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.
InputInput consists of two lines. The first line contains s1 and the second line contains s2. You may assume all letters are in lowercase.OutputOutput 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
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<vector>
#include <sstream>
#include<string>
#include<cstring>
#include<list>
using namespace std;
#define MAXN 51000
#define INF 0x3f3f3f3f
typedef long long LL;
/*
求两个串的最长相同前缀后缀匹配
那么可以将两个串连接起来用求Next数组的方法找出所有匹配，选可行（小于两个串长度的）的最大值
*/
char a[MAXN*2],b[MAXN*2];
int Next[MAXN*2];
void kmp_pre(char t[])
{
int m = strlen(t);
int j,k;
j = 0;k = Next[0] = -1;
while(j<m)
{
if(k==-1||t[j]==t[k])
Next[++j] = ++k;
else
k = Next[k];
}
}
int main()
{
while(scanf("%s%s",a,b)!=EOF)
{
int l1 = strlen(a),l2 = strlen(b),L = l1+l2;
stringstream ss;
ss<<a<<b;
ss>>a;
kmp_pre(a);
int ans = Next[L],k = L;
if(ans>l1||ans>l2)
ans = min(l1,l2);
if(ans>0)
{
for(int i=0;i<ans;i++)
printf("%c",a[i]);
printf(" %d\n",ans);
}
else
printf("0\n");
}
return 0;
}