NYoj -Binary String Matching-优快云博客

本文介绍了一种解决二进制字符串A作为子串在字符串B中出现次数的问题，通过示例展示了如何使用C++中的find函数进行搜索，并提供了一个完整的代码实现。

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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



使用find函数直接搜索。
int index = b.find(a, 0);
while(index != b.npos){
index = b.find(a, index + 1);
}

#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
#include<algorithm>
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);
while(index != b.npos){

n++;
index = b.find(a, index + 1);
}
cout << n << endl;
}
system("pause");
return 0;
}

转载于:https://www.cnblogs.com/hethatpeople/p/8696436.html