哈诺塔非递归beta版本

哈诺塔非递归要很认真的去找他的循环数 循环条件等
如果你对汉诺塔找规律 你可以发现 每个盘子都是有规律的在动 当盘子总数与当前盘子的奇偶性一样的时候
盘子会沿着A-C-B-A的方式重复移动 如果奇偶不同则反过来 所以盘子移动的周期T为3
此外 你可以发现 最小的盘子 盘子①占据了所有的单数步骤 这里我们定义盘子①的步骤是2k-1
以此类推 盘子②是4k-2 盘子③是8k-4 盘子④是16k-8
盘子n是 (2^n)*k-(2(n-1))

#include<iostream>
#include <string>
#include<math.h>
using namespace std;
int main(){
	int n;
	cin >> n;
	int count = 0;				 //用来与下面变量max联动
	int max = pow(2, n) - 1;     //记录最多需要几步 2^n-1
	//cin >> n;
	//string arr[180] = {0};
	string arr[2000] = { "" };   //存入怎么移动的步骤如果不够可以改成动态数组 
	int ans[2000] = {};          //存入移动步骤的是第几个盘子
	for (int i = 1; ; i ++){    //本代码中i只出现这里一次 意思就是目前的盘子编号
			int nums = pow(2, i);  //代表当前算式的倍数
			int num = pow(2, i - 1);
			if (n % 2 == i % 2) {
				for (int k = 1; k < max/2+1; k=k+3) {
					if (count >= max)break;
					if (arr[nums * (k + 0) - (num)] == "" && (nums * (k + 0) - (num)) <= max)
					{
						arr[nums * (k + 0) - (num)] = ("A->C"); 
						ans[nums * (k + 0) - (num)] = i; 
						count++;
					}
					if (arr[nums * (k + 1) - (num)] == "" && (nums * (k + 1) - (num)) <= max)
					{
						arr[nums * (k + 1) - (num)] = ("C->B"); 
						ans[nums * (k + 1) - (num)] = i; 
						count++;
					}
					if (arr[nums * (k + 2) - (num)] == "" && (nums * (k + 2) - (num)) <= max)
					{
						arr[nums * (k + 2) - (num)] = ("B->A"); 
						ans[nums * (k + 2) - (num)] = i;
						count++;
					}
				}
			}else {
				for (int k = 1; k < max/2+1; k=k+3) {
					if (count >= max)break;
					if (arr[nums * (k + 0) - (num)] == "" && (nums * (k + 0) - (num)) <= max)
					{
						arr[nums * (k + 0) - (num)] = ("A->B"); 
							ans[nums * (k + 0) - (num)] = i; 
					count++;
					}
					if (arr[nums * (k + 1) - (num)] == "" && (nums * (k + 1) - (num)) <= max)
					{
						arr[nums * (k + 1) - (num)] = ("B->C"); 
						ans[nums * (k + 1) - (num)] = i; 
						count++;
					}
					if (arr[nums * (k + 2) - (num)] == "" && (nums * (k + 2) - (num)) <= max)
					{
						arr[nums * (k + 2) - (num)] = ("C->A"); 
						ans[nums * (k + 2) - (num)] = i;
						count++;}
				}
			}
			 if (count >= max)break;
	}
	for (int i = 1; i <= max; i++) {
		printf("第%d步是移动第",i);
		printf("%d个盘子 从", ans[i]);
		cout << arr[i] << endl;
	}
	return 0;
}

由于本人实力有限 实验的时候盘子数是12346都没问题 但是其他情况 会报错 还得改进