题目：
A bracketed sequence is called correct (regular) if by inserting “+” and “1” you can get a well-formed mathematical expression from it. For example, sequences “(())()”, “()” and “(()(()))” are correct, while “)(”, “(()” and “(()))(” are not.
The teacher gave Dmitry’s class a very strange task — she asked every student to come up with a sequence of arbitrary length, consisting only of opening and closing brackets. After that all the students took turns naming the sequences they had invented. When Dima’s turn came, he suddenly realized that all his classmates got the correct bracketed sequence, and whether he got the correct bracketed sequence, he did not know.
Dima suspects now that he simply missed the word “correct” in the task statement, so now he wants to save the situation by modifying his sequence slightly. More precisely, he can the arbitrary number of times (possibly zero) perform the reorder operation.
The reorder operation consists of choosing an arbitrary consecutive subsegment (substring) of the sequence and then reordering all the characters in it in an arbitrary way. Such operation takes l nanoseconds, where l is the length of the subsegment being reordered. It’s easy to see that reorder operation doesn’t change the number of opening and closing brackets. For example for “))((” he can choose the substring “)(” and do reorder “)()(” (this operation will take 2 nanoseconds).
Since Dima will soon have to answer, he wants to make his sequence correct as fast as possible. Help him to do this, or determine that it’s impossible.

Input
The first line contains a single integer n (1≤n≤106) — the length of Dima’s sequence.
The second line contains string of length n, consisting of characters “(” and “)” only.

Output
Print a single integer — the minimum number of nanoseconds to make the sequence correct or “-1” if it is impossible to do so.

样例：

输入：
8
))((())(
输出：
6
输入：
3
(()
输出：
-1

题意：
有一个长度为n的只含有左右括号的字符串，现在你可以选定长度为l的子串进行任意排序，才是需要花费l精力。问你，最少可以花费多少精力使它满足题意，若果可以，输出最小花费，否则输出-1.

题解：
遍历一遍字符串，每次判断、查找、排序，使其符合条件，将结果累加即可。

#include<iostream>
#include<string>
using namespace std;
int n;
string str;
int solve(){
    int num1 = 0, res = 0, p = -1;
    for(int i = 0; i < n; i++){
        num1 = (str[i] != ')' ? num1 + 1 : num1 - 1);
        if(p == -1 && num1 < 0)
            p = i;
        if(p != -1 && !num1){
            res = res + (i - p + 1);
            p = -1;
        }
    }
    return num1 == 0 ? res : -1;
}
int main()
{
    std::ios::sync_with_stdio(false);
    std::cin.tie(0);
    std::cout.tie(0);
    cin >> n >> str;
    int res = solve();
    cout << res << endl;
    return 0;
}