Simpsons’ Hidden Talents （HDU-2594）

Simpsons’ Hidden Talents （HDU-2594）

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

题意：给你两个字符串，在第一个字符串中找到一个最长的前缀，同时也是第二个字符串的后缀；找不到的话输出0

思路：把两个串连起来，那么next[len]就是要求的前缀长度了。但是连起来之后因为next[len]很有可能大于两个原来的串的最小的长度，因此，需要再处理一步

AC代码：

#include <bits/stdc++.h>
const int maxx=100010;
const int inf=0x3f3f3f3f;
using namespace std;
char s1[maxx];
char s2[maxx];
int nex[maxx];
int s1len;
void Getnextval()
{int i=0,j=-1;nex[0]=-1;while(i<s1len){while(j!=-1 && s1[i]!=s1[j])j=nex[j];nex[++i]=++j;}
}
int main()
{while(~scanf("%s%s",s1,s2)){strcat(s1,s2);//字符串连接函数s1len=strlen(s1);int s2len=strlen(s2);Getnextval();if(nex[s1len]==0)printf("0\n");else{int cnt=s1len-s2len>s2len?s2len:s1len-s2len;while(cnt<nex[s1len])nex[s1len]=nex[nex[s1len]];printf("%s %d\n",s1+s1len-nex[s1len],nex[s1len]);}}return 0;
}

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