2017年上海金马五校程序设计竞赛（网上资格赛）Problem A : Corn's new language

最新推荐文章于 2018-06-21 19:56:51 发布

原创最新推荐文章于 2018-06-21 19:56:51 发布 · 620 阅读

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本文介绍了一种新颖的编程语言及其语法规则，并提供了一个用于计算该语言程序最大深度的C++实现。通过具体示例，文章解释了如何确定一个程序是否有效及如何计算其最大深度。

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Description

Corn is going to promote programming in the campus, so he wants to add a lot of interesting ideas to make programming more attractive. One task he is working on is to develop a new programming language because he thinks all existing ones are too simple to him. The syntax rules of the language are:

1. () is a valid program

2. (P) is a valid program, if P is a valid program

3. PQ is a valid program, if both P and Q are valid programs

Corn wants a compiler for the language. For a program, if it is valid, the compiler should print the max depth in the program. For example "(())" has a depth of 2, while "((())())" has a depth of 3. Formally, the max depth is the max number of unmatched open brackets in any prefix.

Input

Each case is a non-empty string in a single line, the string will only contain '(' or ')', its length will be no more than 100.

Output

For each case, output "YES" and the integer required above if the program is valid, separated with a space. Or "NO" if the program is invalid.

这个题真的挺简单的，但是我被卡了好久，，记得注意这种情况。 ())( 是no

#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
int main(){
	char str[105];
	while(scanf("%s",&str)!=EOF){
		bool is=true;
		int higt = 0, left = 0,right = 0,flag = 0;
		int u = strlen(str);
		for(int i = 0;i < u; i++){
			if(str[i]=='('){
				left++;
				flag++;
			}else if(str[i]==')'){
				right++;
				flag--;
			}
			if(flag<0) is=false;
			if(flag>higt) higt = flag;
		}
		if(is)
		{
			if(left == right){
			printf("YES %d\n",higt);
		}else if(left!=right){
			printf("NO\n");
		}
		}
		else printf("NO\n");
	}
}