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HDU 2594 Simpsons’ Hidden Talents

2024-08-12 23:36:06   219人阅读

题目链接：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): 7354    Accepted Submission(s): 2620

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
clinton
homer
 
 
riemann
marjorie
 
Sample Output
0
rie 3
题目大意： 每次输入两个字符串s1，s2 然后找到一个字符串使他满足既是s1的前缀又是s2的后缀   如果能找到输出该字符串及它的长度 否则输出0
解题思路：KMP算法  找出s1的next数组然后将s2数组跟s1数组进行逐个比较  
技术分享
 1 #include <stdio.h> 2 #include <string.h> 3 char s1[50050],s2[50050]; 4 int next[50050]; 5 void getnext() 6 { 7     int i = 0,j = -1; 8     next[0] = -1; 9     int len = strlen(s1);10     while (i < len)11     {12         if (j == -1 || s1[i] == s1[j])13         {14             i ++;15             j ++;16             if (s1[i] == s1[j])17                 next[i] = next[j];18             else19                 next[i] = j;20         }21         else22             j = next[j];23     }24 }25 int kmp()26 {27     int i = 0,j = 0;28     int len1 = strlen(s1);29     int len2 = strlen(s2);30     while (i < len2)  //特别注意 跟模板有少许的差别31     {32         if (j == -1 || s1[j] == s2[i])33         {34             i ++;35             j ++;36         }37         else38             j = next[j];39     }40     return j;41 }42 int main()43 {44     while (~scanf("%s",s1))45     {46         scanf("%s",s2);47         getnext();48         int len = kmp();49         50         for (int i = 0; i < len; i ++)51             printf("%c",s1[i]);52         if (len)53             printf(" ");54         printf("%d\n",len);55     }56     return 0;57 }
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HDU 2594 Simpsons’ Hidden Talents


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