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 

思路：这题就是就检测，用到for循环。不难

AC代码：

#include
#include
#include
using namespace std;
int main()
{
	int n,i,j,al,bl,f,s;
	char a[11],b[1000];
	while(scanf("%d",&n)!=EOF)
	{
		while(n--)
		{
			s=0;
			scanf("%s%s",a,b);
			al=strlen(a);
			bl=strlen(b);
            for(i=0;i

  标程给我的收获是：1、#include中的find()函数的应用。另外，m!=string::npos  意思是：m不等于字符串的尾部。 标程：

#include
#include
using namespace std;
int main()
{
	string s1,s2;
	int n;
	cin>>n;
	while(n--)
	{
		cin>>s1>>s2;
		unsigned int m=s2.find(s1,0);
		int num=0;
		while(m!=string::npos)
		{
			num++;
			m=s2.find(s1,m+1);
		}
		cout<