Problem:
Score : 700 points

Problem Statement

You are given N integers A_1, A_2, ..., A_N.

Consider the sums of all non-empty subsequences of A. There are 2^N - 1 such sums, an odd number.

Let the list of these sums in non-decreasing order be S_1, S_2, ..., S_{2^N - 1}.

Find the median of this list, S_{2^{N-1}}.

Constraints

1 \leq N \leq 2000

1 \leq A_i \leq 2000

All input values are integers.

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 ... A_N

Output

Print the median of the sorted list of the sums of all non-empty subsequences of A.

Sample Input 1

3
1 2 1

Sample Output 1

2

In this case, S = (1, 1, 2, 2, 3, 3, 4). Its median is S_4 = 2.

Sample Input 2

1
58

Sample Output 2

58

In this case, S = (58).

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;

int main(int argc, char *argv[])
{
    int n;
    cin >> n;

    vector<int> A(n);
    for(int& x : A) cin >> x;

    map<int, int> c;

    for(int i = 0; i < n; ++i)
    {
        map<int, int> next;
        next[A[i]]++;

        for(auto& x : c)
        {
            if(x.second >= 1)
            {
                next[x.first + A[i]]++;
            }
        }

        for(auto& x : next)
        {
            c[x.first] += x.second;
        }
    }

    auto i = c.begin();
    auto j = c.rbegin();

    while(true)
    {
        while(i != c.end() && i->second == 0) ++i;
        while(j != c.rend() && j->second == 0) ++j;

        if(i->first == j->first)
        {
            cout << i->first << endl;
            break;
        }

        int m = min(i->second, j->second);
        i->second -= m;
        j->second -= m;
    }
}