Oulipo
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3809 Accepted Submission(s): 1504
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
题意:给定a串,b串,统计a串在b串中出现多少次。
解题思路:kmp,根据next数组统计,纠结了好久。
代码:
#include <iostream> #include <stdio.h> #include <string> using namespace std; int next[10005], num; void get_next(char *str) { int len=strlen(str); int i,j=0,k=-1; next[0]=-1; while(j<len) { if(k==-1||str[j]==str[k])next[++j]=++k; else k=next[k]; } } int kmp(char *x,char *y) { int len1=strlen(x); int len2=strlen(y); int i=0,j=0,ans=0; get_next(x); while(i<len2) { while(j!=-1&&y[i]!=x[j])j=next[j]; i++;j++; if(j>=len1) { ans++; j=next[j]; } } return ans; } char a[10005], b[1000050]; int main() { //freopen("data.in","r",stdin); //freopen("data.out","w",stdout); int T; scanf("%d",&T); while(T--) { scanf("%s%s",a,b); printf("%d\n",kmp(a,b)); } return 0; }