数值转换和回文

数值转换和回文 熟制转换 #include "stdio.h" #include "malloc.h" #include "stdlib.h" 100 //存储空间初始分配量 #define STACK_INIT_SIZE #define STACKINCREMENT 10 //存储空间分配增量 #define N 8 //转换成 N 进制 /***************************************** 友情提示:请用C++方式编译 *****************************************/ typedef struct { int *base ; //栈底 int *top ; //栈顶 int stacksize ; //栈容量 }SqStack ; int InitStack ( SqStack &S ) { //构造一个空栈 S.base = (int * ) malloc ( STACK_INIT_SIZE * sizeof( int ) ) ; if ( ! S.base ) exit(0) ; S.top = S.base ; S.stacksize = STACK_INIT_SIZE ; return 0 ; } int GetTop ( SqStack S , int &e ) { //若栈不为空，则用e返回S的栈顶元素，并返回0，否则返回1 if ( S.top == S.base ) return 1 ; e = * ( S.top - 1 ) ; return 0 ; } int Push ( SqStack &S , int e ) { //插入元素e为新的栈顶元素 if( S.top - S.base >= S.stacksize ) //栈满，追加存储空间 { S.base = ( int * ) realloc ( S.base , (S.stacksize + STACKINCREMENT ) * sizeof ( int ) ) ; if (!S.base) return 1 ; //存储分配失败 S.top = S.base + S.stacksize ; S.stacksize += STACKINCREMENT ; } *S.top++ = e ; return 0 ; } int Pop ( SqStack &S , int &e ) { //若栈不空，则删除S的栈顶元素，用e返回其值，并返回0，否则返回1 if ( S.top == S.base ) return 1 ; e = * --S.top ; return 0 ; } int StackEmpty (SqStack S) { //若栈空，则返回1，否则返回0 if ( S.top == S.base ) return 1 ; return 0 ; } int DestroyStack (SqStack &S ) { //销毁栈S，栈S不再存在 free( S.base ) ; S.base = S.top = NULL ; //防止程序后面不小心使用了它 return 0 ; } int main() { int n , e ; //n为要转换的十进制数，e 为临时用的的int型变量 SqStack S ; InitStack( S ) ; //构造一个空栈 scanf("%d",&n) ; while(n) { Push( S , n % 8 ) ; //压入栈 n /= N ; //转换成 N 进制 } while( ! StackEmpty( S ) ) { Pop( S , e ) ; //弹出栈 printf("%d", e ) ; } printf("\n") ; DestroyStack( S ) ; //销毁栈S，栈S不再存在 return 0 ; } #include using namespace std; /*//******************************************************************* 链栈 **********************************************************************/ template class LinkStack { public: struct Node { LSELEM data; Node* next; }; LinkStack (); //构造函数 ~LinkStack(); //析构函数 void SetEmpty(); //置空堆栈 bool IsEmpty(); //判断堆栈是否为空 bool Push(LSELEM); //入栈 bool Pop(LSELEM&); //出栈 bool GetTop(LSELEM&); //获得栈顶元素 private: Node* top; }; template LinkStack ::LinkStack() { top=NULL; } template LinkStack ::~LinkStack() { SetEmpty(); top=NULL; } template void LinkStack ::SetEmpty() { //释放堆栈占用的内存 Node* temp; while (top!=NULL) { temp=top; top=top->next; delete temp; } } template bool LinkStack ::IsEmpty() { return (top==NULL); } template bool LinkStack ::Push(LSELEM element) { Node* temp=new Node(); if (temp ==NULL) return false ; temp->data=element; temp->next=top; top=temp; return true; } template bool LinkStack ::Pop(LSELEM& element) { if ( IsEmpty()) return false; Node* q = top; element = top->data; top=top->next; delete q; return true; } template bool LinkStack ::GetTop(LSELEM& element) { if ( IsEmpty()) return false; element = top->data; return true; } /*//******************************************************************* **********************************************************************/ void main(){ LinkStack stack; int N,M; int tmp; cout<<"输入要转换的十进制数字。\n"; cin>>N; cout<<"选择要转换的制式。\n"; cout<<"1.二进制。\n"; cout<<"2.八进制。\n"; cout<<"3.十六进制。\n"; do{ cin>>M; if(M<0 || M>3) cout<<"输入错误，请重新选择。\n"; }while(M<0 || M>3); cout<<"转换结果:\n"; switch(M){ case 1: while(N){ stack.Push(N%2); N=N/2; } while(!stack.IsEmpty()){ stack.Pop(tmp); cout< using namespace std; /*//******************************************************************* 链栈 **********************************************************************/ template class LinkStack { public: struct Node { LSELEM data; Node* next; }; LinkStack (); //构造函数 ~LinkStack(); //析构函数 void SetEmpty(); //置空堆栈 bool IsEmpty(); //判断堆栈是否为空 bool Push(LSELEM); //入栈 bool Pop(LSELEM&); //出栈 bool GetTop(LSELEM&); //获得栈顶元素 private: Node* top; }; template LinkStack ::LinkStack() { top=NULL; } template LinkStack ::~LinkStack() { SetEmpty(); top=NULL; } template void LinkStack ::SetEmpty() { //释放堆栈占用的内存 Node* temp; while (top!