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.