思路：對於陣列中的每一個數，觀察它們取或不取對最後結果的影響。並且記錄下若干數字的和與M的差的絕對值最小時所取到的若干數字。

/*
 * 微軟100， 9月28題， 輸入和接近M
 * sum 即為M值
 * num排序的陣列
 * len陣列長度
 * vec所取到若干數字構成的向量
 * currentSum vec中的數字的和
 */
int diff = INT_MAX;
vector<int> minVec;
void FindAppM(int sum, int currentSum,int num[], int len, int i, vector<int>& vec)
{
	if(i >= len)
		return;

	//
	if(abs(currentSum - sum) < diff)
	{
		diff = abs(currentSum - sum);
		minVec = vec;
	}

	vec.push_back(num[i]);
	FindAppM(sum, currentSum+num[i],num, len, i+1, vec);
	vec.pop_back();
	FindAppM(sum, currentSum,num, len, i+1, vec);
}


int main() {
	vector<int> sumVec1;
	int num123[5] = {1,2,6,12,51};
	FindAppM(4, 0, num123, 5, 0, sumVec1);
	for(int i=0; i<minVec.size(); i++)
	{
		cout<<minVec[i]<<"*";
	}
	printf("\n***********\n");