A1061 Dating (20 minutes | string processing, annotated in detail, logic analysis)

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EDITORIAL

• Ideas analysis
  • According to the method found equal to the character determination subject, the output time of less than 2 bits to be added in front of 0, i.e., the output% 02d
• 10 minutes a title

Test Case

• input:
  3485djDkxh4hhGE 
  2984akDfkkkkggEdsb 
  s&hgsfdk 
  d&Hyscvnm
  
  output:
  THU 14:04
  

ac Code

• #include <iostream>
  #include <cctype>
  
  using namespace std;
  int main()
  {
  
      string a, b, c, d;
      cin >> a >> b >> c >> d;
      char t[2];
      int pos, i=0, j=0;
      while(i < a.length() && i<b.length())
      {
          if(a[i]==b[i] && (a[i]>= 'A' && a[i]<='G'))
          {
              t[0] = a[i];
              break;
          }
          i++;
      }
      i = i+1;
      while(i<a.length() && i<b.length())
      {
          if(a[i]==b[i] && ((a[i]>= 'A' && a[i]<='N') || isdigit(a[i])))
          {
              t[1] = a[i];
              break;
          }
          i++;
      }
  
      while(j<c.length() && j<d.length())
      {
          if(c[j]==d[j] && isalpha(c[j]))
          {
              pos = j;
              break;
          }
          j++;
      }
      string week[7] = {"MON ", "TUE ", "WED ", "THU ", "FRI ", "SAT ", "SUN "};
      int m = isdigit(t[1]) ? t[1]-'0' : t[1]-'A' + 10;
      cout << week[t[0]-'A'];
      printf("%02d:%02d", m, pos);
  
      return 0;
  }