如何将一个数分成偶数个组？

Question

如果我得到一个数字，以及我需要将该数字分成多少组的数字，我如何将它分成尽可能接近偶数的块？

8 into 3 chunks -> 3, 3, 2

12 into 2 chunks -> 6, 6

9 into 2 chunks -> 4, 5

9 into 4 chunks -> 2, 2, 2, 3

11 into 5 chunks -> 2, 2, 2, 2, 3

Answer 1

这似乎或多或少有效：

int[] GetGroups(int number, int chunks) =>
    Enumerable
        .Range(0, chunks)
        .Select(x => number / chunks + (number % chunks > x ? 1 : 0))
        .ToArray();

var inputs = new[]
{
    new { n = 8, c = 3, },
    new { n = 12, c = 2, },
    new { n = 9, c = 2, },
    new { n = 9, c = 4, },
    new { n = 11, c = 5, },
};

foreach (var input in inputs)
{
    Console.WriteLine($"{input.n} into {input.c} chunks -> {String.Join(", ", GetGroups(input.n, input.c))}");
}

这给了我：

8 into 3 chunks -> 3, 3, 2
12 into 2 chunks -> 6, 6
9 into 2 chunks -> 5, 4
9 into 4 chunks -> 3, 2, 2, 2
11 into 5 chunks -> 3, 2, 2, 2, 2

Answer 2

int[,] input = new int[,] { { 8, 3 }, { 12, 2 }, { 9, 2 }, { 9, 4 }, { 11, 5 } };
int uBound0 = input.GetUpperBound(0);

int num = 0;
int split = 0;
for (int i = 0; i <= uBound0; i++)
{
    num = input[i, 0];
    split = input[i, 1];
    int dev = num / split;
    int mod = num % split;
    string output = string.Empty;
    //int slice = num - mod;
    if (dev <= mod)
        dev++;
    for (int k = 1; k < split; k++)
    {
        output += dev + ",";
    }
    output += (num - (dev * (split - 1)));
    Console.WriteLine(num + " into " + split + " chunks -> " + output);
}

output: 
8 into 3 chunks -> 3,3,2
12 into 2 chunks -> 6,6
9 into 2 chunks -> 4,5
9 into 4 chunks -> 2,2,2,3
11 into 5 chunks -> 2,2,2,2,3

如何将一个数分成偶数个组？

How do I divide a number into an even number of groups?

logic