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
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?
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 ‘)’.
4
())(
4
()()
6
)))(((
Yes
No
For the second sample input, Rikka can choose (1,3) or (2,4) to swap. But do nothing is not allowed.
/*
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=,b=;
int k=s.size();
if(n%==)
{
printf("No\n");
continue;
}
for(int i=;i<n;i++)
{
if(s[i]=='(') a++;
else if(s[i]==')')
{
if(a==)
b++;
else a--;
}
}
if(a==&&b==)
printf("Yes\n");
else if(a==&&b==&&n!=)
printf("Yes\n");
else if(a==&&b==)
printf("Yes\n");
else printf("No\n");
}
return ;
}