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.
Examples
Input
8
))((())(
Output
6
Input
3
(()
Output
-1
Note
In the first example we can firstly reorder the segment from first to the fourth character, replacing it with “()()”, the whole sequence will be “()()())(”. And then reorder the segment from the seventh to eighth character, replacing it with “()”. In the end the sequence will be “()()()()”, while the total time spent is 4+2=6 nanoseconds.
思路：一开始的思路是，看有多少个括号匹配，然后看匹配的括号之外的括号，再加一些判断。结果wa到了第六个样例。。重新整理思路，对于右括号）来说，如果没有左括号（与之匹配的话，就说明从这一点开始就要重新排序了。这样我们记录一下最起始的这个位置，然后往后找直到左右括号数目匹配了，就直接排序就好了。这一段是必须要排序的。因此这么贪心不会多。
代码如下：
#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int maxx=1e6+100;
int n;
string s;
inline void solve()
{
int l=0,r=0,rk=-1,ans=0;
for(int i=0;i<n;i++)
{
if(s[i]=='(') l++;
else
{
if(l) l--;
else
{
r++;
if(rk==-1) rk=i;
}
}
if(l==r&&l)
{
ans+=(i-rk+1);
l=0,r=0,rk=-1;
}
}
cout<<ans<<endl;
}
int main()
{
cin>>n>>s;
int num1=0,num2=0;
for(int i=0;i<n;i++)
{
if(s[i]=='(') num1++;
else num2++;
}
if(num1!=num2) cout<<"-1"<<endl;
else solve();
return 0;
}
努力加油a啊，(o)/~
