Oulipo
http://acm.hdu.edu.cn/showproblem.php?pid=1686
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4041 Accepted Submission(s): 1595Tout 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<stdio.h>
#include<string.h>
int next[10004];
char t[10004],s[1000004];
void get_nextval(int len)
{
int i=0,k=-1;
next[0]=-1;
while(i<len)
if(k==-1||t[i]==t[k])
{
++i;
++k;
if(t[i]!=t[k])
next[i]=k;
else
next[i]=next[k];
}
else
k=next[k];
}
int KMP(int ls,int lt)
{
get_nextval(lt);
int i=0,j=0,ans=0;
while(i<ls)
if(s[i]==t[j])
{
i++;
j++;
if(j==lt)
{
ans++;
j=next[j];
}
}
else
{
if(next[j]!=-1)
j=next[j];
else
{
j=0;
i++;
}
}
return ans;
}
int main()
{
int cas,lt,ls,ans,i;
scanf("%d",&cas);
while(cas--)
{
memset(next,0,sizeof(next));
scanf("%s%s",t,s);
lt=strlen(t);
ls=strlen(s);
ans=KMP(ls,lt);
printf("%d\n",ans);
}
return 0;
}