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: Time Limit Exceeded)

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

#define EPS (1e-7)
#define INF (1e9)
#define PI (acos(-1))
//const ll mod = 1000000007;

int main() {
    //cout.precision(10);
    int n;
    cin >> n;
    vector<int> a(n);
    int sum = 0;
    for(int i = 0; i < n; i++){
        //cin >> a[i];
        scanf("%d", &a[i]);
        sum += a[i];
    }
    sort(a.begin(), a.end());
    int target = sum / 2;
    int dp[2001000];
    for(int i = 1; i <= 2000000; i++){
        dp[i] = INF;
    }
    dp[0] = -1;
    for(int value = 0; value < target; value++){
        for(int index = dp[value] + 1; index < n; index++){
            if(value + a[index] > target) break;
            dp[value + a[index]] = min(dp[value + a[index]], index);
        }
    }
    for(int i = target; i >= 0; i--){
        if(dp[i] < INF){
            //cout << sum - i << endl;
            printf("%d\n", sum - i);
            return 0;
        }
    }
    return 0;
}