Simpsons’ Hidden Talents【KMP模板题】

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

  这道题就是输出前一个元素与后一个元素的最大前后缀匹配，就是先求一遍前缀的nex[]再对应比对最后一个串的最后一个元素的对应的K的值即可。

#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <limits>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#define lowbit(x) ( x&(-x) )
#define pi 3.141592653589793
#define e 2.718281828459045
#define INF 0x3f3f3f3f
using namespace std;
typedef unsigned long long ull;
typedef long long ll;
const int maxN = 5e4 + 7;
char s1[maxN], s2[maxN];
int nex[maxN], pre[maxN];
void cal_nex(int len)
{
    nex[0] = nex[1] = 0;
    int k = 0;
    for(int i=2; i<=len; i++)
    {
        while(k > 0 && s1[k+1] != s1[i]) k = nex[k];
        if(s1[k+1] == s1[i]) k++;
        nex[i] = k;
    }
}
int solve(int len)
{
    int k = 0;
    for(int i=1; i<=len; i++)
    {
        while(k > 0 && s1[k+1] != s2[i]) k = nex[k];
        if(s1[k+1] == s2[i]) k++;
        if(i == len) return k;
    }
    return 0;
}
int main()
{
    while(scanf("%s%s", s1 + 1, s2 + 1)!=EOF)
    {
        int len = (int)strlen(s1 + 1);
        cal_nex(len);
        int ans = solve((int)strlen(s2 + 1));
        if(ans == 0) { printf("0\n");   continue; }
        for(int i=1; i<=ans; i++) printf("%c", s1[i]);
        printf(" %d\n", ans);
    }
    return 0;
}