PTA 7-2 符号配对

直接用栈模拟即可，数组可做，但因为这节数据结构是栈，为了期末考试还是手写一下栈的操作，值得注意的是，这道题用gets函数在PTA上会编译错误，用scanf("%[^ ]", str)会有一个样例无法通过，最后我使用了string读入数据，应该是我对scanf格式化读入不够了解，有知道的朋友可以评论区告诉我，非常感谢

附上代码：

#include <stdio.h>
#include <malloc.h>
#include <stdlib.h>
#include <string.h>

#include <iostream>
#include <string>

using namespace std;

//函数状态码定义
#define TRUE        1
#define FALSE       0
#define OK          1
#define ERROR       0
#define INFEASIBLE -1
#define OVERFLOW   -2

typedef int Status;
typedef char SElemType;

#define STACK_INIT_SIZE 500
#define STACKINCREMENT 10

typedef struct {
    SElemType *base;
    SElemType *top;
    int stacksize;
}SqStack;

Status InitStack(SqStack &S) {
    S.base = (SElemType * )malloc(STACK_INIT_SIZE * sizeof(SElemType));
    if(!S.base) exit(OVERFLOW);
    S.top = S.base;
    S.stacksize = STACK_INIT_SIZE;
    return OK;
}//InitStack

Status GetTop(SqStack &S, SElemType &e) {
    if(S.top == S.base)
        return ERROR;
    e = *(S.top - 1);
    return OK;
}//GetTop

Status Push(SqStack &S, SElemType e) {
    if(S.top - S.base >= S.stacksize) {
        S.base = (SElemType * )realloc(S.base, (S.stacksize + STACKINCREMENT) * sizeof(SElemType));
        if(!S.base)
            exit(OVERFLOW);
        S.top = S.base + S.stacksize;
        S.stacksize += STACKINCREMENT;
    }
    *S.top++ = e;
    return OK;
}//Push

Status Pop(SqStack &S, SElemType &e) {
    if(S.top == S.base)
        return ERROR;
    e = *--S.top;
    return OK;
}//Pop

char change(char ch){
    if(ch == '(')
        return ')';
    if(ch == '{')
        return '}';
    if(ch == '[')
        return ']';
    if(ch == 'a')
        return 'b';
    if(ch == ')')
        return '(';
    if(ch == '}')
        return '{';
    if(ch == ']')
        return '[';
    if(ch == 'b')
        return 'a';
};

int main()
{
    SqStack sta;
    InitStack(sta);

    char ch, ptr[200] = {0}, cnt = 0;

    char ans = ' '; // ( -> 1, { -> 2, [ -> 3, /* -> 4

    string str;

    while(getline(cin, str)) {
        if(str == ".")
            break;
        for(int i = 0; i < str.size(); ++i) {
            if(str[i] == '(' || str[i] == '{' || str[i] == '[' || str[i] == ')' || str[i] == '}' || str[i] == ']')
                ptr[cnt++] = str[i];
            else if(str[i] == '/' && str[i + 1] == '*')
                ptr[cnt++] = 'a', ++i;
            else if(str[i] == '*' && str[i + 1] == '/')
                ptr[cnt++] = 'b', ++i;
        }
    }

    for(int i = 0; i < cnt; ++i) {
        if(ptr[i] == '(' || ptr[i] == '{' || ptr[i] == '[' || ptr[i] == 'a')
            Push(sta, ptr[i]);
        else {
            if(GetTop(sta, ch) == OK) {
                if(ptr[i] == change(ch)) {
                    Pop(sta, ch);
                }
                else {
                    ans = ch;
                    break;
                }
            }
            else {
                ans = ptr[i];
                break;
            }
        }
    }

    if(ans == ' ' && GetTop(sta, ch) == OK) {
        ans = ch;
    }

    if(ans == ' ')
        printf("YES");
    else {
        printf("NO
");
        if(ans == '(' || ans == '{' || ans == '[')
            printf("%c-?",ans);
        else if(ans == 'a')
            printf("/*-?");
        else if(ans == ')' || ans == '}' || ans == ']')
            printf("?-%c",ans);
        else
            printf("?-*/");
    }

    return 0;
}