HDU 2594 Simpsons’ Hidden Talents KMP

                                             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

clintonhomerriemannmarjorie

 

Sample Output

0rie 3
题意 求A字符串的最长前缀是B字符串的最长后缀
思路： 利用next数组的性质 123|123对称
       将两个字符串连接起来 为了防止粘连即 aaa aa 这种情况 在A字符后添加特俗字符
ACcode：
#include<bits/stdc++.h>using namespace std;const int maxn = 1e5+10;char A[maxn],B[maxn];int nextx[maxn];int lenA,lenB,len;int ans=0;void Get_nextx(){    int j=0,k= -1;    nextx[0] = -1;    while(j<len){        if(k == -1 || A[j] == A[k]){            nextx[++j]=++k;        }        else k=nextx[k];    }}int main(){    while(scanf("%s%s",A,B)!=EOF){        lenA=strlen(A);        lenB=strlen(B);        A[lenA]='|';        A[lenA+1]='\0';//在Discuss里面看到的 不知道为什么不加这一句会TLE        strcat(A,B);   //这个函数不稳定 出现莫名其妙的错误 阳哥叫我下次直接用for循环写        len=strlen(A);        Get_nextx();        ans = nextx[len];        if(ans == 0) puts("0");        else {            for(int i=0;i<ans;i++) putchar(A[i]);            printf(" %d\n",ans);        }    }    return 0;}