Hanoi塔问题

采用递归分治的策略来解决汉诺塔问题

#include <iostream>
using namespace std;
void Hanoi(int n,char a,char b,char c);
void move(int n,char a,char b);
int cnt=0;
int main()
{
    cout<<"plase enter n:"<<endl;
    int n;
    cin>>n;
    cout << "Hanoi "<<n<<endl;
    Hanoi(n,'A','B','C');
    cout<<"total move times: "<<cnt<<endl;
    return 0;
}

void Hanoi(int n,char a,char b,char c){
    if(n==1){
        move(n,a,c);cnt++;
    }
    else{
        Hanoi(n-1,a,c,b);
        move(n,a,c);cnt++;
        Hanoi(n-1,b,a,c);
    }
}
void move(int n,char a, char b){
    cout<<"move number "<<n<<" block from "<<a<<" to "<<b<<endl;
}