漢諾塔的遞歸實現方法

//n-num  a-from  b-temp  c-to void hano(int n, char a, char b, char c){	if (n == 1)		cout << a << "=>" << c << endl;	else	{		hano(n - 1, a, c, b);		hano(1, a, b, c);		hano(n - 1, b, a, c);	}}int main(){	hano(2, 'A', 'B', 'C');	cout << "漢諾塔移動完成！" << endl;	cout << endl;		return 0;}