题目意思:找到上串在下串中有多少个
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.
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.
3 BAPC BAPC AZA AZAZAZA VERDI AVERDXIVYERDIAN
1 3 0
#include<stdio.h> #include<string.h> int next[10005],lenw,lent; char W[10005],T[1000005]; void set_next() { int i=0,j=-1; next[0]=-1; lenw=strlen(W); while(i<lenw) { if(j==-1||W[i]==W[j]) { i++;j++; next[i]=j; } else j=next[j]; } } int KMP() { int k=0,i=0,j=0; lent=strlen(T); while(i<lent) { if(T[i]==W[j]||j==-1) { i++;j++; } else j=next[j]; if(j==lenw) { k++; j=next[j];//表示W的第j个以前都与T的从i-next[j]到第i-1这段已经匹好了 } //next[j]的值是在W[j]前面的串中最长前缀和最长后缀相等,那么 //当前的匹配成功,就说明W的第j个以前都与T的从i-next[j]到第i-1这段也已经匹配好了 } return k; } int main() { int t; scanf("%d",&t); while(t--) { getchar(); scanf("%s %s",W,T); set_next(); printf("%d\n",KMP()); } }