在等效数组之间分配元素以实现平衡和

通过分配一个简单的序列，您可以依次迭代地将 min 和 max 元素添加到每个列表中。有一些终止细节需要修复，但这是一般的想法。应用于您的示例，输出将如下所示：

john-schultzs-macbook-pro:~ jschultz$ ./a.out
10 21 13 18  = 62
11 20 14 17  = 62
12 19 15 16  = 62

当num_procs平均分配num_elems时，像这样的简单反射分配将是最佳的。它将是次优的，但仍然不错，当它没有：

#include <stdio.h>

int compute_dist(int lower, int upper, int num_procs)
{
  if (lower > upper || num_procs <= 0)
    return -1;

  int num_elems                = upper - lower + 1;
  int num_elems_per_proc_floor = num_elems / num_procs;
  int num_elems_per_proc_ceil  = num_elems_per_proc_floor + (num_elems % num_procs != 0);
  int procs[num_procs][num_elems_per_proc_ceil];
  int i, j, sum;

  // assign pairs of (lower, upper) to each process until we can't anymore

  for (i = 0; i + 2 <= num_elems_per_proc_floor; i += 2)
    for (j = 0; j < num_procs; ++j)
    {
      procs[j][i]   = lower++;
      procs[j][i+1] = upper--;
    }          

  // handle left overs similarly to the above
  // NOTE: actually you could use just this loop alone if you set i = 0 here, but the above loop is more understandable

  for (; i < num_elems_per_proc_ceil; ++i)
    for (j = 0; j < num_procs; ++j)
      if (lower <= upper)
        procs[j][i] = ((0 == i % 2) ? lower++ : upper--);
      else
        procs[j][i] = 0;

  // print assignment results

  for (j = 0; j < num_procs; ++j)
  {
    for (i = 0, sum = 0; i < num_elems_per_proc_ceil; ++i)
    {
      printf("%d ", procs[j][i]);
      sum += procs[j][i];
    }
    printf(" = %d\n", sum);
  }

  return 0;
}

int main()
{
  compute_dist(10, 21, 3);

  return 0;
}

这是我使用 deque 编写的 O（n） 解决方案（双端队列，不需要 deque，可以使用简单的数组，但由于 popRight 和 popLeft，deque 使代码干净）。代码是 Python，而不是伪代码，但它应该很容易理解（因为它是 Python）。

def balancingSumProblem(seqStart = None, seqStop = None, numberOfArrays = None):
    from random import randint
    from collections import deque

    seq = deque(xrange(seqStart or randint(1, 10), 
                        seqStop and seqStop + 1 or randint(11,30)))
    arrays = [[] for _ in xrange(numberOfArrays or randint(1,6))]

    print "# of elements: {}".format(len(seq))
    print "# of arrays: {}".format(len(arrays))
    averageNumElements = float(len(seq)) / len(arrays)
    print "average number of elements per array: {}".format(averageNumElements)

    oddIteration = True
    try:
        while seq:
            for array in arrays:
                if len(array) < averageNumElements and oddIteration:
                    array.append(seq.pop()) # pop() is like popright()
                elif len(array) < averageNumElements:
                    array.append(seq.popleft())
            oddIteration = not oddIteration
    except IndexError:
        pass

    print arrays
    print [sum(array) for array in arrays]

balancingSumProblem(10,21,3) # Given Example
print "\n---------\n"
balancingSumProblem() # Randomized Test

基本上，从迭代到迭代，它在抓取大元素并将它们均匀分布在数组中和抓取小元素并将它们均匀分布在数组中之间交替进行。它从外到内（尽管你可以从内到外），并尝试使用每个数组的平均元素数来进一步平衡它。

它不是所有测试的 100% 准确，但它在大多数随机测试中都做得很好。您可以尝试在此处运行代码： http://repl.it/cJg

我使用了这个实现，我在本报告中提到过（实现适用于我用于测试 （1-15K） （1-30K） 和 （1-100K） 数据集的情况。我并不是说它对所有情况都有效）：

int aFunction(long lower, long upper, int payload_size, int processes)
{
    long result, i, j;
    MPI_Status status;


    long payload[payload_size];
    int m = 0;
    int k = (payload_size/2)+(payload_size%2)+1;
    int lastAdded1 = 0;
    int lastAdded2 = 0;
    int p = 0;
    int substituted = 0;
    int allowUpdate = 1;
    int s;
    int times = 1;
    int times2 = 0;
    for (i = lower; i <= upper; i=i+payload_size){
        for(j = i; j<(i+payload_size); j++){
            if(j <= upper){
                if(k != 0){
                    if((j-i) >= k){
                        payload[(j-i)] = j- (m);
                        lastAdded2 = payload[(j-i)];
                    }else{
                        payload[(j-i)] = upper - (p*payload_size) - (m++) + (p*payload_size);

                        if(allowUpdate){
                            lastAdded1 = payload[(j-i)];
                            allowUpdate = 0;
                        }
                    }
                }else{

                    int n;
                    int from = lastAdded1 > lastAdded2 ? lastAdded2 : lastAdded1;
                    from = from + 1;
                    int to = lastAdded1 > lastAdded2 ? lastAdded1 : lastAdded2;


                    int tempFrom = (to-from)/payload_size + ((to-from)%payload_size>0 ? 1 : 0);
                    for(s = 0; s < tempFrom; s++){

                        int restIndex = -1;


                        for(n = from; n < from+payload_size; n++){
                            restIndex = restIndex + 1;
                            payload[restIndex] = '\0';
                            if(n < to && n >= from){
                                payload[restIndex] = n;
                            }else{
                                payload[restIndex] = '\0';
                            }
                        }

                        from = from + payload_size;
                    }

                    return 0;

                }
            }else{ payload[(j-i)] = '\0'; }
        }
        p++;
        k=(k/2)+(k%2)+1;
        allowUpdate = 1;

    }

    return 0;
}