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

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．

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排在我前面被hack了１００多人...

 1 /*************************************************************************
 2     > File Name: code/bc/#57/1001.cpp
 3     > Author: 111qqz
 4     > Email: rkz2013@126.com 
 5     > Created Time: 2015年09月26日 星期六 19时04分34秒
 6  ************************************************************************/
 7 
 8 #include<iostream>
 9 #include<iomanip>
10 #include<cstdio>
11 #include<algorithm>
12 #include<cmath>
13 #include<cstring>
14 #include<string>
15 #include<map>
16 #include<set>
17 #include<queue>
18 #include<vector>
19 #include<stack>
20 #include<cctype>
21 #define y1 hust111qqz
22 #define yn hez111qqz
23 #define j1 cute111qqz
24 #define ms(a,x) memset(a,x,sizeof(a))
25 #define lr dying111qqz
26 using namespace std;
27 #define For(i, n) for (int i=0;i<int(n);++i)  
28 typedef long long LL;
29 typedef double DB;
30 const int inf = 0x3f3f3f3f;
31 const int N=1E3+7;
32 char str[N];
33 int len;
34 int main()
35 {
36   #ifndef  ONLINE_JUDGE 
37    freopen("in.txt","r",stdin);
38   #endif
39     int T;
40     cin>>T;
41     while (T--)
42     {
43     scanf("%s",str);
44     len = strlen(str);
45     int cnt = 0;
46     int ans = 0 ;
47     for ( int i = 0 ; i < len ; i++)
48     {
49         if (str[i]=='(')
50         {
51         cnt++;
52         }
53         else
54         {
55         cnt--;
56         if (cnt>=0)
57         {
58             ans++;
59         }
60         if (cnt<0)
61             cnt = 0 ;
62         }
63     }
64     printf("%d
",ans);
65     }
66    
67  #ifndef ONLINE_JUDGE  
68   fclose(stdin);
69   #endif
70     return 0;
71 }

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