Scaena Felix

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 101    Accepted Submission(s): 49

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

题解：水题。。。。匹配括号的个数。。。

代码：

 1 #include<stdio.h>
 2 #include<stack>
 3 using namespace std;
 4 const int MAXN=1100;
 5 char m[MAXN];
 6 int main(){
 7     int T;
 8     scanf("%d",&T);
 9     while(T--){
10         stack<char>st;
11         scanf("%s",m);
12         int k=0;
13         for(int i=0;m[i];i++){
14             if(m[i]==‘(‘)st.push(m[i]);
15             else if(!st.empty())st.pop(),k++;
16         }
17         printf("%d\n",k);
18     }
19     return 0;
20 }