一个整数数组，长度为n,将其分为m份，使得各分的和相等

//题意：
//一个整数数组，长度为n,将其分为m份，使得各分的和相等
//比如：{3,4,2,3,6}可以分成{3，2,4，3,6} m=1;或者 {3,6},{2,4,3} m=2；
//{3,3}，{2,4}{6} m=3，所以m最大值为3
//思路:使用多重背包
//首先遍历一次数组arr[n]，把数组中最大值maxN和总和sum记录下来
//那么可能的等分为m = sum / maxN
//在这m份中，假设为p份，每份都为sum/m

//然后利用多重背包求每份sum/m;

#include <iostream>
#include <vector>
#include <map>
#include <algorithm>
using namespace std;

#define MSIZ 1000
#define NSIZ 10000

int dp[NSIZ];
int v, n;
int path[NSIZ][MSIZ];
int num[NSIZ];
typedef struct Node_
{
	int data;
	int index;
}Node;
Node arr[MSIZ];
vector<vector<int>> vec(MSIZ);

int cmp(int a, int b)
{
	return a > b;
}

void zeroPacket(int cost, int value, int index,  int id)
{
	int i = 0;
	for (i = v; i >= cost; i--)
	{
		if (dp[i] < dp[i - cost] + value)
		{
			dp[i] = max(dp[i], dp[i - cost] + value);
			memcpy(path[i], path[i - cost], sizeof(path[i - cost]));
			path[i][index] = id;
		}

	}
}

void completePacket(int cost, int value, int index, int id)
{
	int i = 0;
	for (i = cost; i <= v; ++i)
	{
		if (dp[i] < dp[i - cost] + value)
		{
			dp[i] = max(dp[i], dp[i - cost] + value);
			memcpy(path[i], path[i - cost], sizeof(path[i - cost]));
			path[i][index] = id;
		}
	}
}

void multiPacket(int cost, int value, int m,int index,  int id)
{
	if (m * cost >= v)
	{
		completePacket(cost, value,index, id);
		return;
	}

	int k = 1;
	while(k <= m)
	{
		zeroPacket(k * cost, k * value,index, id);
		m -= k;
		k *= 2;
	}

	if (m)
	{
		zeroPacket(m * cost, m * value, index, id);
	}
}

int main()
{
	int t, i, j, p, q, k;
	int sum = 0;
	int maxN = INT_MIN;
	path[0][0] = true;
	j = 0;
	int row;

	while (scanf("%d", &t) != EOF)
	{
		memset(num, 0, sizeof(num));
		memset(dp, 0, sizeof(dp));
		memset(path, -1, sizeof(path));
		sum = 0;
		for (i = 0;i < t; ++i)
		{
			scanf("%d", &arr[i].data);
			arr[i].index = i;
			sum += arr[i].data;

			if (maxN < arr[i].data)
			{
				maxN = arr[i].data;
			}
		}

		int m = sum / maxN;
		for (p = m; p >= 2; --p)
		{
			v = sum / p;
			q = 0;
			for (i = 0;i < p; ++i)
			{
				vec[i].clear();
			}
			j = 0;
			do
			{
				memset(dp, 0, sizeof(dp));
				for (i = 0;i < t; ++i)
				{

					if (arr[i].index != -1)
					{
						multiPacket(arr[i].data, arr[i].data, 1, arr[i].index, q);
					}
				}

				if (dp[v] != v)
				{
					break;
				}

				for (i = 0, k = 0;i < t; ++i)
				{
					if (path[v][arr[i].index] == q)
					{
						vec[j].push_back(arr[i].data); 
						arr[i].index = -1;
					}
				}
				++j;
				row = j;


			}while(++q < p);

			if (q == p)
			{
				for (i = 0;i < row;++i)
				{
					for (j = 0; j < vec[i].size(); ++j)
					{
						printf("%d ", vec[i][j]);
					}
					printf("\n");
				}
				break;
			}

		}

		if (q != p)
		{
			for (i = 0;i < t; ++i)
			{
				printf("%d ", arr[i].data);
			}

			printf("\n");
		}
		else
		{
			printf("\n");
		}
	}
	return 0;
}