A1+A3;
A1+A2 = A0+A3 - 2(C1+C2);
=>    A1+A2 =  A0+A3 - 2(C1+C2)
同理可得：
A1+A1 =  A0+A2 - 2(C1)
A1+A2 =  A0+A3 - 2(C1+C2)
A1+A3 =  A0+A4 - 2(C1+C2+C3)
A1+A4 =  A0+A5 - 2(C1+C2+C3+C4)
...
A1+An = A0+An+1 - 2(C1+C2+...+Cn)
----------------------------------------------------- 左右求和
(n+1)A1+(A2+A3+...+An) = nA0 +(A2+A3+...+An) + An+1 - 2(nC1+(n-1)C2+...+2Cn-1+Cn)

=>   (n+1)A1 = nA0 + An+1 - 2(nC1+(n-1)C2+...+2Cn-1+Cn)

=>   A1 = [nA0 + An+1 - 2(nC1+(n-1)C2+...+2Cn-1+Cn)]/(n+1)
*/
//！！！！用cin和cout会超时
#include <iostream>
#include <cstdio>
using namespace std;
double a[3010],c[3010];
int main()
{
int n;
while(~scanf("%d",&n)){
double sum=0;
scanf("%lf %lf",&a[0],&a[n+1]);
for(int i=n;i>=1;i--){
scanf("%lf",&c[i]);
sum+=i*c[i];
}
a[1]=(a[0]*n+a[n+1]-2*sum)/(n+1);
printf("%.2f\n",a[1]);
}
return 0;
}