HDU 5479(栈的应用)

Scaena Felix

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 426    Accepted Submission(s): 186

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

102

 

//判断（）的个数行

// 利用栈的先进后出的特点模拟

#include <stdio.h>#include <string.h>#include <stack>using namespace std;char str[1000+10];int main(){    int t;    scanf("%d",&t);    while(t--)    {        getchar();        scanf("%s",str);        stack <char> s;        int len=strlen(str);        int cnt=0;        for(int i=0;i<len;i++)        {            if(str[i]=='(')                s.push(str[i]);            else if(str[i]==')')            {                if(!s.empty())  //判断是否为空                {                    s.pop();                    cnt++;                }            }        }        printf("%d\n",cnt);    }    return 0;}