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. Ilaurait 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.
kmp模板题。
#include<cstdio>
#include<cstring>
char a[10010],b[1000010];
int next[10010];
int main()
{int i,j,k,m,n,x,y,z,T,ans;scanf("%d",&T);while (T--){ans=0;scanf("%s%s",a+1,b+1);m=strlen(a+1);n=strlen(b+1);for (i=2,j=0;i<=m;i++){while (j>0&&a[i]!=a[j+1]) j=next[j];if (a[i]==a[j+1]) j++;next[i]=j;}for (i=1,j=0;i<=n;i++){while (j>0&&(j==m||b[i]!=a[j+1])) j=next[j];if (b[i]==a[j+1]) j++;if (j==m) ans++;}printf("%d\n",ans);}
}