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
来源
网络
上传者
naonao
循环AC

#include <iostream>
#include <cstring> 
#include <string>
#include <cstdlib>
#include <algorithm>
#include <cstdio>
using namespace std;

#define mem(a) memset(a, 0, sizeof(a))

char a[100], b[1100];

int main() {
    int n;
    cin >> n;
    while (n --) {
        mem(a); mem(b);
        cin >> a >> b;
        int res = 0, p1 = strlen(a), p2 = strlen(b);
        int num = 0;
        while (num <= p2-p1) {
            int flag = 1;
            for (int i = num, j = 0; j<p1; j++,i++) {
                if (a[j] != b[i])   flag = 0;
            }
            if (flag)   res ++ ;
            num ++ ;
            //cout << res << endl;
        }
        cout << res << endl;
    }
    return 0;
}

标程：#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。

#include<iostream>
#include<string>
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<<num<<endl;
    }
}