组合数学题目的编程求解的效率问题

Question

这么一道题：A={1,2,3,...,2015},B是A的一个子集，对B中任意三个元素x,y,z，都有x+y不等于z.
求B中元素最多时，集合B.
Card(B)的最大值是1008，Card(B)=1008时，B集合有且只有四个（应该很容易写出来：所有奇数集，{1008，……2015}，{1007，1009，……，2015}，{1007，……，2014}）
只有4个的证明还没想出来，想编程序找找反例。
用回溯法写了这个程序：算到N=101时，就悲剧了，更别提N=2015了。想想应该是2^N型的问题，大虾们有没有什么好办法？我自己感觉自己的回溯写得太烂了，有兴趣和时间的大虾能不能以此题写一个范例供我学习下。

#define _CRT_SECURE_NO_WARNINGS
#include <vector>
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstdlib>

using namespace std;
typedef vector<int> Set;
typedef vector<vector<int> > ResSet;
int getMinChoice(Set status, Set b)
{
    unsigned int i,k;
    k = b.size();
    i = (k == 0)? 0 : b.back();
    while (1){  
        if (i+1 > status.size())
            break;
        else if (status[i] == 0)
            break;
        i++;
    }
    return i;
}

int main()
{
    int num, temp, tmpindex,flag=0;
    unsigned int k;

    Set w;
    ResSet res;
    Set b;
    Set status;
    Set reslen;

    //freopen("in.txt", "r", stdin);
    freopen("out.txt", "w", stdout);
    scanf("%d", &num);

    for (int i = 0; i<num; i++){
        w.push_back(i + 1);
        status.push_back(0);
    }

    while (1)
    {
        temp = getMinChoice(status, b);
        if (temp < num)
        {
            b.push_back(temp + 1);
            flag = 1;
            status[temp] = -1;
            for (k = 0; k+1 < b.size(); k++)
            {
                tmpindex = b[k] + temp;
                if (tmpindex < num)
                status[tmpindex]++;
            }
        }
        else 
        {
            if (flag == 1)
            {
                if (b.size() == (num + 2) / 2)
                {
                    res.push_back(b);
                }   
                //reslen.push_back(b.size());
                temp = b.back();
                b.pop_back();
                for (k = 0; k < b.size(); k++)
                {
                    tmpindex = b[k] + temp - 1;
                    if (tmpindex < num)
                    status[tmpindex]--;
                }
                flag = 0;
            }
            else if (flag == 0)
            {
                if (b.size() == 0)
                    break;
                temp = b.back();
                b.pop_back();
                for (k = 0; k < b.size(); k++)
                {
                    tmpindex = b[k] + temp - 1;
                    if (tmpindex < num)
                    status[tmpindex]--;
                }
                for (k = temp; k < num; k++)
                {
                    if (status[k] == -1)
                    status[k] = 0;
                }
            }
        }
    }
    cout << res.size() << endl;
        for (unsigned i = 0; i< res.size(); i++)
        {
            Set b = res[i];
            for (unsigned j = 0; j<b.size(); j++)
            {
                cout << b[j] << ",";
            }
            cout << endl;
        }
    return 0;
}