问题形容
给定三根柱子A，B，C，柱子A上依照大小程序放着n个大小不同的盘子，最上面的最大，最下面的最小，输出盘子的个数，当初要把柱子A上的盘子全副挪动到C上。
问：
① 起码要挪动多少次，输入每一步挪动的形式。
② 输出开始计数的步骤和完结技术的步骤，统计在此期间每一个步骤呈现的次数。
代码
package Ring1270.pra.java01;import java.sql.SQLOutput;import java.util.Scanner;/** * https://www.lanqiao.cn/courses/2786/learning/?id=67029 */public class F_TowerOfHanoi { //Constants static int STEP = 0; static int AB = 0; static int AC = 0; static int BC = 0; static int BA = 0; static int CA = 0; static int CB = 0; //Method of count public static void count(String s) { if ("A-->B".equals(s)) AB++; if ("A-->C".equals(s)) AC++; if ("B-->A".equals(s)) BA++; if ("B-->C".equals(s)) BC++; if ("C-->A".equals(s)) CA++; if ("C-->B".equals(s)) CB++; } //Method of output public static void prin() { System.out.println("A-->B : " + AB); System.out.println("A-->C : " + AC); System.out.println("B-->A : " + BA); System.out.println("B-->C : " + BC); System.out.println("C-->A : " + CA); System.out.println("C-->B : " + CB); } //Method of tower's moving public static void move(int n, String from, String pass, String to, int x, int y) { if (n == 1) { System.out.println("Step" + ++STEP + ": " + "第1个盘子从" + from + "挪动到" + to + "，记为：" + from + "-->" + to); if (STEP >= x) { String s = from + "-->" + to; count(s); } } else { //当有两个盘子的时候,这里相当于把第一个挪动到B柱子上 move(n - 1, from, to, pass, x, y); System.out.println("Step" + ++STEP + ": " + "将" + n + "个盘子从" + from + "挪动到" + to + "，记为：" + from + "-->" + to); if (STEP >= x) { String s = from + "-->" + to; count(s); } //而后把第二个盘子挪动，相当于把两头挪动到C下面去 move(n - 1, pass, from, to, x, y); } } //Method of main public static void main(String[] args) throws Exception { Scanner scanner = new Scanner(System.in); System.out.println("Input the number of work's time:"); int n = scanner.nextInt(); System.out.println("Input the step of start:"); int x = scanner.nextInt(); System.out.println("Input the step of end:"); int y = scanner.nextInt(); if (x <= 0 || x > Math.pow(2, n) - 1 || x >= y) { throw new Exception("The start number is error , Runing is over......"); } if (y <= 0 || y > Math.pow(2, n) - 1 || y <= x) { throw new Exception("The end number is error, Runing is over......"); } int p = 0; move(n, "A", "B", "C", x, y); prin(); }}