NSWOJ ~ 1104 ~ BinaryStringMatching（strstr函数）

题意：先给你一个数代表有几组测试数据，然后每组测试数据给你两个字符串A,B，求A在B中出现的次数；

1104 - BinaryStringMatching

时间限制：3秒 内存限制：128兆

题目描述
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

提示

来源
NYOJ

思路：用到一个strstr（string a，string b）函数，函数意思为找到a字符串在b中第一次出现的位置然后返回；用这个函数就很好写了找到第一次出现的位置然后这个坐标加1，出现次数加一，在找，直到找不到；

#include<stdio.h>
#include<string.h>
int strstrcount( char *str1, char *str2 )
{
char *str = str1;
int c = 0;

while( (str = strstr( str, str2 )) != NULL )
{
c++;
str++;
}
return c;
}
main()
{
char str[10001];
int n,i;
scanf("%d",&n);
while(n--)
{
int count=0;
char a[10001];
scanf("%s",str);
scanf("%s",a);
count=strstrcount(a,str);
printf("%d\n",count);
}
}

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