网址：http://acm.nyist.net/JudgeOnline/problem.php?pid=5
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 第一个字符串是子串，所以我们要在第二个字符串中找第一个字符串，又因为数组很小,所以我们完全可以用循环结构来解决这个问题，实际上我也确实只用了0ms #include<cstdio> #include<cstring> using namespace std; const int maxn=1000+10; char s1[maxn],s2[maxn]; int cnt; void match() { int pos=0; int len1=strlen(s1); int len2=strlen(s2); while(pos<len2) { bool flag=true; for(int i=0;i<len1;i++) if(s1[i]!=s2[pos+i]) { flag=false; break; } if(flag) cnt++; pos++; } } int main() { int Test; scanf("%d",&Test); while(Test--) { cnt=0; scanf("%s%s",s1,s2); match(); printf("%d\n",cnt); } return 0; }