Lab 1: Quizer#29
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DudkoAndrei wants to merge 24 commits into
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Rrenkens
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Dec 7, 2022
| this.answer = answer; | ||
| } | ||
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| @Override |
Owner
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Вот этот метод можно было объединить с MathTask и сделать какой-то абстрктный класс.
| * исключение, то и тут выбрасывается исключение. | ||
| */ | ||
| public Task generate() { | ||
| Collections.shuffle(generators); |
Owner
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Можно было 1 раз зашафлить и ходить по кругу. Shuffle дорогая операция и явно здесь не требуется больше одного раза.
| this.allowDuplicate = allowDuplicate; | ||
| this.tasks = new ArrayList<>(tasks); | ||
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| if (!allowDuplicate) { |
Owner
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Надо еще проверить, что в изначальных задачах нету дубликатов.
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| private Task currentTask = null; | ||
| private final int taskCount; | ||
| private boolean answeredOnCurrentTask = false; |
Owner
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Мне кажется, что лучше хранить последний результат.
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| double getRandomDouble() { | ||
| return truncate(rnd.nextDouble(minNumber, maxNumber + Double.MIN_VALUE), precision); |
| super(operation.compute(firstNumber, secondNumber), precision); | ||
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| df.setMaximumFractionDigits(precision); | ||
| df.setMinimumFractionDigits(precision); |
| boolean isFirstNumberUnknown = rnd.nextBoolean(); | ||
| Operation operation = getRandomOperation(); | ||
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| if (operation == Operation.DIVIDE && isDoubleEqual(secondNumber, 0.0, eps)) { |
Owner
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Вот 0 / x = 0 релевантный пример.
| if (operation == Operation.DIVIDE && isDoubleEqual(secondNumber, 0.0, eps)) { | ||
| operation = Operation.MULTIPLY; | ||
| } | ||
| if (operation == Operation.MULTIPLY && isDoubleEqual(firstNumber, 0.0, eps)) { |
Owner
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а тут чем 0.0 * x = 0.0 не подойдет?
Owner
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Ну пока что такая оценка:
Допы:
В итоге 9.25. В остальном супер-пупер. +5 за допы. |
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Сделаны все допы, кроме чтения из json-ов