HDU 2086 A1 = ?

题意：有如下方程：Ai = (Ai-1 + Ai+1)/2 - Ci (i = 1, 2, 3, .... n).
若给出A0, An+1, 和 C1, C2, .....Cn.
请编程计算A1 = ?

解题思路

1.分析输入，只知道A0、An+1以及C1...Cn，如果想要迭代丘A1，肯定是无法求到的
2.必须通过数学公式推导

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)
3.由推导可知， A1 = [nA0 + An+1 - 2(nC1+(n-1)C2+...+2Cn-1+Cn)]/(n+1)

解题遇到的问题

无

后续需要总结学习的知识点

无

##解法
import java.text.DecimalFormat;
import java.util.Scanner;

public class Main {

    public static void main(String[] args) {
        Scanner mScanner = new Scanner(System.in);
        while (mScanner.hasNext()) {
            int n = mScanner.nextInt();
            // A1 = [nA0 + An+1 - 2(nC1+(n-1)C2+...+2Cn-1+Cn)]/(n+1)
            double a1 = n * mScanner.nextDouble() + mScanner.nextDouble();
            for (int i = n; i >= 1; i--) {
                a1 -= 2 * i * mScanner.nextDouble();
            }
            DecimalFormat format = new DecimalFormat("#0.00");
            System.out.println(format.format(a1 / (n + 1)));

        }
        mScanner.close();
    }
}