数组交换

有两个序列a,b，大小都为n,序列元素的值任意整数，无序；
要求：通过交换a,b中的元素，使[序列a元素的和]与[序列b元素的和]之间的差最小。
例如:
var a=[100,99,98,1,2, 3];

var b=[1, 2, 3, 4,5,40];

// exchange_array.cpp : 定义控制台应用程序的入口点。
//

#include "stdafx.h"
#include<iostream>
#include<cmath>
using namespace std;

#define INFTY 2147483647

int sumof(int*in, int len);
int exchange(int*aa, int*bb, int len);
int _tmain(int argc, _TCHAR* argv[])
{
	const int len = 6;
	int a[len] = { 100, 99, 98, 1, 2, 3 };
	int b[len] = { 1, 2, 3, 4, 5, 40 };
	
	int aa[6], bb[6];
	cout << sumof(a, len) << endl;
	cout << sumof(b, len) << endl;
	cout << abs(sumof(a, len) - sumof(b, len))<<endl;
	int k1 = exchange(a, b, len);
	cout << k1 << endl;
	int k2 = exchange(a, b, len);
	cout << k2<<endl;
	while (k1 > k2)
	{
		for (int i = 0; i < len; i++)
		{
			aa[i] = a[i];
			bb[i] = b[i];
		}
		k1 = k2;
		k2 = exchange(a, b, len);
	}

	for (int i = 0; i < len; i++)
		cout << aa[i] << endl;
	cout << k1<<endl;
	system("pause");
	return 0;
}
//求和
int sumof(int*in, int len)
{
	int sum=0;
	for (int i = 0; i < len; i++)
		sum += in[i];
	return sum;
}
//交换一次，得到当前最好的结果
int exchange(int*aa, int*bb, int len)
{
	int ii = 0, jj = 0;
	int best = INFTY;
	
	int re = sumof(bb, len) - sumof(aa, len);
	if (re == 0)
		return 0;
	
		for (int i = 0; i < len; i++)
			for (int j = 0; j < len; j++)
		{
			int x;
			if (re < 0)
				x = abs((2 * (aa[i] - bb[j]) + re));
			else
				x = abs((2 * (bb[j] - aa[i]) + re));
			if (x < best)
			{
				best = x;
				ii = i;
				jj = j;
			}
		}
		int ss = aa[ii];
		aa[ii] = bb[jj];
		bb[jj] = ss;
		return best;
}