Description
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 AVERDXIVYERDIANSample Output
1 3 0这道题标解是KMP，但是用hash就可以过，下面是程序：
#include<stdio.h> #include<string.h> #include<iostream> #define ll unsigned long long using namespace std; ll hash[1000005],x,cf[1000005]; char a[1000005]; void read(ll &s,int &l){ s=l=0; char c=getchar(); while(!(c>='A'&&c<='Z')){ c=getchar(); } while(c>='A'&&c<='Z'){ l++; s*=27; s+=c-'A'+1; c=getchar(); } } ll ask(int i,int j){ return hash[j]-hash[i-1]*cf[j-i+1]; } void work(){ int i,l1,l2,s=0; read(x,l1); scanf("%s",a+1); l2=strlen(a+1); hash[1]=a[1]-'A'+1; for(i=2;i<=l2;i++){ hash[i]=hash[i-1]*27+a[i]-'A'+1; } for(i=1;i<=l2-l1+1;i++){ if(ask(i,i+l1-1)==x){ s++; } } printf("%d\n",s); } int main(){ int t,i; for(cf[0]=1,i=1;i<=1000000;i++){ cf[i]=cf[i-1]*27; } scanf("%d",&t); while(t--){ work(); } return 0; }