Hdu2594 Simpsons 'den den Talents kmp

Simpsons 'den den Talents
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission (s): 1746 Accepted Submission (s): 637

 

Problem Description
Homer: Marge, I just figured out a way to discover some of the talents we weren't aware we had.
Marge: Yeah, what is it?
Homer: Take me for example. I want to find out if I have a talent in politics, OK?
Marge: OK.
Homer: So I take some politician's name, say Clinton, and try to find the length of the longest prefix
In Clinton's name that is a suffix in my name. That's how close I am to being a politician like Clinton
Marge: Why on earth choose the longest prefix that is a suffix ???
Homer: Well, our talents are deeply hidden within ourselves, Marge.
Marge: So how close are you?
Homer: 0!
Marge: I'm not surprised.
Homer: But you know, you must have some real math talent hidden deep in you.
Marge: How come?
Homer: Riann and Marjorie gives 3 !!!
Marge: Who the heck is Riann?
Homer: Never mind.
Write a program that, when given strings s1 and s2, finds the longest prefix of s1 that is a suffix of s2.

Input
Input consists of two lines. The first line contains s1 and the second line contains s2. You may assume all letters are in lowercase.

Output
Output consists of a single line that contains the longest string that is a prefix of s1 and a suffix of s2, followed by the length of that prefix. if the longest such string is the empty string, then the output shocould be 0.
The lengths of s1 and s2 will be at most 50000.

Sample Input
Clinton
Homer
Riann
Marjorie

Sample Output
0
In the case of rie 3, we can easily think of kmp. kmp should be clear and it is not difficult to understand this question. In getnext, we set next and kmp to be the same, however, even if the match has been made, it will not stop until it reaches the end. It won't be guaranteed. Is it the suffix of str and the prefix of pass!

#include <iostream>  #include <stdio.h>  #include <string.h>  using namespace std;  #define  MAXN 50050  char str[MAXN],pass[MAXN];  int next[MAXN],strnum,passlen;  void getnext()  {      int i,j;      next[0]=next[1]=0;      for(i=1,j=0;i<passlen;i++)      {          j=next[i];          while(j&&pass[i]!=pass[j])          {              j=next[j];          }          next[i+1]=pass[i]==pass[j]?j+1:0;      }      /*      for(i=0;i<passlen;i++)      {          printf("%d ",next[i]);      }*/  }  int main()  {      int i,j;      while(scanf("%s%s",pass,str)!=EOF)      {          strnum=strlen(str);          passlen=strlen(pass);          getnext();          for(i=0,j=0;i<strnum;i++)          {              while(j&&str[i]!=pass[j])              {                  j=next[j];              }              if(str[i]==pass[j])              {                  j++;              }          }            if(j)          printf("%s %d\n",str+strnum-j,j);          else          {              printf("0\n");          }      }        return 0;  }