hdu 2594 kmp Simpsons’ Hidden Talents

                   Simpsons’ Hidden Talents

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

0

rie 3

题意：

给出两个字符串，寻找前一个字符串的假前缀和后一的字符串的假后缀，输出最长相等的部分及其长度。

解题思路：

水题一道，刚开始看的时间以为题目中的前后缀是真前后缀，结果是假前后缀。这里的真前后缀不包括字符串本身，而假前后缀包括字符串本身。这一题中就是假前后缀。思路是将两个字符串连接起来，运用next数组的信息，输出结果。

这一题可以加深对next数组的理解。

代码：

#include<stdio.h>                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                #include<string.h>int main(){int i,j,m,lena,lenb;char a[100010];char b[50010];int next[100010];while(scanf("%s%s",a,b)!=EOF){lena=strlen(a);lenb=strlen(b);strcat(a,b);i=1;j=0;m=0;memset(next,0,sizeof(next));while(i<lena+lenb){ if(j==0&&a[i]!=a[j])    {next[i]=0;i++;}  else if(j>0&&a[i]!=a[j])     j=next[j-1];  else   {    next[i]=j+1;    i++;j++;   }    }//这里next数组中最后存入的数字就是所求最长的前后缀。m=next[lena+lenb-1];if(m>=lena) m=lena;if(m>=lenb)  m=lenb;//这两个判断是防止出现aaa  aaaaa，和aaaaa  aaa，这两种情况。if(m>0){a[m]='\0';printf("%s",a);   printf(" ");} printf("%d\n",m); memset(a,0,sizeof(a)); memset(b,0,sizeof(b));}return 0;}