Simpsons’ Hidden Talents
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4543 Accepted Submission(s): 1648Problem 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
marjorieSample Output
0
rie 3Source
题意：找出两个字符串s1与s2中，s1的前缀与s2的后缀的最长公共部分，并输出该串及其长度，如果没有，则直接输出0;
话说自己真是想不到把两个字符串拼接再用next函数～
1 #include <iostream>
2 #include <cstdio>
3 #include <cmath>
4 #include <cstdlib>
5 #include <cstring>
6 #include <cmath>
7 #include <algorithm>
8 #include <cctype>
9 #define N 500015
10 #define INF 1000000
11 #define ll long long
12 using namespace std;
13 int nexts[50050*2];
14 char aa[50050*2],bb[50050];
15 void get_next(char a[])
16 {
17 int j = 0, k = -1;
18 nexts[0] = -1;
19 while(a[j])
20 {
21 if(k == -1 ||a[j] == a[k])
22 nexts[++j] =++k;
23 else
24 k = nexts[k];
25 }
26 }
27 int main(void)
28 {
29 while(cin>>aa>>bb)
30 {
31 int la = (int)strlen(aa);
32 int lb = (int)strlen(bb);
33 strcat(aa,bb);//连接两个字符串，通过找匹配部分，找出两个串最长公共部分
34 get_next(aa);
35 int len = la+lb;
36 while(nexts[len] > la || nexts[len] > lb) len = nexts[len];//如果匹配部分超过两个串中任意一个的长度，就取小的那个
37 len = nexts[len];//取匹配
38 for(int i = 0; i < len; i++) printf("%c",aa[i]);//输出
39 if(len) printf(" ");
40 printf("%d\n",len);
41 }
42 return 0;
43 }