Binary String Matching

描述Given two strings A and B, whose alphabet consist only ‘0’ and ‘1’. Your task is only to tell how many times does A appear as a substring of B? For example, the text string B is ‘1001110110’ while the pattern string A is ‘11’, you should output 3, because the pattern A appeared at the posit

 

输入

The first line consist only one integer N, indicates N cases follows. In each case, there are two lines, the first line gives the string A, length (A) <= 10, and the second line gives the string B, length (B) <= 1000. And it is guaranteed that B is always longer than A.

输出

For each case, output a single line consist a single integer, tells how many times do B appears as a substring of A.

样例输入

3
11
1001110110
101
110010010010001
1010
110100010101011 

样例输出

3
0
3 

//Binary String Matching  length (A) <= 10   length (B) <= 1000
#include <stdio.h>
#include <string.h>

#define TRUE 1
#define FALSE 0

#define SIZE_A 10                        //A的最长长度为10
#define SIZE_B 1000                        //B的最长长度为1000

int compare(char *p,char *q,int);            //比较两个字符串是否相等

int main()
{
    char array_a[SIZE_A];
    char array_b[SIZE_B];
    int num;                            //待处理数据的组数
    int count = 0;                        //A串在B中出现的次数
    char *p = 0,*q = 0;

    scanf("%d", &num);
    while(num > 0)                        //处理每一组串
    {
        int len_a,len_b;
        int i = 0;

        count = 0;            
        scanf("%s", array_a);
        fflush(stdin);
        scanf("%s", array_b);
        //gets(array_a);                    //先接收子串a
        //fflush(stdout);
        //gets(array_b);                    //接收字符串b
        
        len_a = strlen(array_a);        //a和b的长度
        len_b = strlen(array_b);
        
        p = array_a;
        q = array_b;

        for(; i <= len_b - len_a; i ++)
        {
            if(compare(p,q,len_a) == TRUE)
                count ++;
            q ++;
        }
        printf("%d\n",    count);            //输出结果

        num --;
    }
}
//比较两个字符串是否相等
int compare(char *p, char *q,int length)
{
    int flag = TRUE;
    int i = 0;

    for(; i < length; i ++)
    {
        if((*p) != (*q))
        {
            flag = FALSE;
            break;
        }
        p ++;
        q ++;
    }

    return flag;
}

这里可以使用strcmp来做。