Oulipo
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
题意概括:给定两个字符串,输出第一个字符串在第二个字符串中出现的次数。
#include <stdio.h>
#include <string.h>
#define N 1000010
int next[N],f;
void get_next(char p[])
{
int j=0;
int k=-1;
next[0]=-1;
int l=strlen(p);
while(j<l)
{
if(k==-1||p[k]==p[j])
{
k++;
j++;
next[j]=k;
}
else
k=next[k];
}
}
void kmp(char s[],char p[])
{
int j=0;
int i=0;
int l1=strlen(s);
int l2=strlen(p);
while(i<l1)
{
if(s[i]==p[j]||j==-1)
{
j++;
i++;
}
else
j=next[j];//当模式串p读完的时候不会跳出 ,因为while的条件是i<l1,然后读到最后一位的下一位的时候虽然模式串为空了,但是next数组中存的还有一个数就是整个模式串p的最大公共前后缀的大小,所以还会返回继续判断
if(j==l2)
f++;
}
}
int main()
{
int t;
char s[N],p[N];
scanf("%d",&t);
getchar();
while(t--)
{
f=0;
gets(p);
gets(s);
get_next(p);
kmp(s,p);
printf("%d\n",f);
}
return 0;
}