Description
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 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.
Input
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:
- 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.
Output
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.
Sample Input
3 BAPC BAPC AZA AZAZAZA VERDI AVERDXIVYERDIAN
Sample Output
1 3 0
![](https://images2015.cnblogs.com/blog/892758/201610/892758-20161016173405311-1832175163.png)
hash_AC代码:
/* Problem: 3461 User: bbsh Memory: 5296K Time: 110MS Language: G++ Result: Accepted */ #include<cstdio> #include<cstring> using namespace std; const int N=1e6+5; typedef int i64; i64 S,chk,p,P,hash_key[N]; char s1[N],s2[N]; int cas,l1,l2,ans; inline i64 fpow(i64 a,i64 p){ i64 res=1; for(;p;p>>=1,a=a*a) if(p&1) res=res*a; return res; } inline void get_s2_key(){ p=fpow(P,l2);S=0; for(int i=1;i<=l2;i++) S=S*P+s2[i]-'A'; } inline void get_s1_key(){ hash_key[0]=0; for(int i=1;i<=l1;i++) hash_key[i]=hash_key[i-1]*P+s1[i]-'A'; } inline i64 query(int x,int y){ return hash_key[y]-hash_key[x-1]*p; } int main(){ P=29;//prime_num for(scanf("%d",&cas);cas--;){ scanf("%s%s",s2+1,s1+1);ans=0; l1=strlen(s1+1);l2=strlen(s2+1); get_s2_key(); get_s1_key(); for(int i=l2;i<=l1;i++){ chk=query(i-l2+1,i); if(chk==S) ans++; } printf("%d ",ans); } return 0; }
KMP_AC代码:
/* Problem: 3461 User: bbsh Memory: 1424K Time: 110MS Language: G++ Result: Accepted Source Code */ #include<cstdio> #include<cstring> using namespace std; const int N=1e6+5; char s1[N],s2[N]; int cas,l1,l2,fail[N]; void get_next(){ int p=0;fail[1]=0; for(int i=2;i<=l2;i++){ while(p>0&&s2[i]!=s2[p+1]) p=fail[p]; if(s2[i]==s2[p+1]) p++; fail[i]=p; } } void kmp(){ int p=0,ans=0; for(int i=1;i<=l1;i++){ while(p>0&&s1[i]!=s2[p+1]) p=fail[p]; if(s1[i]==s2[p+1]) p++; if(p==l2) ans++,p=fail[p]; } printf("%d ",ans); } int main(){ for(scanf("%d",&cas);cas--;){ scanf("%s%s",s2+1,s1+1); l1=strlen(s1+1);l2=strlen(s2+1); get_next(); kmp(); } return 0; }
附KMP算法详细流程讲解