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

KMP板子題

代碼：

#include<cstdio> #include<iostream> #include<cstring> #include<algorithm>using namespace std; void kmp_pre(char x[],int m,int next[]) {int i,j;j=next[0]=-1;i=0;while(i<m) {while(-1!=j && x[i]!=x[j])j=next[j];next[++i]=++j;} } int next[1000010]; int KMP_Count(char x[],int m,char y[],int n) { //x 是模式串，y 是主串int i,j;int ans=0; //preKMP(x,m,next);kmp_pre(x,m,next);i=j=0;while(i<n) {while(-1!=j && y[i]!=x[j])j=next[j];i++;j++;if(j>=m) {ans++;j=next[j];}}return ans; }char a[10005]; char b[1000005]; int main() {int T;cin>>T;while(T--) {scanf("%s",a);scanf("%s",b);printf("%d\n",KMP_Count(a,strlen(a),b,strlen(b)));}return 0; }

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轉載于:https://www.cnblogs.com/Staceyacm/p/10781820.html

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