给定一个实数数组，按序排列（从小到大）,从数组从找出若干个数，使得这若干个数的和与M最为接近，描述一个算法，并给出算法的复杂度。

#define M 8
#define N 20

int minDif = INT_MAX;
vector<int> minvct;

void findCloestNums(int* arr, int step, int len, vector<int> vct, int goalSum,
int curSum) {
if (!arr || len < 0) {
return;
}
if (minDif > abs(curSum - goalSum)) {
minDif = abs(curSum - goalSum);
minvct = vct;
if (minDif == 0) {
return;
}
}
if (step > len) {
return;
}
vct.push_back(arr[step]);
findCloestNums(arr, step + 1, len, vct, goalSum, curSum + arr[step]);
vct.pop_back();
findCloestNums(arr, step + 1, len, vct, goalSum, curSum);
}