Question

我如何检查一个给定的数字是否是＃34;完美的方形＆＃34; if 比较？

Answer 1

假设我理解你在问什么，你可以先通过Math.sqrt请求数字的平方根，然后在除法后得到其余部分以查看它是否等于0，在这种情况下，数字可以是除以平方根，所以是一个“完美”的方形

public void setAlarm () throws java.text.ParseException {
    RealmResults<Reminder> realmResults = getAllLatest();

    AlarmManager alarmMgr = (AlarmManager) getSystemService(Context.ALARM_SERVICE);

    Reminder reminder =  getLatestReminder(realmResults);
    int hour = convertTimeToHoursAndMinutes(reminder.getReminderTime()).getHours();
    int minutes = convertTimeToHoursAndMinutes(reminder.getReminderTime()).getMinutes();
    PendingIntent pendingIntent  = getPendingIntentFromReminder(reminder);
    Log.v("Reminder is: ", String.valueOf(reminder));
    Calendar calendar = Calendar.getInstance();
    //calendar.setTimeZone(TimeZone.getTimeZone());
    calendar.setTimeInMillis(System.currentTimeMillis());
    calendar.set(Calendar.HOUR_OF_DAY, hour);
    calendar.set(Calendar.MINUTE, minutes);


    alarmMgr.setRepeating(AlarmManager.RTC_WAKEUP, calendar.getTimeInMillis(), AlarmManager.INTERVAL_DAY , pendingIntent);
}
private PendingIntent getPendingIntentFromReminder(Reminder reminder) throws ParseException {
    int hour = convertTimeToHoursAndMinutes(reminder.getReminderTime()).getHours();
    int minutes = convertTimeToHoursAndMinutes(reminder.getReminderTime()).getMinutes();
    Intent intent = new Intent(this, MyReceiver.class);
    intent.setAction(reminder.getName());
    Uri data = Uri.withAppendedPath(Uri.parse("TYS"),
            String.valueOf(reminder.getId()));
    intent.setData(data);
    Calendar calendar = Calendar.getInstance();
    //calendar.setTimeZone(TimeZone.getTimeZone());
    calendar.setTimeInMillis(System.currentTimeMillis());
    calendar.set(Calendar.HOUR_OF_DAY, hour);
    calendar.set(Calendar.MINUTE, minutes);
    //Log.v("Test", calendar.getTimeZone().toString());
    // Log.v("Hours to show", String.valueOf(hour));
    // Log.v("Minutes to show", String.valueOf(minutes));
    intent.putExtra("reminderTitle", reminder.getName());
    //  Log.v("Reminder Name", reminder.getName());
    intent.putExtra("notificationType", 1);
    intent.putExtra("reminderId", String.valueOf(reminder.getId()));
    intent.putExtra("reminderSubtitle", reminder.getSubTitle());
    intent.setAction(Long.toString(System.currentTimeMillis()));
    Log.v("set Alarm", String.valueOf(intent));
    PendingIntent alarmIntent = PendingIntent.getBroadcast(this, 0, intent, 0);
    return alarmIntent;
}

public void deleteReminderIntent(Reminder reminder) throws ParseException {
    Log.v("deleteReminderIntent", String.valueOf(reminder));
    AlarmManager alarmMgr = (AlarmManager) getSystemService(Context.ALARM_SERVICE);
    PendingIntent pendingIntent =getPendingIntentFromReminder(reminder);
    alarmMgr.cancel(pendingIntent);
    pendingIntent.cancel();
    //alarmMgr.cancel(temp);
   // alarmMgr.cancel(AlarmManager.RTC_WAKEUP, calendar.getTimeInMillis(), AlarmManager.INTERVAL_DAY, alarmIntent);
}

要求用户年龄，这实际上取决于您的用例。提示是一种可能的方法，虽然您也可以通过提供表单来实现，这样可以获得html5输入类型function isSquare( number ) { // get the square root of the number in question var sqrt = Math.sqrt(number); // check if the modulo (rest after dividing is 0) return ((number % sqrt) === 0); } console.log(isSquare(9)); // true console.log(isSquare(5)); // false的好处，如下所示：

number

function checkAge() {
  var inputElement = document.getElementById('ageInput'),
    value = parseInt( inputElement.value ),
    outputElement = document.getElementById('output');
  if ( Number.isNaN( value ) ) {
    output.innerHTML = '<error>Please enter a valid number</error>';
    return;
  }
  if (value % 2 === 0) {
    output.innerHTML = 'Your age is even';
  } else {
    output.innerHTML = 'Your age is odd';
  }
}

error {
  color: red;
}

Answer 2

＆＃34; best＆＃34;检查数字是否为奇数的方法是主观的。显示的方式肯定会奏效。但是，在实际代码中，您需要确保捕获无效年龄，方法是检查age % 2是NaN还是使用parseInt(prompt("what is your age?"), 10)之类的内容，然后检查Number.isNaN中的输出}。

Answer 3

if(age % Math.sqrt(age) === 0) {
  console.log("Your age is a perfect square!");
}

我找到答案..谢谢大家（：

JavaScript比较＆＃34; if＆＃34;关键词

3 个答案: