(C语言)实验二:十进制转换R进制、回文、栈逆置队、括号匹配-洋葱先生-杨少通
发布日期:2021-10-03 07:58:42 浏览次数:1 分类:技术文章

本文共 4570 字,大约阅读时间需要 15 分钟。

注:本程序由Visual Studio 2015编写,与VC++6.0稍有区别,复制到VC++6.0注释掉“#include “stdafx.h””即可运行,复制到VS可直接运行。

#include "stdafx.h"#include 
#include
#include
using namespace std;#define OK 1#define ERROR 0#define OVERFLOW -1#define UNDERFLOW -2#define STACK_INIT_SIZE 80#define STACKINCREMENT 10typedef int status;#define ElemType chartypedef struct {
ElemType *base; ElemType *top; int stacksize;}SqStack;SqStack S;#define MAXQSIZE 80typedef struct {
ElemType *base; int front; int rear;}SqQueue;SqQueue Q;status InitStack(SqStack &S) //初始化栈{
S.base = (ElemType*)malloc(STACK_INIT_SIZE * sizeof(ElemType)); if (!S.base)exit(OVERFLOW); S.top = S.base; S.stacksize = STACK_INIT_SIZE; return OK;}status InitQueue(SqQueue &Q) {
//初始化队列 Q.base = (ElemType*)malloc(MAXQSIZE * sizeof(ElemType)); if (!Q.base)exit(OVERFLOW); Q.front = Q.rear = 0; return OK;}status Push(SqStack &S, ElemType e) {
//入栈 if (S.top-S.base == S.stacksize) {
S.base = (ElemType*)realloc(S.base, (S.stacksize + STACKINCREMENT) * sizeof(ElemType)); if (!S.base)exit(OVERFLOW); S.top = S.base + S.stacksize; S.stacksize += STACKINCREMENT; } *S.top++ = e; return OK;}status EnQueue(SqQueue &Q, ElemType e) {
//入队 if ((Q.rear + 1) % MAXQSIZE == Q.front) return ERROR;//队满 Q.base[Q.rear] = e; Q.rear = (Q.rear + 1) % MAXQSIZE; return OK;}status Pop(SqStack &S, ElemType &e) {
//出栈 if (S.top == S.base)exit(UNDERFLOW); e = *(S.top=S.top-1); return OK;}status DeQueue(SqQueue &Q, ElemType &e) {
//出队 if (Q.front == Q.rear)return ERROR; //队空 e = Q.base[Q.front]; Q.front = (Q.front + 1) % MAXQSIZE; return OK;}status StackEmpty(SqStack S) {
//是否为空栈 return S.base == S.top;}status QueueEmpty(SqQueue Q) {
//是否为空队 return Q.rear == Q.front;}void algo(SqQueue &Q) {
//栈实现队的首尾逆置 ElemType e; InitStack(S); cout << " 栈逆置队列前元素如下:"; while (!QueueEmpty(Q)) {
DeQueue(Q, e); cout << e; Push(S, e); } cout << endl; while (!StackEmpty(S)) {
Pop(S, e); EnQueue(Q, e); }}status match(char exp[50]) {
//括号是否匹配 InitStack(S); ElemType e; for (int i = 0; i < strlen(exp); i++) {
switch (exp[i]) {
case '(': case '[': Push(S, exp[i]); break; case ')': if (!StackEmpty(S)) {
Pop(S, e); if (e == '(') break; else return ERROR; } else return ERROR; case ']': if (!StackEmpty(S)) {
Pop(S, e); if (e == '[') break; else return ERROR; } else return ERROR; } } if (StackEmpty(S)) return OK; return ERROR;}char inttochar(int n) {
//整型转化为字符型 switch (n) {
case 0: return '0'; case 1: return '1'; case 2: return '2'; case 3: return '3'; case 4: return '4'; case 5: return '5'; case 6: return '6'; case 7: return '7'; case 8: return '8'; case 9: return '9'; case 10: return 'A'; case 11: return 'B'; case 12: return 'C'; case 13: return 'D'; case 14: return 'E'; case 15: return 'F'; }}void conversion(int data, int r) {
//进制转换 ElemType e; InitStack(S); while (data) {
Push(S, inttochar(data%r)); data = data / r; } while (!StackEmpty(S)) {
Pop(S, e); cout << e; } cout << endl << endl;}int main() {
InitQueue(Q); ElemType e; int num, data, r; char str[30] = ""; char exp[50]; cout << "\t\t\t\t*\t\t\t\t\t*"; cout << endl << "\t\t\t\t*\t计科1512-02210151232-杨少通\t*" << endl; cout << "\t\t\t\t*****************************************" << endl << endl; cout << "****************栈逆置队列元素****************" << endl << endl; cout << " 请输入初始化队列元素个数:"; cin >> num; for (int k= 0; k < num; k++) {
cout << " 请输入第" << k + 1 << "个元素:"; cin >> e; EnQueue(Q, e); } algo(Q); cout << " 栈逆置队列后元素如下:"; while (!QueueEmpty(Q)) {
DeQueue(Q, e); cout << e; } cout << endl << endl; cout << "**************表达式括号是否匹配**************" << endl << endl; cout << " 请输入表达式:"; cin >> exp; if (match(exp)) cout << " 表达式括号匹配!" << endl << endl; else cout << " 表达式括号不匹配!" << endl << endl; cout << "**************十进制转换为R进制***************" << endl << endl; cout << " 请输入十进制数:"; cin >> data; cout << " 请输入要转换的进制数(如:2、8、16):"; cin >> r; cout << " " << data << "转换为" << r << "进制数为:"; conversion(data, r); cout << "***********判断是否为中心对称字符串***********" << endl << endl; cout << " 请输入字符串:"; cin >> str; for (int i = 0; i < strlen(str); i++) Push(S, str[i]); for (int j = 0; j < strlen(str); j++) {
Pop(S, e); if (str[j] != e) {
cout << " “" << str << "”不是中心对称的字符串!" << endl << endl; return 0; } } cout << " “" << str << "”是中心对称的字符串!" << endl << endl; return 0;}

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