Simpsons’ Hidden Talents
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 10068 Accepted Submission(s): 3486
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.
The lengths of s1 and s2 will be at most 50000.
#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 10;
int t, n, m;
int la, lb;
char a[N], b[N];
int nex[N];
void get_nex(){
int i = 0, j = -1;
nex[0] = -1;
while(i < la){
if(j == -1 || a[i] == a[j]){
i++, j++;
nex[i] = j;
}
else j = nex[j];
}
}
int main(){
while(~scanf("%s%s", a, b)){
la = strlen(a);
a[la] = '#';
a[la+1] = '\0';
strcat(a, b);
la = strlen(a);
get_nex();
if(nex[la]){
for(int i = 0; i < nex[la]; i++){
printf("%c", a[i]);
}
printf(" %d\n", nex[la]);
}
else printf("0\n");
}
}