=NULL) { temp=top; top=top->next; delete temp; } } template bool LinkStack ::IsEmpty() { return (top==NULL); } template bool LinkStack ::Push(LSELEM element) { Node* temp=new Node(); if (temp ==NULL) return false ; temp->data=element; temp->next=top; top=temp; return true; } template bool LinkStack ::Pop(LSELEM& element) { if ( IsEmpty()) return false; Node* q = top; element = top->data; top=top->next; delete q; return true; } template bool LinkStack ::GetTop(LSELEM& element) { if ( IsEmpty()) return false; element = top->data; return true; } /*//******************************************************************* **********************************************************************/ void main(){ LinkStack stack; int N,M; int tmp; cout<<"输入要转换的十进制数字。\n"; cin>>N; cout<<"选择要转换的制式。\n"; cout<<"1.二进制。\n"; cout<<"2.八进制。\n"; cout<<"3.十六进制。\n"; do{ cin>>M; if(M<0 || M>3) cout<<"输入错误，请重新选择。\n"; }while(M<0 || M>3); cout<<"转换结果:\n"; switch(M){ case 1: while(N){ stack.Push(N%2); N=N/2; } while(!stack.IsEmpty()){ stack.Pop(tmp); cout< #include Status InitStatus(SqStack &S){ ////构造一个空栈 S.base=(SElemType *)malloc(STACK_INIT_SIZE * sizeof(ElemType)); if (!S.base)exit(OVERFLOW); S.top=S.base; S.stacksize=STACK_INIT_SIZE; return OK; } Status Push(SqStack &S,SElemType e){ ////插入元素e为新的栈顶 if(S.top - S.base >=S.stacksize){////栈满，追加空间 S.base=(ElemType *)realloc(S.base,(S.stacksize + STACKINCREMET) * sizeof(ElemType)); if (!S.base) {exit(OVERFLOW);////存储失败 } S.top=S.base+S.stacksize; S.stacksize+=STACKINCREMET; } *S.top++=e; return OK; }////PUSH bool StackEmpty(SqStack &S){ if (S.top==S.base)return true; else return false; } Status Pop(SqStack &S,SElemType &e){ ////若栈不空，则删除栈顶元素，返回到e，ok if (S.top==S.base)return ERROR; e=*(--S.top); return OK; }//////POP void Convert(){//////将一个10进制转化成八进制数 SqStack S; int N; int e; InitStatus(S); scanf("%d",&N); while (N) { Push(S,N % 8); N=N/8; } while(!StackEmpty(S)){ Pop(S,e); printf("%d",e); } } int main(int argc, char* argv[]) { printf("Hello World!\n"); Convert(); return 0; } ///////////////define.h #define TRUE 1 #define FALSE 0 #define OK 1 #define ERROR 0 #define INFEASIBLE -1 #define OVERFLOW -2 #define ElemType int #define SElemType int #define STACK_INIT_SIZE 100/////存储空间初始分量 #define STACKINCREMET 10////存储空间增量 typedef int Status; typedef struct{ SElemType *base;///栈构造前和销毁以后base为null SElemType *top;///栈顶指针 int stacksize;////当前已分配大小 }SqStack; 回文 typedef struct Node { char data; struct Node *next; }Node,*LinkStack; #define NULL 0 #include int InitStack(); void main() { InitStack() ? printf("Yes!\n") :printf("No!\n"); getch(); } int InitStack() { int i, j = 0; char *p; LinkStack top; Node *r, *s, *temp; top = (Node *)malloc(sizeof(Node)); if(top == NULL) { printf("overflow!\n"); exit(0); } top -> next = NULL; s = top; printf("please input the strint:\n"); gets(p); for(i = 0; i < strlen(p); i++) { r = (LinkStack)malloc(sizeof(Node)); r -> data = p[j++]; r -> next = s -> next; s -> next = r; } j = 0; for(i = strlen(p) - 1; i >= 0; i--) { temp = s -> next; if(temp -> data == p[j++]) { s -> next = temp -> next; free(temp); } else { break; } } if(i == -1) return 1; else return 0; } 给定一字符串，试判断它是否为回文串。例如输入12344321，该串为回文串，程序输出yes，否则输出no 程序代码: #include #include #include #define MAX 30 typedef struct stack { char data[MAX]; int top; }*STACK,node; void initstack(STACK st) { st->top=-1; } int push(STACK st,char *stringin) { if(st->top==MAX-1) return 0; while(*stringin) st->data[++st->top]=*(stringin++); return 1; } int pop(STACK st,char *stringout) { if(st->top==-1) return 0; while(st->top!=-1) *(stringout++)=st->data[st->top--]; *stringout='\0'; return 1; } int compare(char *stringin,char *stringout) { if(!strcmp(stringin,stringout)) return 1; return 0; } void showanswer(int judge) { if(judge) printf("yes\n"); else printf("no\n"); } void main() { char *stringin=NULL,*stringout=NULL; int judge; STACK st=NULL; stringin=(char *)malloc(MAX*sizeof(char)); stringout=(char *)malloc(MAX*sizeof(char)); gets(stringin); st=(STACK)malloc(sizeof(node)); initstack(st); push(st,stringin); pop(st,stringout); judge=compare(stringin,stringout); showanswer(judge); } 运行环境:vc++