Oulipo

Description

The French author Georges Perec (1936–1982) once wrote a book, La disparition, without the letter ‘e’. He was a member of the Oulipo group. A quote from the book:

Tout avait Pair normal, mais tout s’affirmait faux. Tout avait Fair normal, d’abord, puis surgissait l’inhumain, l’affolant. Il aurait voulu savoir où s’articulait l’association qui l’unissait au roman : stir son tapis, assaillant à tout instant son imagination, l’intuition d’un tabou, la vision d’un mal obscur, d’un quoi vacant, d’un non-dit : la vision, l’avision d’un oubli commandant tout, où s’abolissait la raison : tout avait l’air normal mais…

Perec would probably have scored high (or rather, low) in the following contest. People are asked to write a perhaps even meaningful text on some subject with as few occurrences of a given “word” as possible. Our task is to provide the jury with a program that counts these occurrences, in order to obtain a ranking of the competitors. These competitors often write very long texts with nonsense meaning; a sequence of 500,000 consecutive ‘T’s is not unusual. And they never use spaces.

So we want to quickly find out how often a word, i.e., a given string, occurs in a text. More formally: given the alphabet {‘A’, ‘B’, ‘C’, …, ‘Z’} and two finite strings over that alphabet, a word W and a text T, count the number of occurrences of W in T. All the consecutive characters of W must exactly match consecutive characters of T. Occurrences may overlap.

Input

The first line of the input file contains a single number: the number of test cases to follow. Each test case has the following format:

One line with the word W, a string over {‘A’, ‘B’, ‘C’, …, ‘Z’}, with 1 ≤ |W| ≤ 10,000 (here |W| denotes the length of the string W).
One line with the text T, a string over {‘A’, ‘B’, ‘C’, …, ‘Z’}, with |W| ≤ |T| ≤ 1,000,000.
Output

For every test case in the input file, the output should contain a single number, on a single line: the number of occurrences of the word W in the text T.

题解
KMP模板。

代码

var
  n,m,t:longint;
  a,b:ansistring;
  next:array [0..10001] of longint;
procedure getnext;
var
  i,j:longint;
begin 
  next[1]:=0;
  j:=0;
  for i:=2 to m do
    begin
      while (j>0) and (b[j+1]<>b[i]) do
        j:=next[j];
      if b[j+1]=b[i] then inc(j);
      next[i]:=j;
    end;
end;

function kmp:longint; 
var
  i,j,cnt:longint;
begin
  cnt:=0; j:=0;
  for i:=1 to n do
    begin
      while (j>0) and (b[j+1]<>a[i]) do
        j:=next[j];
      if b[j+1]=a[i] then inc(j);
      if j=m then
        begin
          inc(cnt);
          j:=next[j];
        end;
      end;
  exit(cnt);
end;

begin
  readln(t);
  while t>0 do
    begin
      fillchar(next,sizeof(next),0);
      readln(b);
      readln(a);
      m:=length(b);
      n:=length(a);
      getnext;
      writeln(kmp);
      dec(t);
    end;
end.