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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
题目意思:求a串在b串出现的次数
直接kmp
#include<iostream> #include<cstdio> #include<cstring> #define N 1000010 using namespace std; int fail[N],f[N],kmp; int main() { freopen("oulipo.in","r",stdin); freopen("oulipo.out","w",stdout); int T; scanf("%d",&T); for (int I= 1;I<=T;I++) { int ans = 0; char s1[N],s2[N]; scanf("%s%s",s2+1,s1+1); int l1=strlen(s1+1),l2=strlen(s2+1); // cout<<l1<<endl<<l2<<endl; fail[1] = 0; for (int i=2;i<=l2;i++) { int p = fail[i-1]; while (p && s2[p+1] !=s2[i]) p=fail[p]; if (s2[p+1] == s2[i]) p++; fail[i] = p; } int p=0; for (int i=1;i<=l1;i++) { while (p && s2[p+1] != s1[i]) p=fail[p]; if (s2[p+1] == s1[i]) p++; f[i] = p; if (p == l2) { ans++; // printf("%d",i-l2+1); //break; } } // for (int i=1;i<=l2;i++) // printf("%d ",fail[i]);printf (" \n"); // for (int i=1;i<=l1;i++) printf("%d ",f[i]); printf("%d\n",ans); } }
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原文地址:http://www.cnblogs.com/liumengyue/p/5494631.html