Binary String Matching

Binary String Matching

时间限制：3000 ms  |  内存限制：65535 KB

难度：3

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 

#include string.h
#include stdio.h

#define MAX_DATA_LEN 1000
#define MAX_VALUE_LEN 10

void BinaryStringMatching(char *data, char* value)
{
    int di;
    int vj;
    int matchTime = 0;
    int dataLen = strlen(data);
    int valueLen = strlen(value);

    for ( di = 0; di < dataLen; ++di )
    {
       for ( vj = 0; vj < valueLen; ++vj )
       {
            if (value[vj] != data[di+vj])
                break;
       }
       if (vj == valueLen) // match
       {
           ++matchTime;
       }
    }

    printf( "%d\n", matchTime );
}

    int
main( int argc, char **argv )
{
    char data[MAX_DATA_LEN];
    char value[MAX_VALUE_LEN];
    int dataLen;
    int valueLen;
    int i;
    int n;

    scanf("%d", &n);
    for ( i = 0; i < n; ++i )
    {
        scanf("%s", value);
        scanf("%s", data);
        BinaryStringMatching(data, value);
    }

    return 0;
}