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  • NYOJ5——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 
    #include<iostream>//我用模式匹配算子串出现的次数
    #include<string>
    using namespace std;
    
    int Match(string pat,string sat)
    {
        int count=0;
        int i=0; 
        int m=pat.length(),n=sat.length();
          while(i<=(n-m))
          {
               int j=0;
               while((sat[i]==pat[j])&&(j<pat.length()))
               {
                  i++;
                  j++;
    
               }
               if(j==pat.length())
               {
                  count++;
               }
               i=i-j+1;
          }
          
      
       return count;
    }
    int main()
    {
      
       int num;   
       cin>>num;
       string  pat,sat;
       while(num--)
       {
          int p=0,s=0;
          int count;
          cin>>pat;
          cin>>sat;
    
          count=Match(pat,sat);
          cout<<count<<endl;
       }
          
       return 0;
    }
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  • 原文地址:https://www.cnblogs.com/bxyan/p/4635809.html
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