POJ

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

题意：输入测试样例次数t，输入子串W，输入主串T，输出子串在主串中的重复次数。子串在主串中的出现位置可以重叠。

直接套KMP模板即可，适于作为KMP学习的入门题目

#include<stdio.h>
#include<string.h>
int ans;
void kmp_pre(char x[],int m,int next[])
{
    int i,j;
    j=next[0]=-1;
    i=0;
    while(i<m)
    {
        while(j!=-1&&x[i]!=x[j])j=next[j];
        next[++i]=++j;
    }
}
int kmp_count(char x[],int m,char y[],int n)
{
    int i=0,j=0,next[10004];
    kmp_pre(x,m,next);
    while(i<n)
    {
        while(j!=-1&&x[j]!=y[i])j=next[j];
        i++;j++;
        if(j>=m)
        {
            ans++;
            j=next[j];
        }
    }
    return ans;
}
char S[1000006],T[10004];
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        ans=0;
        scanf("%s %s",T,S);
        printf("%d
",kmp_count(T,strlen(T),S,strlen(S)));
    }
}