Question

我试图随机生成一个方程式，该方程式也有50％的错误机会，并显示错误的答案。不正确的答案应该有-2，-1，+ 1或+2的错误。

有时我的代码打印出这样的分割方程式（我无法发布图像）： 2/10 = 13 1/5 = 43 等

我无法弄清楚为什么等式显示未经检查的数字组合？

（首先在我的 onCreateView 方法中调用 generateNumbers（）

public void generateNumbers() {
    //randomly generate 2 numbers and an operator
    number1 = (int) (Math.random() * 10) + 1;
    number2 = (int) (Math.random() * 10) + 1;
    operator = (int) (Math.random() * 4) + 1;
    //50% chance whether the displayed answer will be right or wrong
    rightOrWrong = (int) (Math.random() * 2) + 1;
    //calculate the offset of displayed answer for a wrong equation (Error)
    error = (int) (Math.random() * 4) + 1;
    generateEquation();
}

public void generateEquation() {
    StringBuilder equation = new StringBuilder();
    //append the first number
    equation.append(number1);
    //generate/append the operator and calculate the real answer
    if (operator == 1) {
        equation.append(" + ");
        actualAnswer = number1 + number2;
    } else if (operator == 2) {
        equation.append(" - ");
        actualAnswer = number1 - number2;
    } else if (operator == 3) {
        equation.append(" x ");
        actualAnswer = number1 * number2;
    } else if (operator == 4) {
        if ((number1%number2==0) && (number1>number2)) {
            actualAnswer = number1 / number2;
        } else {
            generateNumbers();
        }
        equation.append(" / ");

    }
    //append the second number and the equals sign
    equation.append(number2 + " = ");

    //we will display the correct answer for the equation
    if (rightOrWrong == 1) {
        displayedAnswer = actualAnswer;
        equation.append(displayedAnswer);
    }
    //we will display an incorrect answer for the equation
    //need to calculate error (-2, -1, +1, +2)
    else {
        if (error == 1) {
            displayedAnswer = actualAnswer - 1;
        } else if (error == 2) {
            displayedAnswer = actualAnswer - 2;
        }else if (error == 3) {
            displayedAnswer = actualAnswer + 1;
        }else {
            displayedAnswer = actualAnswer + 2;
        }
        //append the displayed answer with error
        equation.append(displayedAnswer);

    }
    questionNumber.setText("You have answered " + count + " out of 20 questions");
    finalEquation.setText(equation.toString());
}

Answer 1

我认为您需要在调用return之后添加generateNumbers声明

} else if (operator == 4) {
    if ((number1%number2==0) && (number1>number2)) {
        actualAnswer = number1 / number2;
    } else {
        generateNumbers();
    }
    equation.append(" / ");

}

因为这将重新启动整个过程，而不是继续使用更多数字。

Answer 2

更改您的代码以解除低机会分裂问题：

public void generateNumbers() {
    number1 = (int) (Math.random() * 10) + 1;
    number2 = (int) (Math.random() * 10) + 1;

    //don't get operator here
    //operator = (int) (Math.random() * 4) + 1;

    rightOrWrong = (int) (Math.random() * 2) + 1;
    error = (int) (Math.random() * 4) + 1;

    //don't generate equation here, start (in the View with generateEquation instead of generate numbers
    //generateEquation();
}

public void generateEquation() {
    StringBuilder equation = new StringBuilder();
    generateNumbers();

    //determine operator here
    operator = (int) (Math.random() * 4) + 1;


    equation.append(number1);
    if (operator == 1) {
        equation.append(" + ");
        actualAnswer = number1 + number2;
    } else if (operator == 2) {
        equation.append(" - ");
        actualAnswer = number1 - number2;
    } else if (operator == 3) {
        equation.append(" x ");
        actualAnswer = number1 * number2;
    } else if (operator == 4) {
        equation.append(" / ");
        // generate new numbers if they are not suiteable
        while((number1%number2!=0) && (number1<number2))
        {
            generateNumbers();
        }
        actualAnswer = number1 / number2;
    }
    ......

随机生成方程和答案

2 个答案: