居然根據遞迴寫出非遞迴形式，有本事把漢諾塔的非遞迴寫一個？抱歉，我是菜雞，不會！

遞迴形式：（很簡單啦）

 1 #pragma GCC optimize(3)
 2 #include<iostream>
 3 using namespace std;
 4 int akm(int m,int n){
 5     if(m==0) return n+1;
 6     if(m>0&&n==0) return akm(m-1,1);
 7     if(m>0&&n>0) return akm(m-1,akm(m,n-1));
 8  }
 
 9 int main(){
10     ios::sync_with_stdio(false); 
11     int m,n;
12     cin>>m>>n;
13     cout<<akm(m,n)<<endl; 
14     
15 }
16 
17 /*
18 Ack(2,1)=Ack(1,Ack(2,0))
19         =Ack(1,Ack(1,1))
20         =Ack(1,Ack(0,Ack(1,0)))
21         =Ack(1,Ack(0,Ack(0,1)))
22         =Ack(1,Ack(0,2))
 
23         =Ack(1,3)
24         =Ack(0,Ack(1,2))
25         =Ack(0,Ack(0,Ack(1,1)))
26         =Ack(0,Ack(0,Ack(0,Ack(1,0))))
27         =Ack(0,Ack(0,Ack(0,Ack(0,1))))
28         =Ack(0,Ack(0,Ack(0,2)))
29         =Ack(0,Ack(0,3))
30         =Ack(0,4)
31         =5
32 */

非遞迴阿克馬，純純牛馬

：

 1 #include<bits/stdc++.h>//
 
stl大法 
 2 using namespace std;
 3 int asker(int m, int n)
 4 {
 5     stack<int> s1;//STL 
 6     stack<int> s2;//STL 
 7     s1.push(m);//STL 放到棧頂 
 8     s2.push(n);//STL 放到棧頂 
 9     while (!s1.empty())//判斷是否棧空 
10     {
11         while (m != 0)
12         {
13             if (n == 0)
14             {
15                 m = m - 1;
16                 n = 1;
17                 s1.push(m);
18                 s2.push(n);
19             }
20             else
21             {
22                 n = n - 1;
23                 s1.push(m - 1);
24                 s2.push(-1);
25             }
26         }
27         n = n + 1;
28         while ((!s1.empty())&& (s2.top() != -1))//top獲得棧頂元素 
29         {
30             s1.pop();//刪除棧頂元素 
31             s2.pop();
32         }
33         if(!s1.empty())
34         {
35             m = s1.top();
36             s2.pop();
37             s2.push(n);
38         }
39     }
40     return n;
41 }
42 int main()
43 {
44     ios::sync_with_stdio(false);
45     int n,m; 
46     cin>>n>>m;
47     cout<<asker(m,n)<<endl;
48 }