hdoj 5479 || bestcoder #57 div 2 A Scaena Felix(模拟)

Scaena Felix

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 182    Accepted Submission(s): 85

Problem Description

Given a parentheses sequence consist of '(' and ')', a modify can filp a parentheses, changing '(' to ')' or ')' to '('.

If we want every not empty <b>substring</b> of this parentheses sequence not to be "paren-matching", how many times at least to modify this parentheses sequence?

For example, "()","(())","()()" are "paren-matching" strings, but "((", ")(", "((()" are not.

 

Input

The first line of the input is a integer T, meaning that there are T test cases.

Every test cases contains a parentheses sequence S only consists of '(' and ')'.

1≤|S|≤1,000.

 

Output

For every test case output the least number of modification.

 

Sample Input

3 () (((( (())

 

Sample Output

1 0 2

 

Source

BestCoder Round #57 (div.2)

 

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模拟．

直接搞...

并不明白坑在哪里．．．

排在我前面被hack了１００多人...

/*************************************************************************
    > File Name: code/bc/#57/1001.cpp
    > Author: 111qqz
    > Email: rkz2013@126.com 
    > Created Time: 2015年09月26日 星期六 19时04分34秒
 ************************************************************************/

#include<iostream>
#include<iomanip>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
#include<map>
#include<set>
#include<queue>
#include<vector>
#include<stack>
#include<cctype>
#define y1 hust111qqz
#define yn hez111qqz
#define j1 cute111qqz
#define ms(a,x) memset(a,x,sizeof(a))
#define lr dying111qqz
using namespace std;
#define For(i, n) for (int i=0;i<int(n);++i)  
typedef long long LL;
typedef double DB;
const int inf = 0x3f3f3f3f;
const int N=1E3+7;
char str[N];
int len;
int main()
{
  #ifndef  ONLINE_JUDGE 
   freopen("in.txt","r",stdin);
  #endif
    int T;
    cin>>T;
    while (T--)
    {
    scanf("%s",str);
    len = strlen(str);
    int cnt = 0;
    int ans = 0 ;
    for ( int i = 0 ; i < len ; i++)
    {
        if (str[i]=='(')
        {
        cnt++;
        }
        else
        {
        cnt--;
        if (cnt>=0)
        {
            ans++;
        }
        if (cnt<0)
            cnt = 0 ;
        }
    }
    printf("%d
",ans);
    }
   
 #ifndef ONLINE_JUDGE  
  fclose(stdin);
  #endif
    return 0;
}