HDU 2086 A1 = ?

A1 = ?

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 6492    Accepted Submission(s): 4038

 

 

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A₀, A_n+1, 和 C₁, C₂, .....C_n.

请编程计算A₁ = ?

 

 

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a₀, 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

 

 

Source

2006/1/15 ACM程序设计期末考试

 

 

/* 
    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;
}