HDU5831

Rikka with Parenthesis II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 857    Accepted Submission(s): 442

Problem Description

As we know, Rikka is poor at math. Yuta is worrying about this situation, so he gives Rikka some math tasks to practice. There is one of them:

Correct parentheses sequences can be defined recursively as follows:
1.The empty string "" is a correct sequence.
2.If "X" and "Y" are correct sequences, then "XY" (the concatenation of X and Y) is a correct sequence.
3.If "X" is a correct sequence, then "(X)" is a correct sequence.
Each correct parentheses sequence can be derived using the above rules.
Examples of correct parentheses sequences include "", "()", "()()()", "(()())", and "(((())))".

Now Yuta has a parentheses sequence S, and he wants Rikka to choose two different position i,j and swap Si,Sj.

Rikka likes correct parentheses sequence. So she wants to know if she can change S to a correct parentheses sequence after this operation.

It is too difficult for Rikka. Can you help her?

 

 

Input

The first line contains a number t(1<=t<=1000), the number of the testcases. And there are no more then 10 testcases with n>100

For each testcase, the first line contains an integers n(1<=n<=100000), the length of S. And the second line contains a string of length S which only contains ‘(’ and ‘)’.

 

 

Output

For each testcase, print "Yes" or "No" in a line.

 

 

Sample Input

3 4 ())( 4 ()() 6 )))(((

 

 

Sample Output

Yes Yes No

Hint

For the second sample input, Rikka can choose (1,3) or (2,4) to swap. But do nothing is not allowed.

 

 

Author

学军中学

 

 

Source

2016 Multi-University Training Contest 8

 

/*
找出几个特殊情况，剩下的就好办了，))((也是可以的，()不可以。从左向右，(,a++,如果是)并且a==0,b++;a!=0,a--;
*/
#include<iostream>
#include<string>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<vector>
#include<iomanip>
#include<queue>
#include<stack>
using namespace std;
int t,n;
string s;
int main()
{
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        cin>>s;
        int a=0,b=0;
        int k=s.size();
        if(n%2==1)
        {
            printf("No\n");
            continue;
        }
        for(int i=0;i<n;i++)
        {
            if(s[i]=='(') a++;
            else if(s[i]==')')
            {
                if(a==0)
                b++;
                else a--;
            }
        }
        if(a==1&&b==1)
        printf("Yes\n");
        else if(a==0&&b==0&&n!=2)
        printf("Yes\n");
        else if(a==2&&b==2)
        printf("Yes\n");
        else printf("No\n");
    }
    return 0;
}