Rikka with Parenthesis II

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 136 Accepted Submission(s): 97

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

Recommend

wange2014 | We have carefully selected several similar problems for you:

题意：给你一个由'('和')'组成的字符串，首先你必须交换其中两个不同位置的字符，然后判断是否有一种方案使得最后形成的字符串合法；

#include <iostream>

#include <cstdio>

#include <cstring>

#include <cstdlib>

#include <cmath>

#include <vector>

#include <queue>

#include <stack>

#include <map>

#include <algorithm>

#include <set>

using namespace std;

typedef long long ll;

typedef unsigned long long Ull;

#define MM(a,b) memset(a,b,sizeof(a));

const double eps = 1e-10;

const int  inf =0x7f7f7f7f;

const double pi=acos(-1);

const int N=100005;

char s[N];

int main()

{

    int cas,n;

    scanf("%d",&cas);

    while(cas--)

    {

        scanf("%d",&n);

        scanf("%s",s);

        int l=0,r=0,len=n;

        if(len%2==1) {printf("No\n");continue;}

        for(int i=0;s[i]!='\0';i++)

        {

            if(s[i]=='(') r++;

            else if(s[i]==')')

            {

                if(r>=1) r--;

                else l++;

            }

        }

        if(len==2)

          {

            if(l==1&&r==1) printf("Yes\n");

            else printf("No\n");

            continue;

          }

         if(l==0&&r==0) printf("Yes\n");

         else if(l+r==2)

         {

             if(l==1&&r==1) printf("Yes\n");

             else printf("No\n");

         }

         else if(l+r==4)

         {

             if(l==2&&r==2) printf("Yes\n");

             else printf("No\n");

         }

         else printf("No\n");

    }

    return 0;

}

　　分析：比赛时只看到这道题是以前做过的题目的简化版，，结果还是太大意了，，，

本来分析出来了 ))(( 这种特殊情况也是可以的，但是草稿纸没打好，一不留神以为是右移成了())(,,,,其实是()()。。悲剧

巴特西

hdu 5831 Rikka with Parenthesis II 括号匹配+交换

Rikka with Parenthesis II

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