Simpsons' Hidden Talents hdu2594 kmp

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.

Solution

一开始连题目都看错了，如果只是求b串最长的前缀同时是a串的子串的话就是傻逼题了

已知next[i]表示1..next[i]与i-next[i]+1..i完全相等，那么把两个字符串连起来对自己求next，next[strlen(str)]就是答案了

字符串来自c++的恶意

Code

var
nex:array[0..100001]of longint;
a,b:ansistring;
procedure getNex;
var
i,j:longint;
begin
nex[1]:=0;
j:=0;
for i:=2 to length(a) do
begin
while (a[i]<>a[j+1])and(j>0) do j:=nex[j];
if (a[i]=a[j+1]) then inc(j);
nex[i]:=j;
end;
if nex[length(a)]>length(a)-length(b) then nex[length(a)]:=length(a)-length(b);
if nex[length(a)]>length(b) then nex[length(a)]:=length(b);
if nex[length(a)]>0 then
write(copy(a,1,nex[length(a)]),' ');
writeln(nex[length(a)]);
end;
begin
while not eof do
begin
readln(a);
readln(b);
a:=a+b;
getNex;
end;
end.