(KMP)Simpsons’ Hidden Talents -- hdu -- 2594
http://acm.hdu.edu.cn/showproblem.php?pid=2594
Simpsons’ Hidden Talents
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4756 Accepted Submission(s): 1732
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.
The lengths of s1 and s2 will be at most 50000.
#include<stdio.h>
#include<string.h>
#include<iostream>
#include<stack>
using namespace std; #define N 100050 int Next[N]; void FindNext(char S[])
{
int i=, j=-;
int Slen = strlen(S); Next[] = -; while(i<Slen)
{
if(j==- || S[i]==S[j])
Next[++i] = ++j;
else
j = Next[j];
}
}
int main()
{
char s1[N], s2[N]; while(scanf("%s%s", s1, s2)!=EOF)
{
int len1=strlen(s1), len2=strlen(s2);
int Min = min(len1, len2);
char S[N], s[N]; strcpy(s, s1);
strcat(s1, s2); FindNext(s1); int len = strlen(s1);
if(Next[len]== && len>)
printf("0\n");
else if(Next[len]>Min)
{
if(len1>len2)
printf("%s %d\n", s2, len2);
else
printf("%s %d\n", s, len1);
}
else
{
memset(S, , sizeof(S));
strncpy(S, s1, Next[len]);
printf("%s %d\n", S, Next[len]);
} }
return ;
}
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