HDU 2594 Simpsons’ Hidden Talents

题目：

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
题意描述：
输入两个串s1和s2
计算并输出s1串的前缀和s2的后缀的最长相似部分和它的长度
解题思路：
KMP模板题，不同在于让KMP函数匹配过程中让s1跑完，最后返回最相似度j即可。
代码实现：

 1 #include<stdio.h>
 2 #include<string.h>
 3 char s1[50010],s2[50010];
 4 int kmp(char s[],char t[]);
 5 void get_next(char t[],int next[],int l);
 6 int main()
 7 {
 8     int i,j,k;
 9     while(scanf("%s%s",s1,s2) != EOF)
10     {
11         if( (k=kmp(s2,s1)) > 0 )//加括号 
12         {
13             for(i=0;i<k;i++)
14                 printf("%c",s1[i]);
15             printf(" %d
",k);
16         }
17         else
18             printf("0
");
19     }
20     return 0;
21 }
22 int kmp(char s[],char t[])
23 {
24     int i,j,l1,l2;
25     l1=strlen(s);
26     l2=strlen(t);
27     int next[50010];
28     get_next(t,next,l2);
29     i=0;
30     j=0;
31     while(i < l1)
32     {
33         if(j==-1 || s[i] == t[j])
34         {
35             i++;
36             j++;
37         }
38         else
39             j=next[j];
40     }
41     //printf("j=%d
",j);
42     return j;
43 }
44 void get_next(char t[],int next[],int l)
45 {
46     int i,j;
47     i=0;j=-1;
48     next[0]=-1;
49     while( i < l)
50     {
51         if(j==-1 || t[i]==t[j])
52         {
53             i++;
54             j++;
55             next[i]=j;
56         }
57         else
58             j=next[j];
59     }
60     /*for(i=0;i<=l;i++)
61         printf("%d ",next[i]);
62     printf("
");*/
63 }

易错分析：

1、注意直接使用函数返回值作为判断时最好将其赋值给一个变量，避免重复调用