HDU2594 Simpsons’ Hidden Talents(kmp,next的性质)

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

clintonhomerriemannmarjorie

Sample Output

0rie 3

思路

给你两个字符串，问你第一个串的前缀和第二个串的后缀最大能匹配多少个，如果没有就输出0，有的话输出这个字符串并且输出长度

代码

#include <cstdio>#include <cstring>#include <string>#include <set>#include <iostream>#include <stack>#include <queue>#include <vector>#include <algorithm>#define mem(a,b) memset(a,b,sizeof(a))#define inf 0x3f3f3f3f#define mod 10007#define debug() puts("what the fuck!!!")#define N 200000+20#define ll long longusing namespace std;int nxt[N];void get_next(string s){    int len=s.length();    int j=0,k=-1;    nxt[0]=-1;    while(j<len)        if(k==-1||s[j]==s[k])            nxt[++j]=++k;        else            k=nxt[k];}int main(){    string s1,s2,s;    while(cin>>s1>>s2)    {        s=s1+"#"+s2;        get_next(s);        int len=s.length();        int ans=nxt[len];        if(!ans)            puts("0");        else            cout<<s.substr(0,ans)<<" "<<ans<<endl;    }    return 0;}