匹配所有连续子串

匹配所有连续子串

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

思想

每在B中遇到与A的第一位相同的字符时 , 如果从这个字符到最后一个字符的长度大于等于A字符的长度 , 就校验该字符后几个字符与A剩下的字符是否一一相等 .

实现

#include <stdio.h>
#include <string.h>
int main()
{
    int T;
    scanf("%d", &T);
    while(T--) {
        char s[11];
        char str[1001];
        scanf("%s", s);
        scanf("%s", str);
        int str_len = strlen(str);
        int s_len = strlen(s);
        int flag, c = 0;
        for(int i = 0; i <= str_len - s_len; i++) {
            if(str[i] == s[0]) {
                flag = 1;
                for(int j = i + 1; j < i + s_len; j++) {
                    if(str[j] != s[j - i]) {
                        flag = 0;
                        break;
                    }
                }
                c += flag;
            }
        }
        printf("%d\n", c);
    }
    return 0;
}