akm(1, 3) akm=5 akm(1 akm =3 =akm(1,2) akmn(0,4)=5 y=2 akm(1,0) 用到一个栈记忆每次递归调用时的实参值,每个结点两个域{m,m}。对以上实例,栈 的变化如下: V vn 1n vn 1 1 vN 改akm(m-1,1) 改akm(m-1,1) 改akn(m-1)改abmn(m-1,y) 0 改akm(m-1,1)v=n+1=2改aAm(m-1,)改akm(m-1,)改aAm(m-1,)栈空,返回y=5 v=n+1=3v=n+1=4y=n+1=5 相应算法如下 #include <iostream. h> #define maxie 3500 unsigned akm( unsigned m, unsigned n)i stack<node> st( maxSize ) node w; unsigned v; while( stGetTop(). vm>0) 川计算akn(m-1,antm,m1)) while( st GetTop().I7>0) 川计算akm(m,m1),直到akm(m,0) w=st. GerTop(: st Pop(): 川计算akm(m-1,1) 1; St Push(w);3 用到一个栈记忆每次递归调用时的实参值，每个结点两个域{vm, vn}。对以上实例，栈 的变化如下： 相应算法如下 #include <iostream.h> #include “stack.h” #define maxSize 3500; unsigned akm ( unsigned m, unsigned n ) { struct node { unsigned vm, vn; } stack<node> st ( maxSize ); node w; unsigned v; w.vm = m; w.vn = n; st.Push (w); do { while ( st.GetTop( ).vm > 0 ) { //计算 akm(m-1, akm(m, n-1) ) while ( st.GetTop( ).vn > 0 ) //计算 akm(m, n-1), 直到 akm(m, 0) { w.vn--; st.Push( w ); } w = st.GetTop( ); st.Pop( ); //计算 akm(m-1, 1) w.vm--; w.vn = 1; st.Push( w ); akm(2, 1) v =akm(2, 0) akm(1, 1) v =akm(1, 0) akm(0, 1) = 2 akm(0, 2) = 3 akm(1, 3) v = akm(1, 2) v = akm(1, 1) v = akm(1, 0) akm(0, 1) = 2 akm(0, 2) = 3 akm(0, 3) = 4 akm(0, 4) = 5 v = 2 v = 2 v = 3 v = 4 v = 3 akm = 5 akm = 5 akm = 3 2 1 2 1 2 1 2 1 2 1 1 3 2 0 1 1 1 1 1 1 0 2 1 0 0 1 改 akm(m-1,1) 改 akm(m-1,1) v = n+1= 2 改 akm(m-1,v) 改 akm(m-1,v) v = n+1 = 3 vm vn vm vn vm vn vm vn vm vn vm vn vm vn vm vn vm vn vm vn vm vn vm vn 1 3 1 3 1 3 1 3 0 4 1 2 1 2 1 2 0 3 1 1 1 1 0 2 1 0 0 1 改 akm(m-1,1) v = n+1 = 2 改 akm(m-1,v) 改 akm(m-1,v) 改 akm(m-1,v) 栈空, 返回 v = 5 v = n+1 = 3 v = n+1 = 4 v = n+1 = 5