Java如何计算3个保龄球得分的平均值

我正在编写一个程序，计算并显示用户输入的每个保龄球手的平均保龄球分数。 我在计算3个分数的平均值时遇到了麻烦，现在我想这是在计算总分数。我如何使它计算平均分数

public static void main (String [] args)
{



//local constants


  //local variables
    String bowler = "";
    int total = 0;
    int average = 0;
    int score1 = 0;
    int score2 = 0;
    int score3 = 0;

  /********************   Start main method  *****************/

  //Enter in the name of the first bowler
  System.out.print(setLeft(40," Input First Bowler or stop to Quit: "));
  bowler = Keyboard.readString();

  //Enter While loop if input isn't q
  while(!bowler.equals("stop"))
  {

      System.out.print(setLeft(40," 1st Bowling Score:"));
      score1 = Keyboard.readInt();
      System.out.print(setLeft(40," 2nd Bowling Score:"));
      score2 = Keyboard.readInt();
      System.out.print(setLeft(40," 3rd Bowling Score:"));
      score3 = Keyboard.readInt();
      if(score1 >= 0 && score1 <= 300 && score2 >= 0 && score2 <= 300 && score3 >= 0 && score3 <= 300)
      {
          total += score1;
          total += score2;
          total += score3;
          System.out.println(setLeft(41,"Total: ")+ total);
          average = score1 + score2 + score3 / 3;
          System.out.println(setLeft(41,"Average: ") + average);


      }
      else
      {
          System.out.println(setLeft(40,"Error"));

      }

除法/运算符的优先级高于加法运算符+，因此需要在除法前用括号括起和：

average = (score1 + score2 + score3) / 3;
// Here --^------------------------^

---

Java的数学运算符遵循标准的数学优先级，因此

   int average = score1 + score2 + (score3 / 3);

然而，你的意图是可能的

   int average = (score1 + score2 + score3) / 3;

最后，您很可能希望使用double或float算法进行此计算，否则将向下舍入

double average = (double)(score1 + score2 + score3) / 3;

---

到目前为止，平均分的输入和输出是什么？如果我为每个分数输入20，它表示平均分是46，不确定为什么p：在代码上随意地加上括号。+1用于指出整数除法，尽管这个问题的平均值也为int，因此也可能是双精度的