分三种情况考虑：
1.不需要交换：即当前sumA 与sumB相同，return0
2.交换一个元素：算出dis = abs(sumA - sumB) / 2；从A数组和B数组分别找到两个元素，使这两个元素的差值最接近于dis，即交换这两个元素就可以使A,B最接近，将交换后的数组和差值记录下来；
3.交换两个元素：构造新数组C,D；C为A数组任意两元素之和，D为B数组任意两元素之和；然后从数组C,D中找到两元素，使这两个元素差值最接近于dis，即交换C，D两个元素可使得A,B最接近。将交换后的数组差值记录下来。新的求和使用：newSumA = sumA + D[j] - C[i];  newSumB = sumB + C[i] - D[j]
最后将2,3中较小的输出即可。 
#include "bits/stdc++.h"
using namespace std;
int GetMinDisByChange(vector<int> &A, vector<int>&B);
int ComputeMinDis(vector<int> &A, vector<int> &B, const int &threshold, const int &sumA, const int &sumB);

int main(){
	/*
	4
	1 3 7 9
	3
	2 10 12*/
	int n, m;
	cin >> n;
	vector<int> A(n);
	for (int i = 0; i < n; i++) cin >> A[i];
	cin >> m;
	vector<int>B(m);
	for (int i = 0; i < m; i++) cin >> B[i];
	cout << GetMinDisByChange(A, B) << endl;
	return 0;
}

int ComputeMinDis(vector<int> &A, vector<int> &B, const int &threshold, const int &sumA, const int &sumB){
	int tDis = INT_MAX;
	int tId_i = 0, tId_j = 0;
	for (int i = 0; i < A.size(); i++)
	{
		for (int j = 0; j < B.size(); j++)
		{
			if (sumA > sumB && A[i] > B[j] ||
				sumA < sumB && A[i] < B[j])
			{
				int Dis = abs(A[i] - B[j]);
				if (Dis && abs(Dis - threshold) < tDis)
				{
					tDis = abs(Dis - threshold);
					tId_i = i;
					tId_j = j;
				}
			}
		}
	}
	tDis = abs(sumA + B[tId_j] - A[tId_i] - (sumB + A[tId_i] - B[tId_j]));
	return tDis;
}

int GetMinDisByChange(vector<int> &A, vector<int>&B){
	int sumA, sumB;
	const int MaxChange = 2;
	int ChangeCount = 0;
	sumA = 0;
	sumB = 0;
	for (int i = 0; i < A.size(); i++) sumA += A[i];
	for (int i = 0; i < B.size(); i++) sumB += B[i];
	int initialDis = abs(sumA - sumB);
	int MinDis = initialDis;
	if (initialDis == 0)
	{
		return 0;
	}
	vector<int> C;
	vector<int> D;
	for (int i = 0; i < A.size(); i++)
		for (int j = i + 1; j < A.size(); j++)
			C.push_back(A[i] + A[j]);
	for (int i = 0; i < B.size(); i++)
		for (int j = i + 1; j < B.size(); j++)
			D.push_back(B[i] + B[j]);
	//Swap One time
	int DisBySwap1 = ComputeMinDis(A, B, initialDis / 2, sumA, sumB);
	//swap Two times
	int DisBySwap2 = ComputeMinDis(C, D, initialDis / 2, sumA, sumB);
	return min(DisBySwap1, DisBySwap2);
}