#include <stdio.h>
#include <string.h>

int next[50005];
char str1[50005],str2[50005];

void build_next(int len2)
{
	int i=0,j=-1;
	next[0] = -1;
	while (i < len2)
	{
		if (j==-1 || str2[i] == str2[j])
		{
			i++;
			j++;
			if (str2[i] != str2[j])
			{
				next[i] = j;
			}
			else
				next[i] = next[j];
		}
		else
			j = next[j];
	}
}

int KMP(int len1,int len2)
{
    build_next(len2);
	int i=0,j=0,cnt=0;

	while (i < len1)
	{
		if (j==-1 || str1[i] == str2[j])
		{
			i++;
			j++;
		}
		else
			j = next[j];
	}
	return j;
}

int main()
{
	int N,n,i;
	while (gets(str2))
	{
		gets(str1);
		n=KMP(strlen(str1),strlen(str2));
		for (i=0;i<n;i++)
		{
			putchar(str2[i]);
		}
		if(n)
		    printf(" ");
		printf("%d\n",n);
	}
	return 0;
}

分析：    此题意思为给出两个字符串，然后比较str1的前缀与str2的后缀相同的最大长度，即为代码中的j

做完此题后，又对KMP算法有了一点掌握，其中next储存的就是下次需要把模式串从哪个位置开始与目标串进行比较。

next又与前缀后缀有关，如果前后缀相同，那么直接令next[]=相同的前后缀的长度即可。

Simpsons’ Hidden Talents

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

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

Source

HDU 2010-05 Programming Contest