产生一系列的随机数字，加起来就是N的C＃的 [英] Generate a series of random numbers that add up to N in c#

问题描述

如何生成30随机数字1-9之间，所有加起来200（或任意N），在C＃？

How do I generate 30 random numbers between 1-9, that all add up to 200 (or some arbitrary N), in C#?

我想产生一个数字字符串，可以加在一起为N。

I'm trying to generate a string of digits that can add together to be N.

推荐答案

我不知道什么是统计上的这一点，但这里的问题是，你不希望随机选择一个数字，使得它不可能总之n，其中的条目，或用过冲或下冲的M号。下面是我会怎么做：

I'm not sure what the statistics are on this but, the issue here is that you don't want to randomly select a number that makes it impossible to sum N with M number of entries either by overshooting or undershooting. Here's how I would do it:

static void Main()
{
    int count = 30;
    int[] numbers = getNumbers(count, 155);
    for (int index = 0; index < count; index++)
    {
        Console.Write(numbers[index]);
        if ((index + 1) % 10 == 0)
            Console.WriteLine("");
        else if (index != count - 1)
            Console.Write(",");
    }
    Console.ReadKey();
}
static int[] getNumbers(int count, int total)
{
    const int LOWERBOUND = 1;
    const int UPPERBOUND = 9;

    int[] result = new int[count];
    int currentsum = 0;
    int low, high, calc;

    if((UPPERBOUND * count) < total ||
        (LOWERBOUND * count) > total ||
        UPPERBOUND < LOWERBOUND)
        throw new Exception("Not possible.");

    Random rnd = new Random();

    for (int index = 0; index < count; index++)
    {
        calc = (total - currentsum) - (UPPERBOUND * (count - 1 - index));
        low = calc < LOWERBOUND ? LOWERBOUND : calc;
        calc = (total - currentsum) - (LOWERBOUND * (count - 1 - index));
        high = calc > UPPERBOUND ? UPPERBOUND : calc;

        result[index] = rnd.Next(low, high + 1);

        currentsum += result[index];
    }

    // The tail numbers will tend to drift higher or lower so we should shuffle to compensate somewhat.

    int shuffleCount = rnd.Next(count * 5, count * 10);
    while (shuffleCount-- > 0)
        swap(ref result[rnd.Next(0, count)], ref result[rnd.Next(0, count)]);

    return result;
}
public static void swap(ref int item1, ref int item2)
{
    int temp = item1;
    item1 = item2;
    item2 = temp;
}

我没有很多时间来测试这个如此道歉，如果有一个在我的逻辑缺陷的地方。

I didn't have a lot of time to test this so apologies if there's a flaw in my logic somewhere.

编辑：

我做了一些测试，一切似乎固体。如果你想要一个漂亮的pretty的小号$ P $垫，它看起来像你想沿着东西线总=计数*（（UPPER + LOWER）/ 2） 。虽然我相当肯定，如上和下之间的差异增加了更灵活的这成为

I did some testing and everything seems solid. If you want a nice pretty spread it looks like you want something along the lines of Total = Count * ((UPPER + LOWER) / 2). Although I'm fairly certain that as the difference between UPPER and LOWER increases the more flexible this becomes.

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