POJ 3977 Subset

#include <cstdio>
#include <map>
#define ll long long
using namespace std;

int n;
ll a[40];

ll ll_abs(ll x)
{
    return x >= 0 ? x : -x;
}

void solve()
{
    map<ll, int> mp;
    map<ll, int>::iterator it;
    pair<ll, int> res(ll_abs(a[0]), 1); //初始化结果为第一个元素
    for(int i = 1; i < 1<<(n/2); ++i){  //枚举区间为[1, 2^n)，当i为0时，子集为空
        ll sum = 0;
        int num = 0;
        for(int j = 0; j < n/2; ++j){   //按位枚举
            if((i>>j)&1){
                sum += a[j];
                ++num;
            }
        }
        res = min(res, make_pair(ll_abs(sum), num));    //子集的元素只取自于A
        it = mp.find(sum);
        if(it != mp.end())
            it->second = min(it->second, num);
        else
            mp[sum] = num;
    }
    for(int i = 1; i < 1<<(n-n/2); ++i){
        ll sum = 0;
        int num = 0;
        for(int j = 0; j < n-n/2; ++j){
            if((i>>j)&1){
                sum += a[n/2+j];
                ++num;
            }
        }
        res = min(res, make_pair(ll_abs(sum), num));    //子集的元素只取自于B
        it = mp.lower_bound(-sum);  //查找与-sum最相近的值
        if(it != mp.end())  //可能在该位置
            res = min(res, make_pair(ll_abs(it->first+sum), it->second+num));
        if(it != mp.begin()){  //可能在该位置的前一个位置
            --it;
            res = min(res, make_pair(ll_abs(it->first+sum), it->second+num));
        }
    }
    printf("%I64d %d\n", res.first, res.second);
}

int main()
{
    while(scanf("%d", &n), n){
        for(int i = 0; i < n; ++i)
            scanf("%I64d", &a[i]);
        solve();
    }
    return 0;
}