事實(shí)上，若有n個(gè)盤子，則移動(dòng)完畢所需之次數(shù)為2^n - 1，所以當(dāng)盤數(shù)為64時(shí)，則所需次數(shù)為：

2⁶⁴- 1 = 18446744073709551615

為5.05390248594782e+16年，也就是約5000世紀(jì)，如果對(duì)這數(shù)字沒什么概念，就假設(shè)每秒鐘搬一個(gè)盤子好了，也要約5850億年左右。 

演算法

Procedure HANOI(n, A, B, C) [
    IF(n == 1) [
        PRINT("Move sheet " n " from " A " to " C);
    ]
    ELSE [
        HANOI(n-1, A, C, B);
        PRINT("Move sheet " n " from " A " to " C);
        HANOI(n-1, B, A, C); 
    ]    
] 

實(shí)作

• C

#include <stdio.h>

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

int main() {
    int n;
    printf("請(qǐng)輸入盤數(shù)：");
    scanf("%d", &n);
    hanoi(n, 'A', 'B', 'C');
    return 0;
} 

• Java

import java.io.*;

public class Hanoi {
    public static void main(String args[]) throws IOException {
        int n;
        BufferedReader buf;
        buf = new BufferedReader(new InputStreamReader(System.in));

        System.out.print("請(qǐng)輸入盤數(shù)：");
        n = Integer.parseInt(buf.readLine());

        Hanoi hanoi = new Hanoi();
        hanoi.move(n, 'A', 'B', 'C');
    }

    public void move(int n, char a, char b, char c) {
        if(n == 1)
            System.out.println("盤 " + n + " 由 " + a + " 移至 " + c);
        else {
            move(n - 1, a, c, b);
            System.out.println("盤 " + n + " 由 " + a + " 移至 " + c);
            move(n - 1, b, a, c);
        }
    }
}