8.11 B - Simpsons’ Hidden Talents

                            B - Simpsons’ Hidden Talents

Homer: Marge, I just figured out a way to discover someof 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 thelength of the longest prefix 
in Clinton’s name that is a suffix in my name. That’s how close I am to being apolitician 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 ofs1 that is a suffix of s2.

Input

Input consists of two lines. The first line contains s1and the second line contains s2. You may assume all letters 

are in lowercase.

Output

Output consists of a single line that contains thelongest string that is a prefix of s1 and a suffix of s2, followed 

by thelength of that prefix. If the longest such string is the empty string, then theoutput 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

 

题意：判断第一个字符串前缀和第二个字符串后缀相等的最大长度，输出这个相同的前后缀，并输出长度。

思路：将第一个字符串处理成一个next数组，然后用第一个字符串把第二个字符串完全跑一边，记录最后一次比较时

的变量。

 

#include<stdio.h>
#include<string.h>
int next[1110000],extend[1110000];
char s1[1110000],s2[1110000];
int main()
{
	int i,j,k,T,m,n,max,f,c;

	while(scanf("%s%s",s1,s2)!=EOF)
	{
		memset(next,0,sizeof(next));
		memset(extend,0,sizeof(extend));
		m=strlen(s1);n=strlen(s2);
		j=0;
		for(i=1;i<m;)
		{
			
			if(j==0&&s1[i]!=s1[j])
			{
				i++;
			}
			else if(j>0&&s1[i]!=s1[j])
			{
				j=next[j-1];
			}
			else
			{
				next[i]=j+1;
				i++;
				j++;
			}
			
		}
		
/*		for(i=0;i<n;i++)
		{
			printf("%d ",next[i]);
		}
		printf("\n");
*/		
		i=0;j=0;	
		while(i < n)
		{
			
			if(j==0&&s2[i]!=s1[j])
			{
				i++;
			}
			else if(j>0&&s2[i]!=s1[j])
			{
				j=next[j-1];
			}
			else
			{
				i++;
				j++;
			}
			//printf("%d ",j);
		}
		
		
		//printf("j====%d\n",j);
	
			for(i = 0; i < j; i++)//KMP比较到最后j记录的是第二个字符串后缀与第一个字符串前缀相等的长度
				printf("%c", s1[i]);
		if(j)
			printf(" ");
		printf("%d\n", j);		
	}	
	return 0;
}