nyoj 5-Binary String Matching

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于 2019-06-07 19:12:53 发布

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本文深入探讨了字符串匹配算法，特别是针对由'0'和'1'组成的字符串A和B，如何高效地找出A作为子串在B中出现的次数。通过实例解析，详细介绍了算法的实现过程，并提供了完整的代码示例。

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题目描述:

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

输入描述:

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.

输出描述:

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:

复制

3
11
1001110110
101
110010010010001
1010
110100010101011

样例输出:

3
0
3

#include<iostream>
using namespace std;
int main(){
    int n;
    cin>>n;
    while(n--){
        int m=0;
        string a;
        string b;
        cin>>a;
        cin>>b;
        int j=0;
        for(int i=0;i<=b.size()-1;i++){
            if((int)b.find(a,i)==-1)
                break;
            else{
                 j=(int)b.find(a,i);
                 m++;
                 i=j;
            }
        }
        cout<<m<<endl;
    }
}