NYOJ 5 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

来源网络
上传者
naonao

#include <stdio.h>
#include <string.h>
char tmp[11],
str[1001];

int main()
{
int t, i, j, k;
scanf("%d", &t);
while(t--)
{
int cnt = 0;
scanf("%s %s", tmp, str);
int len1 = strlen(tmp);
int len2 = strlen(str);
for(i = 0; i < len2; i++)
{
if(str[i] == tmp[0])
{
k = i + 1;
j = 1;
while(j < len1)
{
if(str[k++] != tmp[j])
break;
j++;
}
if(j == len1)
cnt++;
}
}
printf("%d\n", cnt);
}
return 0;
}