Binary String Matching string(stl) find()函数的应用 完全不用kmp

题目来源：http://acm.nyist.net/JudgeOnline/problem.php?pid=5

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 


分析： string 中的find(string&,location)  ，在一个字符串中查找指定的单个字符或字符组。如果找到，返回首次匹配的开始位置，如果没有找到匹配的内容，则返回 string::npos。,一般有2个输入参数，一个是待查询的字符串，一个是查询的起始位置，默认起始位置为0.
代码如下：

#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <cstdio>
#include <cmath>
#include <iostream>
#include <vector>
#include<string>
#include<cstring>
#include<string.h>
#include<set>
#include<queue>
using namespace std;
typedef long long ll;


int main()
{

    int t,ans;
    cin>>t;
    while(t--)
    {
        string p,t;
        int num=0;
        cin>>p>>t;
        ans=t.find(p);
        while(ans!=string::npos)
        {
            num++;
            ans=t.find(p,ans+1);
        }
        cout<<num<<endl;

    }
    return 0;
}