Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 17128 | Accepted: 6892 |
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:
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 AVERDXIVYERDIAN
Sample Output
1 3 0
#include<iostream> #include<cstring> #include<string> #include<cstdio> using namespace std; char a[10005],p[1000005]; int next[10005],res,len1,len2; void getnext() { int i,j; next[0]=0; next[1]=0; for(i=1;i<len1;i++) { j=next[i]; while(j&&a[i]!=a[j]) j=next[j]; if(a[i]==a[j]) next[i+1]=j+1; else next[i+1]=0; } } void KMP() { int i,j=0; for(i=0;i<len2;i++) { while(j&&p[i]!=a[j]) j=next[j]; if(p[i]==a[j]) j++; if(j==len1) { j=next[j]; //匹配的时候可以重合. res++; } } } int main() { int n; cin>>n; while(n--) { scanf("%s%s",a,p); res=0; len1=strlen(a),len2=strlen(p); getnext(); KMP(); printf("%d\n",res); } return 0; }