HDOJ 1433 Simply Syntax

Simply Syntax

Problem Description

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 Ns.

3. If s and t are correct sentences, then so are Cst, Dst, Est and Ist.

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 consisting only of characters p through z and N, C, D, E, and I. 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

CpIszNIszCqpq

 

Sample Output

NOYESYESNO

1. Every character from p through z is a correct sentence. //每一个语法都由p到z贯穿，所以最后一个字母需为p到z

2. If s is a correct sentence, then so is Ns. //s语法正确，Ns语法也正确

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int fan(char a[]){int sum=0,i,len;len=strlen(a);for (i=len-1;i>=0;i--)//倒着判断{if (a[i]>='p'&&a[i]<='z')sum++;else if (a[i]=='N'){continue;}else if(a[i]=='C'||a[i]=='D'||a[i]=='E'||a[i]=='I')sum--;elsereturn 0;if (sum<1)return 0;}if (sum==1)return 1;elsereturn 0;}int main(){int i;char str[300];while (scanf ("%s",str)!=EOF){//gets(str);if (fan(str))printf ("YES\n");elseprintf ("NO\n"); } return 0;}