NYOJ 5--Binary String Matching【string】_c. another problem on strin-优快云博客

本文链接：https://blog.youkuaiyun.com/hpuhjh/article/details/44180135

本文介绍了一种解决二进制字符串匹配问题的方法，通过使用C++标准库中的字符串操作函数find()，实现对给定的两个仅由'0'和'1'组成的字符串A和B进行匹配，统计字符串A作为子串在B中出现的次数。

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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

STL真是太强大了，string中find()函数的使用，想学习的朋友可以看一下这篇文章：http://www.jb51.net/article/37560.htm

#include <iostream>
#include <algorithm>
#include <string>
using namespace std;
string str,s;
int main() {
  int n;
  cin>>n;
    while(n--){
      cin>>s>>str;
      int num = 0; unsigned ans = 0;
      while ((ans=str.find(s,ans))!=string::npos){
        num++;
        ans++;
      }
      cout<<num<<endl;
    }
  return 0;
}