hanio(T).rar_hanio_汉诺塔_汉诺塔 VC++_汉诺塔-递归算法

#include<stdio.h> #include<iostream.h> class hanio{ public: int n; //n代表盘子的个数 long count; //count返回move函数调用的次数，即反映出算法时间复杂度问题的参量 char from,to,use; //from,to,use是源桩，目标桩，借用桩 void move(int n,char from, char to); //将第n个盘子从from桩移动到to桩上 void change(int n,char from,char use,char to); void solve(); void set_n(int); int set_n(); long getCount(); hanio(int n, char from ,char use, char to); //带盘子个数n和源桩、目标桩、借用桩的构造函数 hanio(int n); //只带盘子个数的构造函数 }; int hanio::set_n() { cout<<endl<<"enter a number:"; cin>>n; cout<<endl; return n; } long hanio::getCount() { count =0; return count; } /*void hanio::set_n(int num) { n=num; } */ hanio::hanio(int n, char a, char b, char c) { //n=num; from=a; use=b; to=c; } void hanio::move(int n, char from, char to) { count++; cout<<"the "<<count <<"time to call 'move' function"<<endl; //cout<<"::::: move "<<n<<"from "<<from<<"to "<<to<<endl; } void hanio::change(int n, char from, char use,char to) { if(n>0) { change(n-1,from,to,use); move(n,from,to); change(n-1,use,from,to); } } void hanio::solve() { change(n,from,use,to); } void main(void) { hanio ob(5,'A','B','C'); ob.solve(); ob.set_n(); ob.solve(); return ; }