Simply Syntax(思维)
Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 5551 | Accepted: 2481 |
Description
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
Output
Sample Input
Cp
Isz
NIsz
Cqpq
Sample Output
NO
YES
YES
NO
题解:
0,p~z每个字母都是合法的句子
1, 如s是一个合法的句子,那么Ns也是一个合法的句子
2,如果s与t都是合法的一个句子,那么Cst, Dst, Est, and Ist.
3,只有满足以上的才是一个合法的句子
想的太多了,其实就按照所说的递归就好:
代码:
package com.hbc.week3; import java.util.Scanner; public class SimplySyntax {
private static Scanner cin = null;
static{
cin = new Scanner(System.in);
}
static boolean judge(String s){
char c = s.charAt(0);
if(s.length() == 1){
if(c >= 'p' && c <= 'z')
return true;
else
return false;
} if(c == 'N'){
if(judge(s.substring(1))){
return true;
}
return false;
}
if(c == 'C'
|| c == 'D'
|| c == 'E'
|| c == 'I'){
for(int i = 2; i < s.length(); i++){
if(judge(s.substring(1, i)) && judge(s.substring(i, s.length()))){
return true;
}
}
return false;
}
return false;
}
public static void main(String[] args) {
//System.out.println(isSimpleSentence("pqp"));
while(cin.hasNext()){
String s = cin.next();
if(judge(s)){
System.out.println("YES");
}else{
System.out.println("NO");
}
}
}
}
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