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 wordW and a textT, count the number of occurrences ofW inT. All the consecutive characters of W must exactly match consecutive characters ofT. Occurrences may overlap.
InputThe 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.OutputFor every test case in the input file, the output should contain a single number, on a single line: the number of occurrences of the wordW in the textT.
Sample Input3BAPCBAPCAZAAZAZAZAVERDIAVERDXIVYERDIANSample Output130KMP问题。。。。不过和KMP稍有不同,这个题要求计算次数而不是匹配成功就可以,所以next数组多求一位。AC代码:#include <iostream>#include <stdio.h>#include <string.h>#include <math.h>#include <algorithm>using namespace std;int lent,lenp,next[10005];char p[10005],t[1000005];void getnext(){ int i=0,j=-1; next[0]=-1; while(i!=lenp) { if(j==-1||p[i]==p[j]) { next[++i]=++j; } else { j=next[j]; } }}int KMP(){ int i=0,j=0,c=0; while(i!=lent&&j!=lenp) { if(t[i]==p[j]||j==-1) { ++i,++j; } else { j=next[j]; } if(j==lenp) { c++; j=next[j]; } } return c;}int main(){ int n,i,k,j,ans; scanf("%d",&n); getchar(); while(n--) { scanf("%s%s",p,t); lent=strlen(t); lenp=strlen(p); getnext(); ans=KMP(); printf("%d/n",ans); } return 0;}
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