A1 = ? HDU 杭电2086 【数学】

Problem Description

有如下方程：A _i = (A _i-1 + A _i+1)/2 - C _i (i = 1, 2, 3, .... n).
若给出A ₀, A _n+1, 和 C ₁, C ₂, .....C _n.
请编程计算A ₁ = ?

Input

输入包括多个测试实例。
对于每个实例，首先是一个正整数n,(n <= 3000); 然后是2个数a ₀, a _n+1.接下来的n行每行有一个数c _i(i = 1, ....n);输入以文件结束符结束。

Output

对于每个测试实例，用一行输出所求得的a1(保留2位小数).

Sample Input

   
   
   
   

    
    
    
    1
50.00
25.00
10.00
2
50.00
25.00
10.00
20.00
   
   
   
   

Sample Output

   
   
   
   

    
    
    
    27.50
15.00
   
   
   
   

/* 
 
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) 
      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) 
*/  

#include <stdio.h>
int main()
{
	int n;
	double a0,ann,c;
	while(~scanf("%d",&n))
	{
		scanf("%lf%lf",&a0,&ann);
		double sum=0;
		for(int i=1;i<=n;++i)
		{
			scanf("%lf",&c);
			sum+=c*(n-i+1);
		}
		double res=(ann+n*a0-2*sum)/(n+1);
		printf("%.2lf\n",res);
	}
	return 0;
}