因为：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)
=> 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)
这是一道数学题，反正我是做不出来的，虽然代码很简单，但是要推出公式还是不容易。
#include<stdio.h>
#include<iostream>
using namespace std;
const int Max=3500;
int main()
{
double a,b,c[Max];
int n;
while(~scanf("%d",&n))
{
scanf("%lf%lf",&a,&b);
for(int i=1;i<=n;i++)
scanf("%lf",&c[i]);
int m=n;
int k=1;
double ans=0;
while(m)
{
ans +=k*c[m];
m--;
k++;
}
ans=(n*a+b-2*ans)/(n+1);
printf("%.2f\n",ans);
}
return 0;
}