Time Limit: 1000MS
Memory Limit: 65536K
Total Submissions: 45515
Accepted: 18181
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.
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:
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.
3
BAPC
BAPC
AZA
AZAZAZA
VERDI
AVERDXIVYERDIAN
1
3
0
#include
#include
#include
#include
using namespace std;
int n,k,len1,len2;
int next1[];
char s1[];
char s2[];
long long HASH[];
long long E[],M=;
long long EE = ;
int init()
{
long long Key=;
int ans=;
memset(E,,sizeof(E));
memset(HASH,,sizeof(HASH));
E[len2]=;
for (int i=len2-;i>=;i--)
E[i]=E[i+]*EE%M;
for (int i=;i<=len2;i++)
HASH[]=(HASH[]+E[i]*(s1[i-]))%M;
for (int i=;i<=len2;i++)
Key=(Key+E[i]*(s2[i-]))%M;
if (HASH[]==Key) ans++;
for (int i=;i<=len1-len2+;i++)
{
HASH[i]=(HASH[i-]-s1[i-]*E[]%M+M)%M*EE%M+s1[i-+len2];
if (HASH[i]==Key) ans++;
}
printf("%d\n",ans);
}
int main(){
int n;
cin >> n;
for (int i=;i<=n;i++){
scanf("%s",s2) ;
scanf("%s",s1) ;
len1=strlen(s1);
len2=strlen(s2);
init();
}
return ;
}
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