/**
阿克曼(Ackmann)函数

【题目描述】
阿克曼(Ackmann)函数A(m，n)中，m，n定义域是非负整数(m<=3,n<=10)，函数值定义为：
akm(m,n) = n+1;         (m=0时)
akm(m,n) = akm(m-1,1);  (m>0,n=0时)
akm(m,n) = akm(m-1,akm(m, n-1)); （m,n>0时)

【输入】输入m和n。

【输出】函数值。

【输入样例】2 3
【输出样例】9
*/

import java.util.Stack;
 
public class Main2 {
    public static void main( String[] args ) {
        System.out.println(AkmRecur(2, 3));
        System.out.println(AkmNonRecur(2, 3));
    }
    //递归
    public static int AkmRecur(int m, int n)    {
        if (m == 0)
            return n + 1;
        else if (n == 0)
            return AkmRecur(m - 1, 1);
        else
            return AkmRecur(m - 1, AkmRecur(m, n - 1));
    }
    //非递归
    public static int AkmNonRecur(int m, int n)    {
        Stack<Integer> sx = new Stack<Integer>();
        Stack<Integer> sy = new Stack<Integer>();
        int x = 0;
        int y = 0;
        sx.push(m);
        sy.push(n);
        while ((!sx.empty()) && (!sy.empty())) {
             if (sx.peek() != 0 && sy.peek() == 0) {
                     x = sx.peek();
                     y = sy.peek();
                     sx.pop();
                     sy.pop();
                     sx.push(x-1);
                     sy.push(1);
             }else if (sx.peek() != 0 && sy.peek() != 0) {
                     while (sx.peek()!= 0 &&sy.peek()!= 0) {
                         x = sx.peek();
                         y = sy.peek()-1;
                         sx.pop();
                         sy.pop();
                         sx.push(x-1);
                         sy.push(-1);
                         sx.push(x);
                         sy.push(y);
                     }
             } else {
                 y = sy.peek();
                 sx.pop();
                 if (sx.empty()){
                     return y + 1;
                 } else {
                     sy.pop();
                     sy.pop();
                     sy.push(y+1);
                 }
             }
        }
        return -1;
    }
}