NYOJ 5-Binary String Matching

题目链接

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

写这题的时候还不会strstr函数,现在看看strstr好像也能做吧,反正是个很水的题:

#include<stdio.h>
int main()
{
int num , i , j , flag ;
char str[11] , str1[1001];
scanf("%d" , &num);
while(num--)
{
scanf("%s\n%s" , str , str1);
for(i = 0 , flag = 0; str1[i] != 0 ; i ++)
{
if(str[0] == str1[i])
{
for(j = 1 ; str[j] != 0 && str1[i + j] != 0 ; j++)
{
if(str[j] != str1[i + j])
break;
}
if(str[j] == 0)
flag++;
}
}
printf("%d\n" , flag);
}
return 0;
}