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2013
11-11

POJ 2752 Seek the Name, Seek the Fame [解题报告] Java

Seek the Name, Seek the Fame

问题描述 :

The little cat is so famous, that many couples tramp over hill and dale to Byteland, and asked the little cat to give names to their newly-born babies. They seek the name, and at the same time seek the fame. In order to escape from such boring job, the innovative little cat works out an easy but fantastic algorithm:

Step1. Connect the father’s name and the mother’s name, to a new string S.

Step2. Find a proper prefix-suffix string of S (which is not only the prefix, but also the suffix of S).

Example: Father=’ala’, Mother=’la’, we have S = ‘ala’+'la’ = ‘alala’. Potential prefix-suffix strings of S are {‘a’, ‘ala’, ‘alala’}. Given the string S, could you help the little cat to write a program to calculate the length of possible prefix-suffix strings of S? (He might thank you by giving your baby a name:)

输入:

The input contains a number of test cases. Each test case occupies a single line that contains the string S described above.

Restrictions: Only lowercase letters may appear in the input. 1 <= Length of S <= 400000.

输出:

For each test case, output a single line with integer numbers in increasing order, denoting the possible length of the new baby’s name.

样例输入:

ababcababababcabab
aaaaa

样例输出:

2 4 9 18
1 2 3 4 5

解题代码:

//* @author:
import java.util.*;
public class Main{
  static final int  MAX=400001;
  static int next[]=new int[MAX];
  static int a[]=new int[MAX];
 
  public static void main(String args[]){
    int i, j, len, sum;
    Scanner sc=new Scanner(System.in);

    while(sc.hasNext())
    {
        String s=sc.next();
         len=s.length();
         getNext(s);
         sum = 0;
         j = len;
        while(j != 0)
        {
            a[++sum] = j;
            j = next[j];
        }
        for (i = sum; i > 0; i--) System.out.printf("%d ",a[i]);
        System.out.printf("\n");
    }
    
   }
    public static void getNext(String T) {//建立模式串T的next[]表
    int i = 0;
    int j = next[0] = -1;
	
    while (i< T.length())
     if (0 > j || T.charAt(i) == T.charAt(j)) {//匹配
        j++; i++;
        next[i] = j;
     }else//失配
       j = next[j];
  }

}

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  2. 5.1处,反了;“上一个操作符的优先级比操作符ch的优先级大,或栈是空的就入栈。”如代码所述,应为“上一个操作符的优先级比操作符ch的优先级小,或栈是空的就入栈。”