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: Accepted)

#include <iostream>
#include <iomanip>
#include <stdio.h>
#include <set>
#include <vector>
#include <map>
#include <cmath>
#include <algorithm>
#include <memory.h>
#include <string>
#include <sstream>
#include <cstdlib>
#include <ctime>
#include <cassert>
#include <bitset>

using namespace std;

typedef long long LL;
typedef pair<int,int> PII;

#define MP make_pair
#define PB push_back
#define FF first
#define SS second

#define FORN(i, n) for (int i = 0; i <  (int)(n); i++)
#define FOR1(i, n) for (int i = 1; i <= (int)(n); i++)
#define FORD(i, n) for (int i = (int)(n) - 1; i >= 0; i--)

#define DEBUG(X) { cout << #X << " = " << (X) << endl; }
#define PR0(A,n) { cout << #A << " = "; FORN(_,n) cout << A[_] << ' '; cout << endl; }

#define MOD 1000000007
#define INF 2000000000

int GLL(LL& x) {
    return scanf("%lld", &x);
}

int GI(int& x) {
    return scanf("%d", &x);
}

int n;

vector<int> a;

const int MAXS = 4000005;
bitset<MAXS> dp;

int main() {
    GI(n);
    a.resize(n);

    dp[0] = 1;

    int s = 0;

    FORN(i, n)
    {
        GI(a[i]);
        s += a[i];

        bitset<MAXS> newsums = dp;
        newsums <<= a[i];
        dp |= newsums;
    }

    for (int i = (s+1) / 2; i <= s; i++)
    {
        if (dp[i])
        {
            cout << i << "\n";
            return 0;
        }
    }

    return 0;
}