数学：除法问题生成器

Question

我正在制作一个程序，询问您是否想做加法、减法、乘法和除法。然后它会询问多少个问题。除了除法之外，我一切正常。我想确保在做除法问题时不会有余数。我不知道如何使这项工作。

      else if (player==4) {

      System.out.print("How many questions would you like?-->"); player =
      in.nextInt(); numQuestions= player;


      do{

      //question
          do {
              do{
          num1 = (int) (Math.random() * 100);
          num2 = (int) (Math.random() *10);

              }while (num2 > num1);
          } while (num1 % num2 == 0);


              compAnswer = num1 / num2;



                  System.out.println(num1+" / " + num2 + "=");                  
                  System.out.print("What's your answer? -->");
                  player = in.nextInt();
                    if (player == compAnswer) {
                        System.out.println(" That's right, the answer is "
                                + compAnswer);
                        System.out.println("");
                        score++;
                    } else {
                        System.out.println("That's wrong! The answer was "
                                + compAnswer);
                        System.out.println("");                     

                    }
            //x++;


      }while( x < numQuestions + 1 );

      System.out.println("");
      System.out.println("Thanks for playing! Your score was " + score +
      "."); }

Answer 1

您正在专门挑选有 提醒的号码。您可以在有提醒时循环播放：

} while (num1 % num2 != 0);

但是，您根本不需要循环。如果你选择第二个操作数和答案，你可以计算第一个操作数：

num2 = (int) (Math.random() *10);
compAnswer = (int) (Math.random() * 10);

num1 = num2 * compAnswer;