UVA——673 Parentheses Balance

分手后的思念是犯贱 2021-09-26 11:36 190阅读 0赞

## 题目： ##

Parentheses Balance

Problem Description

You are given a string consisting of parentheses () and \[\]. A string of this type is said to be*correct*:

(a)

if it is the empty string

(b)

if A and B are correct, AB is correct,

(c)

if A is correct, (A) and \[A\] is correct.

Write a program that takes a sequence of strings of this type and check their correctness. Your program can assume that the maximum string length is 128.

Input

the file contains a positive integer *n* and a sequence of *n* strings of parentheses() and \[\], one string a line.

Output

A sequence of Yes or No on the output file.

Sample Input

3
    ([])
    (([()])))
    ([()[]()])()

Sample output

Yes
    No
    Yes

## 题目描述： ##

括号匹配问题，用栈来写的，比较坑的是，当读入空行时要判为Yes。当时我用的scanf,然后就用%c读的，再手动把换行符换成\\0。

## 解题代码： ##

#include<stdio.h>  #include<string.h> int f[110][110]; char e[150]; int a,b,c,i,j; typedef struct q { char e[150]; int top; }q; int main() { q w; int t; while(scanf("%d",&a)!=EOF){ getchar(); while(a -- ){ memset(e,'\0',sizeof(e)); i=0; while(1){ scanf("%c",&e[i]); if(e[i] == '\n'){ e[i] = '\0'; break; } i ++; } memset(w.e,'\0',sizeof(w.e)); w.top = 0; t = 0; if(strlen(e) == 0){ printf("Yes\n"); continue; } for(i = 0;i < strlen(e);i ++){ if(e[i] == '(' || e[i] == '['){ w.e[w.top++] = e[i]; } else{ if(w.top && (w.e[w.top-1] == '[' && e[i] == ']' || w.e[w.top-1] == '(' && e[i] == ')')){ w.top --; } else{ t = 1; break; } } } if(t || w.top) printf("No\n"); else printf("Yes\n"); } } return 0; }