nyoj 5 Binary String Matching（string)

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 

#include <iostream>
#include <string>
#include <algorithm>
using namespace std;
int main(){
    int test, j, i;
    string a, b;
    int n;
    cin >> test;
    while(test--){
        n = 0;
        cin >> a >> b;
        int len_a = a.length(), len_b = b.length();
        for(i = 0; i < len_b; i++){
            int k = i;
            for(j = 0; j < len_a; k++,j++){
                if(b[k] != a[j])
                    break;
            }
            if(j == len_a)
                n++;
        }
        cout << n << endl;
    }
    return 0;
}

上面是直接遍历。

以下代码利用find()函数

#include <iostream>
#include <string>
using namespace std;

int main(){
    int t;
    string a, b;
    cin >> t;

    while(t--){
        cin >> a >> b;
        int n = 0;
        int index = b.find(a, 0);//返回从0开始找到子串在串中的位置下标
        while(index != b.npos){//npos表示不存在
            n++;
            index = b.find(a, index + 1);
        }
        cout << n << endl;
    }

    return 0;
}