}
}
} i++;
if(!StackEmpty(s)){ }
else return FALSE;
if(!StackEmpty(s)) return FALSE; return OK;
Pop(s,c1); strcpy(a,c1);
void InitStack(Stack &s) { }
Status Push(Stack &s,ElemType e) { }
Status Pop(Stack &s,ElemType e) { }
Status StackEmpty(Stack s) { }
if(s、size==0) return TRUE; else return FALSE; LinkType p; if(s、top){ }
return OK;
strcpy(e,s、top->data); p=s、top; s、top=p->next; delete p; s、size--; LinkType p; p=new NodeType; if(!p) exit(OVERFLOW); p->next=s、top; s、top=p;
strcpy(p->data,e); s、size++; return OK; s、top=NULL; s、size=0;
Status IsOperator(char c) { }
3、24 试编写如下定义得递归函数得递归算法，并根据算法画出求g(5,2)时栈得变化过程。
解：
int g(int m,int n); int main() { }
int g(int m,int n) { }
假设主函数得返回地址为0，递归函数3条语句得地址分别为1、2、3。
3 3 3 3 3 0
0 1 2 3 4 5
64 32 16 8 4 2
if(m>0)
return(g(m-1,2*n)+n); else return 0; int m,n;
cout<<\请输入m与n得值：\cin>>m>>n;
if(n>=0) cout< char *p=\while(*p){ } return FALSE; if(*p==c) return TRUE; p++; 0m?0,n?0? g?m,n?????gm?1,2n?nm?0,n?0?3、25 试写出求递归函数F(n)得递归算法，并消除递归： 解： #include ??n?1F?n???n?Fn?2???n?0n?0 { } 3、26 求解平方根 int i; int a[N]; int n; cout<<\请输入n:\cin>>n; for(i=0;i cout< if(i<1) a[i]=1; else a[i]=i*a[i/2]; A得迭代函数定义如下： ?p?sqrt?A,p,e????1?A????sqrt?A,2??p?p??,e???????p2?A?ep?A?e2 其中，p就是A得近似平方根，e就是结果允许误差。试写出相应得递归算法，并消除递归。 解： #include double Sqrt(double A,double p,double e); int main() { } double Sqrt(double A,double p,double e) { } 3、27 已知Ackerman函数得定义如下： if((p*p-A)>-e && (p*p-A) return p; return Sqrt(A,(p+A/p)/2,e); else double A,p,e; cout<<\请输入A p e:\cin>>A>>p>>e; cout< n?1m?0??akm?m,n???akm?m?1,1?m?0,n?0 ?akm?m?1,akm?m,n?1??m?0,n?0? (1) 写出递归算法； (2) 写出非递归算法； (3) 根据非递归算法，画出求akm(2,1)时栈得变化过程。 解： unsigned int akm(unsigned int m,unsigned int n) { } 非递归算法： int akm1(int m,int n) { } 0,akm(2,1) 0 2 1 1,akm(2,0) 6 2 0 2,akm(1,1) 4 1 1 3,akm(1,0) 6 1 0 g=akm(2,0) akm=akm(m-1,1)=akm(1,1) g=akm(m,n-1)=akm(1,0) akm=akm(m-1,1)=akm(0,1) Stack s; InitStack(s); ElemType e,e1,d; e、mval=m; Push(s,e); do{ while(e、mval){ } if(StackLength(s)>1){ } e1、nval=e、nval; Pop(s,e); e、mval--; e、nval=e1、nval+1; while(e、nval){ } e、mval--; e、nval=1; e、nval--; Push(s,e); e、nval=n; unsigned int g; if(m==0) return n+1; if(n==0) return akm(m-1,1); else{ } g=akm(m,n-1); return akm(m-1,g); else }while(StackLength(s)!