joj1693

 1693: Simply Syntax

Result	TIME Limit	MEMORY Limit	Run Times	AC Times	JUDGE
	3s	8192K	385	151	Standard

In the land of Hedonia the official language is Hedonian. A Hedonian professor had noticed that many of her students still did not master the syntax of Hedonian well. Tired of correcting the many syntactical mistakes, she decided to challenge the students and asked them to write a program that could check the syntactical correctness of any sentence they wrote. Similar to the nature of Hedonians, the syntax of Hedonian is also pleasantly simple. Here are the rules:

0.

The only characters in the language are the characters  p through  z and  N,  C,  D,  E, and  I.

1.

Every character from  p through  z is a correct sentence.

2.

If  s is a correct sentence, then so is  N s.

3.

If  s and  t are correct sentences, then so are  C st,  D st,  E st and  I st.

4.

Rules 0. to 3. are the only rules to determine the syntactical correctness of a sentence.

You are asked to write a program that checks if sentences satisfy the syntax rules given in Rule 0. - Rule 4.

Input

The input consists of a number of sentences. Each sentence is ended by a new-line character. The collection of sentences is terminated by the end-of-file character. If necessary, you may assume that each sentence has at most 256 characters and at least 1 character.

Output

The output consists of the answers YES for each well-formed sentence and NO for each not-well-formed sentence. The answers are given in the same order as the sentences. Each answer is followed by a new-line character, and the list of answers is followed by an end-of-file character.

Sample Input

Cp
Isz
NIsz
Cqpq

Sample Output

NO
YES
YES
NO

This problem is used for contest: 80  118 

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Problem Set with Online Judge System Version 3.12
Jilin University

Developed by skywind, SIYEE

/*这个题让我纠结了一阵子，首先这道题应该是一个二叉树的摸样，CDEI是内部接点
那些小写字母是外部接点，N不是树德接点可以直接舍掉，如果还有其他的点则直接
返回NO*/


#include<cstdio>
#include<string.h>
int main()
{
char name[257];
while(gets(name))
{
int len=strlen(name);
int count=0;
for(int i=len-1;i>=0;i--)
{
if(name[i]>='p'&&name[i]<='z')
count++;
else if(name[i]=='C'||name[i]=='D'||
  name[i]=='E'||name[i]=='I')
{
if(count>=2)count--;
else
{
count=-1;break;
}
}
else if(name[i]=='N');
else
{
count=-1;break;
}
}
if(count==1)printf("YES\n");
else printf("NO\n");
}
return 0;
}