HDU-ACM2086

A1 = ?

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6895    Accepted Submission(s): 4282

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
   
   
   
   

import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner input=new Scanner(System.in);
        while (input.hasNext()){
            int n=input.nextInt();
            double a,b;
            a=input.nextDouble();
            b=input.nextDouble();
            double c[]=new double[n+1];
            for (int i=1;i<=n;i++){
                c[i]=input.nextDouble();
            }
            double a1;
            a1=1.0*n/(n+1)*a+1.0/(n+1)*b;
            double k=2.0*n;
            for (int i=1;i<=n;i++){
                a1-=k/(n+1)*c[i];
                k-=2;
            }
            System.out.printf("%.2f",a1);
            System.out.println();
        }
    }
}