主打一个枚举找规律...

站台	1	2	3	4	5	6	...
上车	a	b	a+b	a+2b	2a+3b	3a+5b
下车		b	b	a+b	a+2b	2a+3b
人数	a	a	2a	2a+b	3a+2b	4a+4b

上车这一行: 观察a的系数（假设是dx[i]）可知: dx[i] = dx[i-1] + dx[i-2]

b的系数（假设是dy[i]）可知: dy[i] = dy[i-1] + dy[i-2]

人数: （假设为p[i]）p[i] = (dx[i] + 1) * a + (dy[i] -1)* b

import java.util.Scanner;/*** @Author YuLing* @Date 2023-12-17 18:09* @Description: 车站* @Version 1.0*/
public class Main {public static void main(String[] args) {Scanner scanner = new Scanner(System.in);int a = scanner.nextInt();//已知int n = scanner.nextInt();//已知 则需要找到n-1int m = scanner.nextInt();//已知int x = scanner.nextInt();//destination: 从x站开出车上的人数int[] dx = new int[n + 1];//a的系数int[] dy = new int[n + 1];//b的系数dx[1] = 1;dx[2] = 0;for (int i = 3; i < n; i++) {dx[i] = dx[i - 1] + dx[i - 2];}dy[1] = 0;dy[2] = 1;for (int i = 3; i < n; i++) {dy[i] = dy[i - 1] + dy[i - 2];}//(dx[n-1]+1)*a+(dy[n-1]-1)*b=mint b = (m - (dx[n - 1] + 1) * a) / (dy[n - 1] - 1);
//        System.out.println("b = " + b);//从x站开出时车上的人数int px = (dx[x] + 1) * a + (dy[x] - 1) * b;System.out.println(px);}
}