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 SS, and he wants Rikka to choose two different position i,ji,j and swap Si,SjSi,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?
InputThe 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 ‘)’.OutputFor 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.
水题，注意两点1一定要交换所以"()"不行；2记录 l和r 遇到第一次r>l的情况和最后面的'('交换位置，之后就不能交换位置了
#include<iostream>
#include<cstdio>
#include<queue>
#include<vector>
#include<cstring>
#include<string>
#include<algorithm>
#include<map>
#include<cmath>
#include<math.h>
using namespace std; int main()
{
int T,n;
int l,r,sum;
scanf("%d",&T);
char a[];
bool flag1,flag2;
while(T--)
{
scanf("%d",&n);
if(n==)
{
printf("Yes\n");
continue;
}
cin>>a;
l=;r=;sum=;
if(n==&&a[]=='('&&a[]==')')
printf("No\n");
else if(n%!=)
printf("No\n");
else
{
flag1=false;flag2=true;
for(int i=;i<n;i++)
{
if(a[i]=='(')
l++;
else if(a[i]==')')
r++;
if(r>l&&!flag1)
{
for(int j=n-;j>i;j--)
{
if(a[j]=='(')
{
a[i]='(';
a[j]=')';
l++;
r--;
flag1=true;
break;
}
}
}
else if(r>l&&flag1==true)
{
flag2=false;
break;
}
}
if(flag2&&l==r)
printf("Yes\n");
else
printf("No\n");
} }
return ;
}