Oulipo
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4004 Accepted Submission(s): 1580
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.
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.
3 BAPC BAPC AZA AZAZAZA VERDI AVERDXIVYERDIAN
1 3 0
//109MS 1252K
#include<stdio.h>
#include<string.h>
char text[1000007],pattern[10007];
int next[10007],n,m;
void pre()
{
next[0]=-1;
int j=-1;
for(int i=1;i<m;i++)
{
while(j>=0&&pattern[j+1]!=pattern[i])j=next[j];
if(pattern[j+1]==pattern[i])j++;
next[i]=j;
}
}
int kmp()
{
int ans=0,j=-1;
for(int i=0;i<n;i++)
{
while(j>=0&&pattern[j+1]!=text[i])j=next[j];
if(pattern[j+1]==text[i])j++;
if(j==m-1)ans++;
}
return ans;
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
scanf("%s%s",pattern,text);
m=strlen(pattern);
n=strlen(text);
pre();
printf("%d\n",kmp());
}
return 0;
}