编写函数int fun(char *s1,char *s2),在字符串s1中统计字符串s2出现的次数并返回。若s2在s1中未出现,则返回0。例如:
若输入的字符串s1为:abaaAabcaabbabca,字符串s2为:ab ,则程序输出:n=4
若输入的字符串s1为:abaaAabcaabbabca,字符串s2为:abd,则程序输出:No find
答案: int fun(char *s1, char *s2)
{
int i, j, len1 = strlen (s1), len2 = strlen (s2), n=0;
for (i=0; i<=len1-len2; i++)
{
for (j=0; j<len2; j++)
{
if (*(s1+i+j) != *(s2+j))
break;
}
if (j == len2)
{
i += j-1;
n++;
}
}
return n;
}
求每行的注释!!!!!!!!!!1