=1||e、mval!=0); return e、nval+1; 4,akm(0,1) 4 0 1 0,akm(2,1) 0 2 1 1,akm(2,0) 6 2 0 2,akm(1,1) 4 1 1 3,akm(1,0) 6 1 0 0,akm(2,1) 0 2 1 1,akm(2,0) 6 2 0 2,akm(1,1) 4 1 1 3,akm(0,2) 7 0 2 akm=n+1=2 退栈 g=akm(2,0) akm=akm(m-1,1)=akm(1,1) g=akm(m,n-1)=akm(1,0) akm=akm(m-1,1)=akm(0,1)=2 退栈 g=akm(2,0) akm=akm(m-1,1)=akm(1,1) g=akm(m,n-1)=akm(1,0)=2; akm=akm(m-1,g)=akm(0,2);鏍嘔鷓幗礦雏軛。 akm=n+1=3 退栈 0,akm(2,1) 0 2 1 1,akm(2,0) 6 2 0 2,akm(1,1) 4 1 1 顫莖镝。 0,akm(2,1) 0 2 1 1,akm(2,0) 6 2 0 0,akm(2,1) 0 2 1 1,akm(1,3) 6 1 3 2,akm(1,2) 6 1 2 3,akm(1,1) 6 1 1 4,akm(1,0) 6 1 0 5,akm(0,1) 4 0 1 0,akm(2,1) 0 2 1 1,akm(1,3) 6 1 3 2,akm(1,2) 6 1 2 3,akm(1,1) 6 1 1 4,akm(1,0) 6 1 0 0,akm(2,1) 0 2 1 1,akm(1,3) 6 1 3 2,akm(1,2) 6 1 2 3,akm(1,1) 6 1 1 4,akm(0,2) 7 0 2 0,akm(2,1) 0 2 1 1,akm(1,3) 6 1 3 2,akm(1,2) 6 1 2 3,akm(1,1) 6 1 1 0,akm(2,1) 0 2 1 1,akm(1,3) 6 1 3 2,akm(1,2) 6 1 2 3,akm(0,3) 7 0 3 0,akm(2,1) 0 2 1 1,akm(1,3) 6 1 3 2,akm(1,2) 6 1 2 0,akm(2,1) 0 2 1 1,akm(1,3) 6 1 3 g=akm(2,0) akm=akm(m-1,1)=akm(1,1) g=akm(m,n-1)=akm(1,0)=2; akm=akm(m-1,g)=akm(0,2)=3; 退栈肃郧况簡 g=akm(2,0) akm=akm(m-1,1)=akm(1,1)=3 退栈 g=akm(2,0)=3; akm=akm(1,3) g=akm(1,2); akm(m-1,g) g=akm(1,1); akm(m-1,g) g=akm(1,0); akm(m-1,g) akm=akm(0,1) akm(0,1)=2退栈 g=akm(2,0)=3; akm=akm(1,3) g=akm(1,2); akm(m-1,g) g=akm(1,1); akm(m-1,g) g=akm(1,0); akm(m-1,g) akm=akm(0,1)=2退栈 g=akm(2,0)=3; akm=akm(1,3) g=akm(1,2); akm(m-1,g) g=akm(1,1); akm(m-1,g) g=akm(1,0)=2; akm(m-1,g)=akm(0,2)鴉厦纤璦緋羅呕。 akm=n+1=3 退栈 g=akm(2,0)=3; akm=akm(1,3) g=akm(1,2); akm(m-1,g) g=akm(1,1); akm(m-1,g) g=akm(1,0)=2; akm(m-1,g)=akm(0,2)=3退栈冑堅壮攛览個毂。 g=akm(2,0)=3; akm=akm(1,3) g=akm(1,2); akm(m-1,g) g=akm(1,1)=3; akm(m-1,g)=akm(0,3)讕陨鎮鹫锞蒞轎。 akm=n+1=4 退栈 g=akm(2,0)=3; akm=akm(1,3) g=akm(1,2); akm(m-1,g) g=akm(1,1)=3; akm(m-1,g)=akm(0,3)=4 退栈織赇饺漣难娄锷。 g=akm(2,0)=3; akm=akm(1,3) g=akm(1,2)=4; akm(m-1,g)=akm(0,4)龅颠鈹语骧紼秽。