1、题目信息
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
2、题目大致意思
第一次输入一个整数n，代表此时组数，接着输出n组测试用例，每组包含两行a、b字符串，a的长度<=10,b的长度<=1000
输出b中包含多少个a。
3、思路
1）使用c++ 中string类来处理
2）对于遍历字符串b，每次取出长度为a的字串，进行判断
4、代码
#include <iostream>
#include <string>
using namespace std;
int main()
{
string a,b;
int n,m;
cin>>n;
while(n--){
m=0;
cin>>a;
cin>>b;
for(int i=0;i<b.length();i++){
if(b.substr(i,a.length())==a) m++;
}
cout<<m<<endl;
}
return 0;
}