HDU-1039-Easier Done Than Said?(Java && 没用正则表达式是我的遗憾.....)
Easier Done Than Said?
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 9845 Accepted Submission(s): 4784
FnordCom is developing such a password generator. You work in the quality control department, and it‘s your job to test the generator and make sure that the passwords are acceptable. To be acceptable, a password must satisfy these three rules:
It must contain at least one vowel.
It cannot contain three consecutive vowels or three consecutive consonants.
It cannot contain two consecutive occurrences of the same letter, except for ‘ee‘ or ‘oo‘.
(For the purposes of this problem, the vowels are ‘a‘, ‘e‘, ‘i‘, ‘o‘, and ‘u‘; all other letters are consonants.) Note that these rules are not perfect; there are many common/pronounceable words that are not acceptable.
a tv ptoui bontres zoggax wiinq eep houctuh end
<a> is acceptable. <tv> is not acceptable. <ptoui> is not acceptable. <bontres> is not acceptable. <zoggax> is not acceptable. <wiinq> is not acceptable. <eep> is acceptable. <houctuh> is acceptable.
import java.io.*; import java.util.*; public class Main { public static void main(String[] args) { // TODO Auto-generated method stub Scanner input = new Scanner(System.in); while (input.hasNext()) { boolean flag1 = false, flag2 = true, flag3 = true; String str = input.next(); if (str.endsWith("end")) break; char c[] = str.toCharArray(); // 第一个条件:必须包含至少一个元音字母 for (int i = 0; i < c.length; i++) { if (c[i] == 'a' || c[i] == 'e' || c[i] == 'i' || c[i] == 'o' || c[i] == 'u') { flag1 = true; break; } } // 第二个条件:不能包含三个连续的元音字母或者三个连续的辅音字母 int a = 0, b = 0; for (int i = 0; i < c.length; i++) { if (c[i] == 'a' || c[i] == 'e' || c[i] == 'i' || c[i] == 'o' || c[i] == 'u') { a++; if (a >= 3) { flag2 = false; } b = 0; } else { b++; if (b >= 3) { flag2 = false; } a = 0; } } // 第三个条件:不能包含两个连续相同的字母,除了'ee'和'oo'这两种情况除外 int i, j = 0; for (i = 1; i < c.length; i++, j++) { if (c[i] == c[j]) { if (c[i] == 'e' || c[i] == 'o') continue; else { flag3 = false; break; } } } if (flag1 && flag2 && flag3) { System.out.println("<" + str + "> is acceptable."); } else { System.out.println("<" + str + "> is not acceptable."); } } } }
import java.util.Scanner; import java.util.regex.Matcher; import java.util.regex.Pattern; public class Main { public static void main(String[] args) throws Exception { Scanner cin = new Scanner(System.in); while (cin.hasNext()) { String str = cin.next(); if(str.equals("end")) break; Pattern p1 = Pattern.compile("[aeiou]{3}|[^aeiou]{3}"); Pattern p2 = Pattern.compile("([a-df-np-z])\\1"); Pattern p3 = Pattern.compile("[aeiou]+"); Matcher m = p1.matcher(str); boolean flag = false; if(!m.find()) { m = p2.matcher(str); if(!m.find()) { m = p3.matcher(str); if(m.find()) flag = true; } } if(flag) System.out.println("<"+str+"> is acceptable."); else System.out.println("<"+str+"> is not acceptable."); } cin.close(); } }
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