题解归档 - cf2227B

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• 本地编号：cf2227B
• 本地来源：problems/cf2227B/idea.md
• 题目链接：https://codeforces.com/contest/2227/problem/B
• 原始标题：cf2227B - Party Monster

思路

cf2227B - Party Monster

pattern: invariant / full-substring rearrangement.

claim: choosing the whole string as the removed substring allows reinserting all characters in arbitrary order. A regular bracket sequence exists exactly when the counts of '(' and ')' are equal, i.e. n is even and open * 2 == n.

check: this matches )( -> (), rejects odd length or unequal counts.

代码

来源：problems/cf2227B/solution.cpp

/* Author: likely
 * Time: 2026-06-27
**/
#include<bits/stdc++.h>
#define ll long long
using namespace std;

int main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int t;
    cin>>t;
    while(t--){
        int n;
        string s;
        cin>>n>>s;
        int open=0;
        for(char c:s) open+=(c=='(');
        cout<<(open*2==n?"YES":"NO")<<"\n";
    }
    return 0;
}