Oulipo
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7200 Accepted Submission(s): 2867
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
KMP模板题。注意k的赋值,题目要求是包括几个 不是能切割几个。如asdasd asdasdasd 这样的情况下结果应该是2 所以当k==plen后再计算,k的赋值不能从0再開始,而是从pre[k]開始。
#include <stdio.h> #include <string.h> #include <algorithm> #define maxn 1000000+10 using namespace std; int pre[maxn]; char T[maxn]; char P[10010]; void getpre(int plen) { pre[0]=pre[1]=0; for(int i=1;i<plen;i++){ int k=pre[i]; while(k&&P[i]!=P[k])k=pre[k]; pre[i+1]=P[i]==P[k]?k+1:0; } } int search(int slen,int plen) { int num=0; getpre(plen); int k=0; for(int i=0;i<slen;i++){ while(k&&P[k]!=T[i])k=pre[k]; if(P[k]==T[i])k++; if(k==plen){ ++num; k=pre[k]; //从0開始是挪动了plen。题目是找出包括几个,不是能切割几个。所以从pre[k]開始。 } } return num; } int main() { int t; scanf("%d",&t); while(t--){ memset(pre,0,sizeof(pre)); scanf("%s%s",P,T); int slen=strlen(T); int plen=strlen(P); int res=search(slen,plen); printf("%d ",res); } return 0; }