HDU 5831 Rikka with Parenthesis II （模拟）

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水
注意））（（这种情况也是yes
【代码】
/* ***********************************************Author        :angon************************************************ */#include <stdio.h>#include <string.h>#include <iostream>#include <algorithm>#include <stack>#include <vector>#include <queue>#include <set>#include <map>#include <string>#include <math.h>#include <stdlib.h>#include <time.h>using namespace std;#define showtime fprintf(stderr,"time = %.15f\n",clock() / (double)CLOCKS_PER_SEC)#define REP(i,k,n) for(int i=k;i<n;i++)#define REPP(i,k,n) for(int i=k;i<=n;i++)#define scan(d) scanf("%d",&d)#define scanl(d) scanf("%I64d",&d)#define scann(n,m) scanf("%d%d",&n,&m)#define scannl(n,m) scanf("%i64d%I64d",&n,&m)#define mst(a,k)  memset(a,k,sizeof(a))#define LL long long#define N 1005#define mod 1000000007inline int read(){int s=0;char ch=getchar();for(; ch<'0'||ch>'9'; ch=getchar());for(; ch>='0'&&ch<='9'; ch=getchar())s=s*10+ch-'0';return s;}char s[100000+5];stack<char>st;int main(){    //freopen("in.txt","r",stdin);    //freopen("out.txt","w",stdout);    int t,n;    scan(t);    while(t--)    {        while(!st.empty()) st.pop();        scan(n);        scanf("%s",s);        n = strlen(s);        if(n&1 || strcmp(s,"()")==0)        {            printf("No\n");            continue;        }        for(int i=0;i<n;i++)        {            if(st.empty())            {                st.push(s[i]);            }            else            {                if(s[i]=='(')                    st.push('(');                else                {                    if(st.top()=='(')                        st.pop();                    else                    {                        st.push(')');                    }                }            }        }        if(st.empty())        {            printf("Yes\n");        }        else        {            int cnt=0;            while(!st.empty())            {                s[cnt++]=st.top();                if(cnt>4) break;                st.pop();            }            if(cnt==2 && s[0]=='('&&s[1]==')' || cnt==4 && s[0]=='('&&s[1]=='('&&s[2]==')'&&s[3]==')')                printf("Yes\n");            else                printf("No\n");        }    }    return 0;}
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