CodeForces1214C

CodeForces1214C
是个不是很难的题目.
首先考虑如果左右括号数量不匹配那么肯定无论如何都不能通过移动一个括号完成匹配.
否则,我们考虑,将所有匹配的括号都去掉,剩下的括号只要大于(2)个,就不可能,否则就可以.
需要注意的是,剩下的左右括号都要算.
(Code:)

#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <string>
#include <vector>
#include <queue>
#include <cmath>
#include <ctime>
#include <map>
#include <set>
#define MEM(x,y) memset ( x , y , sizeof ( x ) )
#define rep(i,a,b) for (int i = a ; i <= b ; ++ i)
#define per(i,a,b) for (int i = a ; i >= b ; -- i)
#define pii pair < int , int >
#define X first
#define Y second
#define rint read<int>
#define int long long
#define pb push_back

using std::set ;
using std::pair ;
using std::max ;
using std::min ;
using std::priority_queue ;
using std::vector ;
using std::swap ;
using std::sort ;
using std::unique ;
using std::greater ;

template < class T >
    inline T read () {
        T x = 0 , f = 1 ; char ch = getchar () ;
        while ( ch < '0' || ch > '9' ) {
            if ( ch == '-' ) f = - 1 ;
            ch = getchar () ;
        }
       while ( ch >= '0' && ch <= '9' ) {
            x = ( x << 3 ) + ( x << 1 ) + ( ch - 48 ) ;
            ch = getchar () ;
       }
   return f * x ;
}

template < class T >
    inline void write (int x) {
       static T stk[100] , top = 0;
       if (x == 0) { putchar ('0') ; return ; }
       if (x < 0) { x = - x ; putchar ( '-' ) ; }
       while (x) { stk[++top] = x % 10 ; x /= 10 ; }
       while (top) { putchar ( stk[top--] + '0') ; }
    }

const int N = 2e5 + 100 ;

int n , cnt , num ; char s[N] ;

signed main (int argc , char * argv[]) {
    n = rint () ; scanf ("%s" , s + 1 ) ;
    rep ( i , 1 , n ) num += ( s[i] == '(' ) ;
    if ( num != n - num ) { puts ("No") ; return 0 ; }
    num = 0 ;
    rep ( i , 1 , n )
        if ( s[i] == '(' ) ++ cnt ;
        else { if ( cnt ) -- cnt ; else ++ num ; }
    if ( cnt + num > 2 ) puts ("No") ; else puts ("Yes") ;
    return 0 ;
}