猜谜游戏有麻烦

我需要做一个猜谜游戏，程序生成一个随机数，用户必须猜出这个数字。如果用户猜出的数字少于 10 次，程序会祝贺他们，并让他们知道他们猜不到 10 次。如果他们的猜测超过 10 次，那么它会让他们知道它高于 10 次，依此类推。

出该数字，然后决定再次使用另一个全新的数字，这次在 8 次尝试中猜到它，而不是仍然祝贺他们，因为它尝试不到 10 次，它计算前一个游戏的 3 次尝试。然后，程序会告诉他们他们尝试了10多次，即使他们没有。我不知道如何解决这个问题。到目前为止，我完成的代码如下：

int main()
{
   srand(time(0));
   int guess;
   int number;
   char selection = 'y';
   int numberOfGuesses=0;
 while(selection == 'y' || selection == 'Y')
 {
      number = rand() % 1000 + 1;
      cout << "I have a number between 1 and 1000.nCan you guess my number?nPlease type your first guess: ";
      cin >>guess;
      do
      {
         if(number > guess)
         {
            cout << "Too low. Try again: " << endl;
            cin >> guess;
            numberOfGuesses++;
         }
         if (number < guess)
         {
            cout << "Too high. Try again: " << endl;
            cin >> guess;
            numberOfGuesses++;
         }
      } 
      while(number != guess);
      if(numberOfGuesses < 9)
      {
            cout << "You guessed the number in less than 10 guesses!n Would you like to play again (y or n)?";
            cin >> selection;
      }
      else if(numberOfGuesses > 9)
      {
            cout << "You guessed the numbern Would you like to play again (y or n)?";
            cin >> selection;
      }
      else if(numberOfGuesses == 9)
      {
            cout << "You guessed the number.n Would you like to play again (y or n)?";
            cin >> selection;
      }
  }
   return 0;
 }