# UVa 271 Simply Sintax
###### tags: `Algorithm` `Ans`

#include<iostream>
using namespace std;
#define maxN 10000000

int rear;
int stk[maxN];

int main()
{
    ios::sync_with_stdio(0);
    cin.tie(0);

    string s;
    while (getline(cin,s))
    {
        rear=0;
        int len=s.size();
        for( int i=0; i<len; i++)
        {
            if (s[i]=='N')
            {
                stk[rear++] = -1;
            }
            else if (s[i]=='C' || s[i]=='D' || s[i]=='E' || s[i]=='I')
            {
                stk[rear++] = -2;
            }
            else if ('p'<=s[i]&&s[i]<='z')
            {
                stk[rear++] = 1;
            }

            if (rear<2) continue;
            while (stk[rear-1]>0&&stk[rear-2]<0)
            {
                if (stk[rear-2]==-1)
                {
                    stk[(--rear)-1] = 1;
                }
                else if (stk[rear-2]==-2)
                {
                    stk[(--rear)-1] = -1;
                }
            }
        }
        if (rear==1&&stk[0]==1) cout << "YES\n";
        else cout << "NO\n";
    }
}

---

利用for迴圈從後方檢查回來
如果是`p~z`就將句子數量+1
如果是`N`，只要N後面有句子就能連接成一句
如果是`C`,`D`,`E`,`I`，只要後面有2個或2個以上的句子就連接成一句

//by Koios1143
#include<iostream>
#include<memory.h>
using namespace std;
int main(){
//---------------------------------
	#ifdef local
		freopen("UVa271.txt","r",stdin);
		freopen(".txt","w",stdout);
	#endif
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
//---------------------------------
	bool alpha[128];
	memset(alpha,false,sizeof(alpha));
	for(int i='p' ; i<='z' ; i++){
		alpha[i]=true;
	}
	alpha['N']=alpha['C']=alpha['D']=alpha['E']=alpha['I']=true;
	
	string s;
	while(cin>>s){
		int size=s.size();
		bool YN=true;
		int cnt=0;
		for(int i=size-1 ; i>=0 ; i--){
			if(!alpha[s[i]]){
				YN=false;
				break;
			}
			if(s[i]=='N'){
				if(cnt>=1){
					//cnt--;
				}
				else{
					YN=false;
					break;
				}
			}
			else if(s[i]>='A'&&s[i]<='Z'){
				if(cnt>=2)
					cnt--;
				else{
					YN=false;
					break;
				}
			}
			else{
				cnt++;
				
			}
		}
		if(cnt>1)
			YN=false;
		if(YN)
			cout<<"YES\n";
		else
			cout<<"NO\n";
	}
	
	
		
 return 0 ;
}

附註:其實題目有說輸入保證只有那幾個字母，所以其實不用用bool判斷