【KMP】Oulipo
KMP算法
求串内匹配数,计数时返回next[]位置。
Problem Description
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
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
Sample Input
3
BAPC
BAPC
AZA
AZAZAZA
VERDI
AVERDXIVYERDIAN
Sample Output
1
3
0
Source
#include<stdio.h>
#include<memory.h>
#include<string.h>
char a[],b[];
int Next[];
void getn(int s){
int i=,j=-;
memset(Next,,sizeof(Next));
Next[]=-;
while(i<s){
if(j==-||a[i]==a[j]){
i++;
j++;
Next[i]=j;
}
else j=Next[j];
}
}
int getnum(int x,int y){
int i=,j=,k=;
while(i<y&&j<x){ //j<x避免死循环
if(j==-||b[i]==a[j]){ i++; j++; }
else j=Next[j];
if(j==x){
k++;
j=Next[j]; //返回对应next[]位置
}
}
return k;
}
int main()
{
int n,x,y;
scanf("%d",&n);
while(n--){
scanf("%s%s",a,b);
x=strlen(a);
y=strlen(b);
getn(x);
printf("%d\n",getnum(x,y));
}
return ;
}
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