生成数字的5的倍数? [英] Generating numbers to the multiple of 5?

问题描述

你好.

我有一个文本框，可以接受5个字符，代表从00000到99999的数值.

当用户输入一个值(例如12345)时，将调用以下功能:

Hello.

I have a text box that can accept 5 chars, representing a numeric value from 00000 to 99999.

When the user enters a value, say for example, 12345, the following function is called:

if ((NumToValidate % 5) == 0)
        return VALIDATION_SUCCESS;

其中NumToValidate是5个字符输入的数字形式.

该函数检查输入的值(在这种情况下为12345)是5的倍数，如果是，则返回成功.

它运作良好，但是，我不知道如何创建一个函数，该函数可以循环生成从00000到99999的5的倍数的值(所有可能的值)

如果可能，则使用AND生成一个随机值并找到其下一个5的倍数.

例如

现在生成的兰德值= 1834，找到第5个数字(如果有)产生5的倍数.

谢谢
Stephen

Where NumToValidate is the numerical form of the 5 char input.

The function checks that the inputed value, in this case 12345 is a multiple of 5, and if it is it returns as a success.

It works well, however, I have no idea how to create a function that can loop through generateing values that are a multiple of 5, from 00000 to 99999 (All possible values)

AND, if possible, to generate a random value and find its next multiple of 5.

e.g.

Rand value generated = 1834 now find what 5th digit, if any, produces a multiple of 5.

Thank you,
Stephen

推荐答案

循环，用于生成可被5整除的val

Loop for generating vals divisible by 5

for(int val = 0; val < MAX; val+=5)
{
   //Because we are adding 5 and starting at 0 val must be divisible by 5... unless 
   // you modify it inside the loop.
}

要生成0到99999之间的随机值，请使用结合了模运算符的随机生成器.

To generate a random value between 0 and 99999 use the random generator combined with modulus operator.

Random rand = new Random();
int val = random.Next(MIN, MAX);

//Now get to the ''next'' divisible by 5
int leftOver = val % 5;
int newVal = val + leftOver; //Adding the leftover will result in divisible by 5

首先，我们可以生成一个列表，其中包含在0到100000(不包括在内)之间的所有5的多态性.

First we can generate a list with all mutiples of 5 between 0 and 100000 (not included).

List<int> numbers = new List<int>;
for (int i = 0; i < 100000; i += 5)
{
    numbers.Add(i);
}

然后，您可以在任意位置取一个数字并将其从列表中删除，这样每个数字只能获得一次.

Then you can take a number at a random position and remove it from the list so that you will get each number only once.

Random rand = new Random();

while (numbers.Count > 0)
{
    int randomIndex = rand.Next(numbers.Count);

    int randomNumber = numbers[randomIndex];

    // Could validate that the code works...
    System.Diagnostics.Debug.Assert((randomNumer % 5) == 0);

    numbers.RemoveAt(randomIndex);

    DoSomethingWithRandomMultipleOf5(randomNumbr);
}

如果您不希望号码只返回一次，则更加简单.

If you don''t want number to be returned only once, it is even easier.

const int maxReturnedValue = 99995;
const int multipleOf = 5;
const int randLimit = (maxReturnedValue - 1) / multipleOf;

Random rand = new Random();

while (WantAnotherRandomNumber())
{
    int randomNumber = rand.Next(randLimit) * multipleOf;

    // Confirms that the code works. Could be useful when you
    // are not too sure that you code is correct.
    System.Diagnostics.Debug.Assert((randomNumer % 5) == 0);
    System.Diagnostics.Debug.Assert(randomNumer >= 0);
    System.Diagnostics.Debug.Assert(randomNumer <= 99995);

    DoSomethingWithRandomMultipleOf5(randomNumbr);
}

出于完整性考虑，请参见答案 ^ ]我觉得最优雅，但David1987更快(很多).

链接更改为

Just for completeness, see the answer here[^] that I find most elegant, but David1987 was faster (a lot).

The link changed to another post[^].

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