Simpsons’ Hidden Talents（KMP）

Simpsons’ Hidden Talents

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 11684 Accepted Submission(s): 4075

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: Riemann and Marjorie gives 3!!!
Marge: Who the heck is Riemann?
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 should be 0.
The lengths of s1 and s2 will be at most 50000.

Sample Input
clinton
homer
riemann
marjorie

Sample Output
0
rie 3

/*
    题解 :
    本题就是给你两个串 a 、b 求 b 的最长后缀是 a 的最长前缀
    这里直接用了next 数组 将两个串接起来
    然后在求出next 的值 最后在求 next[lena + lenb] 的值 这里还需要判断一下就是
    next的值不能超过a b 的长度
*/

#include
#include
#include
#include

using namespace std;
char ch[50007 *2],p[50007];
int nxt[50007*2];
int lens;
void GetNext(int len)//求next数组
{
    int i =1;
    nxt[0] = nxt[1] = 0;
    while(iint j = nxt[i];
        while( j && p[i] != p[j]) j = nxt[j];
        if(p[i] == p[j])
            nxt[i+1] = j+1;
        else
            nxt[i+1] = 0;
        i++;
    }
}
int main()
{
    while(scanf("%s",p) != EOF)
    {
        scanf("%s",ch);
        int lenp = strlen(p);
        lens = strlen(ch);
        strcat(p,ch);//字符串链接
        GetNext(lenp + lens);
        int indx = nxt[lenp + lens];
        while(indx > lenp || indx > lens)//直到求出来next数组的值均小于 两个串长
            indx = nxt[indx];
        for(int i =0;iprintf("%c",p[i]);
        if(indx)
            printf(" ");
        printf("%d\n",indx);
    }
    return 0;
}