2022 年 11 月 2 日

C, C++, Python, ZJ解題

a227: 河內塔

2022.11.02 PM 11:00 by CBJ

來源 : https://zerojudge.tw/ShowProblem?problemid=a227
出題者 : 2011三龍杯(成附建杯)
標籤 : 遞迴
難易度 : 3

解題想法 : 實作一個函式，功能是把from移到to的位置，以此邏輯下去實作(還需要計算環的編號)

實作技巧 : 寫遞迴(自訂函式)時，先想像它已經是具有完整功能的函式，放心地去用它就對了。

//C language
//solution link(含註解): https://github.com/CBJ0519/CBJsProgramDiary.com/blob/main/zj/a227.c

#include<stdio.h>
void hanoi(char from, char to, char another, int amount, int num){
    if(amount==1){ 
        printf("Move ring %d from %c to %c\n",num,from,to);
        return;
    }
    hanoi(from, another, to, amount-1, 1);
    hanoi(from, to, another, 1, amount);
    hanoi(another, to, from, amount-1, 1);
}
int main(){
    int n;
    while(scanf("%d",&n)!=EOF){ 
        hanoi('A','C','B',n,1);
    }
    return 0;
}

//C++ language
//solution link(含註解): https://github.com/CBJ0519/CBJsProgramDiary.com/blob/main/zj/a227.cpp

#include<iostream>
using namespace std;
void hanoi(char from, char to, char another, int amount, int num){
    if(amount==1){  
        cout<<"Move ring "<<num<<" from "<<from<<" to "<<to<<"\n";
        return;
    }
    hanoi(from, another, to, amount-1, 1);
    hanoi(from, to, another, 1, amount);
    hanoi(another, to, from, amount-1, 1);
}
int main(){
    int n;
    while(cin>>n){ 
        hanoi('A','C','B',n,1);  
    }
    return 0;
}

## Python language
## solution link(含註解): https://github.com/CBJ0519/CBJsProgramDiary.com/blob/main/zj/a227.py

from sys import stdin
def hanoi(From, To, Another, Amount, Num):
    if Amount==1:
        print(f"Move ring {Num} from {From} to {To}")
        return
    hanoi(From, Another, To, Amount-1, 1)
    hanoi(From, To, Another, 1, Amount)
    hanoi(Another, To, From, Amount-1, 1)
for read in stdin:
    n=int(read) 
    hanoi('A','C','B',n,1);

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