> 如题。
> 
> 反过来扫，然后记录当前的句子数，一旦小于1那肯定NO掉，如果最后全处理完了结果句子数还大于1，也NO掉，否则YES之。
> 
> #include <cstdio>
> #include <iostream>
> #include <iomanip>
> #include <algorithm>
> #include <cmath>
> #include <cstring>
> #include <string>
> #include <set>
> #include <vector>
> #include <iterator>
> 
> using namespace std;
> 
> bool
> solve_prob(string &sentence)
> {
> 	int nr_subsent(0);
> 
> 	reverse(sentence.begin(), sentence.end());
> 	for (string::iterator iter = sentence.begin(); iter != sentence.end(); ++iter) {
> 		if (*iter >= 'p' && *iter <= 'z') {
> 			++nr_subsent;
> 		} else if (*iter == 'N') {
> 			// nothing to be done.
> 		} else if (*iter == 'C' || *iter == 'D' || *iter == 'E' || *iter == 'I') {
> 			--nr_subsent;
> 		} else {
> 			return false; // 题上说是p-z，N，CDEI，实际上如果这儿写一句*(int *)0 = 0会RE。。。悲剧- -
> 		}
> 		if (nr_subsent < 1) {
> 			return false;
> 		}
> 	}
> 
> 	return (nr_subsent == 1);
> }
> 
> int main()
> {
> 	string sentence;
> 
> 	while (cin >> sentence) {
> 		cout << (solve_prob(sentence) ? "YES" : "NO") << endl;
> 	}
> 
> 	return 0;
> }
赞，从句子数这个角度解题非常高明！