Oulipo
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3770 Accepted Submission(s): 1485
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.
#include <iostream>
#include <cstring>
using namespace std;
const int MAXN = 1000005;
const int MAXM = 10005;
char text[MAXN];
char pattern[MAXM];
int next[MAXM];
int n,m,t;
void get_next(){
next[0]=next[1]=0;
for(int i=1; i<m; i++){
int j=next[i];
while(j && pattern[i]!=pattern[j]) j=next[j];
next[i+1]=(pattern[i]==pattern[j]? j+1 : 0);
}
}
//void get_next(){
// int i=0, j=-1;
// next[i]=j;
// while(i<m){
// if(j==-1 || pattern[i]==pattern[j]){
// j++; i++;
// next[i]=j;
// }else
// j=next[j];
// }
//}
void kmp(){
int cnt=0,j=0;
for(int i=0; i<n; i++){
while(j && text[i]!=pattern[j]) j=next[j];
if(text[i]==pattern[j]) j++;
if(j==m){
cnt++;
}
}
cout<<cnt<<endl;
}
int main()
{
cin>>t;
while(t--){
cin>>pattern>>text;
m = strlen(pattern);
n = strlen(text);
get_next();
kmp();
}
return 0;
}