KMP(2)--hdu2594

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Simpsons’ Hidden Talents

Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u

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

这题的题意是求s1的前缀和s2的后缀重复度的最大值：输入 s1 s2 问s2的后缀以及s1的前缀中相同的字符最多有多少个是什么

水题，就是将s2作为文本串，s1某个前缀作为模式串，问在s2中能匹配到s1的最大前缀,但是注意它问的是s2的后缀，所以在s2中找到的s1的某个前缀如果不是在s2串的末尾，就继续找,直到找到的s1的某个前缀是在s2的最末尾。

应该说清楚了吧......

#include<stdio.h>#include<string.h>#define MAX 50005int next[MAX];char str1[MAX],str2[MAX];void GetNextval(){int plen=strlen(str1);int i=0;int j=next[0]=-1;while(i<plen){if(j==-1||str1[i]==str1[j]){next[++i]=++j;}else j=next[j];}}int KMP(){int len=strlen(str2);int plen=strlen(str1);int i=0;int j=0;while(i<len){if(j==plen){j=next[j];}if(j==-1||str2[i]==str1[j]){i++;j++;}else j=next[j];}return j;      //返回找到的s1前缀的最末尾的字符的下标}int main(){while(scanf("%s%s",str1,str2)!=EOF){GetNextval();str1[KMP()]='\0';if(KMP())printf("%s %d\n",str1,KMP());else printf("0\n");}return 0;}s[i+len]=s[i];}s[i+len]='\0';GetNextval();if(KMP())printf("yes\n");    //根据下标输出找到的最长的s1的前缀else printf("no\n");}return 0;}

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