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.
Sample Input
3 BAPC BAPC AZA AZAZAZA VERDI AVERDXIVYERDIAN
Sample Output
1 3 0
题目大意:找出主串中有多少子串
注意:每次找到在主串中找到一个子串,就相当于在此位置没有匹配上,因为没有匹配上就会再接着往下匹配
if(j==len2){
count++;
j=next[j];
}
AC代码
#include <stdio.h>
#include <string.h>
char a1[1000100],a2[10100];
int next[10100],len1,len2;
void getnext(){
int j=0,i=-1;
next[0]=-1;
while(j<len2){
if(i==-1||a1[j]==a2[i]){
j++;i++;
next[j]=i;
}
else i=next[i];
}
}
int kmp(){
int count=0;
int i=0,j=0;
while(i<len1){
if(j==-1||a1[i]==a2[j]){
i++;
j++;
}
else j=next[j];
if(j==len2){
count++;
j=next[j];
}
}
return count;
}
int main(){
int w;
scanf("%d",&w);
getchar();
while(w--){
memset(next,0,sizeof(next));
scanf("%s%s",a2,a1); //a1主串
getchar();
len1=strlen(a1),len2=strlen(a2);
getnext();
printf("%d\n",kmp());
}
return 0;
}