汉诺塔问题的打印

[编程题]汉诺塔问题的打印

热度指数：857 时间限制：C/C++ 1秒，其他语言2秒空间限制：C/C++ 256M，其他语言512M
算法知识视频讲解

打印n层汉诺塔从最左边移动到最右边的全部过程

输入描述:

输入一个大于0的正整数

输出描述:

n层汉诺塔从最左边移动到最右边的全部过程

示例1

输入

输出

Move 1 from left to right
Move 2 from left to mid
Move 1 from right to mid
Move 3 from left to right
Move 1 from mid to left
Move 2 from mid to right
Move 1 from left to right

算法知识视频讲解

C/C++

祈嗻

#include <iostream>

using namespace std;

void Hanio(int n, string a, string b, string c){

if(n == 1)

cout << "Move 1 from " << a << " to " << c << endl;

else{

Hanio(n - 1, a, c, b);

cout << "Move " << n << " from " << a << " to " << c << endl;

Hanio(n - 1, b, a, c);

}

int main() {

int n;

string a = "left";

string b = "mid";

string c = "right";

cin >> n;

Hanio(n, a, b, c);

return 0;

}

编辑于 2024-01-06 10:26:25 回复(0)

C/C++

AlbatrossLy

#include <stdio.h>
#include <iostream>
using namespace std;

void move(int n, char* pos1, char* pos3)
{
    printf("Move %d from %s to %s\n", n, pos1, pos3);
}

void Hanoi(int n, char* pos1, char* pos2, char* pos3)
{
    //如果是1个盘子，直接从起始柱A移动到目标柱C
    if (n == 1) 
        move(n, pos1, pos3);
    else
    {
        //如果盘子大于1个，需要把n-1个盘子，从起始柱pos1，通过目标柱pos3，移动到中转柱pos2
        Hanoi(n-1, pos1, pos3, pos2); 
        //此时pos1上的n-1个盘子全部移动pos2上去了，那么可以直接把pos1上剩下的1个盘子，直接移动到pos3上
        move(n, pos1, pos3);
        //把pos2剩下的n-1个盘子，通过中转位置pos1，移动到目标位置pos3
        Hanoi(n-1, pos2, pos1, pos3);
    }
}

int main()
{
    int n;
    while(cin >> n)
    {
        char* pos1 = "left";
        char* pos2 = "mid";
        char* pos3 = "right";
        Hanoi(n, pos1, pos2, pos3);
    }
    return 0;
}

发表于 2022-01-15 09:31:30 回复(0)

牛客212725854号

#include <stdio.h>
#include <string.h>

void hanota(int n, char *A, char *B, char *C)
{
if(1 == n) printf("Move %d from %s to %s\n", n, A, C);
else
{
hanota(n-1, A,C,B);
printf("Move %d from %s to %s\n", n, A, C);
hanota(n-1, B,A,C);
}
}

int main()
{
int n;

scanf("%d", &n);
hanota(n,"left","mid","right");

return 0;
}

发表于 2021-02-09 13:03:22 回复(0)