题目 1168: 简单计算

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题目描述:

有一个n+2个元素a[0], a[1], ..., a[n+1] (n <= 3000, -1000 <= a[i] <=1000)构成的数列.
已知对i=1, 2, ..., n有a[i] = (a[i-1] + a[i+1])/2 - c[i].

给定a0, a[n+1], c[1], ... , c[n]. 写一个程序计算a[1].

解题思路:

2 * a[1] = a[0] + a[2] - 2 * c[1]

   2 * a[2] = a[1] + a[3] - 2 * c[2]

        .......

   2 * a[n-1] = a[n-2] + a[n] - 2 * c[n-1]

   2 * a[n] = a[n-1] + a[n+1] - 2 * c[n]

    

  左边所有项相加等于右边所有项相加, 消除后即可得到：a[1] + a[n] = a[0] + a[n+1] - 2*sum(c{1..n})

     a[n] = a[0] + a[n+1] - 2*sum(c{1..n}) - a[1]

同理： a[n-1] = a[0] + a[n] - 2*sum(c{1..n-1}) - a[1]

        .......

    a[2] = a[0] + a[3] - 2*sum(c{1..2}) - a[1]

    a[1] = a[0] + a[2] - 2*sum(c{1..1}) - a[1]

  左边所有项相加等于右边所有项相加, 消除后即可得到最后的解：

  a[1] = (n * a[0] + a[n+1] - 2*(sum(c{1..1}) + sum(c{1..2}) + .... + sum(c{1..n})))/(n+1)

代码:

package lanqiao;import java.util.*;public class Main {public static void main(String[] args) {Scanner sc = new Scanner(System.in);while(sc.hasNext()){int n = sc.nextInt();double start = sc.nextDouble(), end = sc.nextDouble();double[] c = new double[n+1];for(int i = 1; i <= n; i++)c[i] = sc.nextDouble();double sum = 0;for(int i = 1; i <= n; i++){for(int j = 1; j <= i; j++){sum += c[j];}}System.out.printf("%.2f\n", (n*start + end - 2*sum) / (n+1));}sc.close();}
}

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