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)
注意事项:
参考代码:
import java.util.Scanner; public class C1168 { 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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