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<iostream>
#include<cstdio>
#include<vector>
#include<string>
#include<cstring>
#include<functional>
#include<stack>
#include<queue>
#include <iomanip>
#include<map>
#include<limits>
#include<cmath>
#include<algorithm>
#include<bitset>
#include<deque>
#include<utility>
#include<complex>
#include<cstdlib>
#include<set>
#include<cctype>

using namespace std;

#define DBG cerr << '!' << endl;
#define REP(i,n) for(ll (i) = (0);(i) < (n);++i)
#define rep(i,s,g) for(ll (i) = (s);(i) < (g);++i)
#define rrep(i,s,g) for(ll (i) = (s);i >= (g);--(i))
#define PB push_back
#define MP make_pair
#define FI first
#define SE second
#define SHOW1d(v,n) {for(int W = 0;W < (n);W++)cerr << v[W] << ' ';cerr << endl << endl;}
#define SHOW2d(v,i,j) {for(int aaa = 0;aaa < i;aaa++){for(int bbb = 0;bbb < j;bbb++)cerr << v[aaa][bbb] << ' ';cerr << endl;}cerr << endl;}
#define ALL(v) v.begin(),v.end()
#define Decimal fixed<<setprecision(10)

typedef long long ll;
typedef vector<int> iv;
typedef vector<iv> iiv;
typedef vector<string> sv;

int main()
{
	int n;cin >> n;
	vector<ll> v(n);
	
	REP(i,n)cin >> v[i];
	
	if(n%2 == 1)
	{
	//	cout << n/2 + 1 << endl;
		sort(ALL(v));
		ll sum = 0;
		REP(i,n/2 + 1)sum += v[i];
		cout << sum << endl;
		return 0;
	}
	else
	{
		sort(ALL(v),greater<ll>());
		vector<ll> a;
		vector<ll> b;
		ll suma=0,sumb=0;
		
		REP(i,n)
		{
			if(suma>sumb&&b.size()<n/2)
			{
				b.PB(v[i]);
				sumb+=v[i];
			}
			else
			{
				a.PB(v[i]);
				suma+=v[i];
			}
		}
		cout << max(suma,sumb) << endl;
		return 0;
	}
	
	return 0;
}