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>
#define pii pair<int,int>
#define pll pair<ll,ll>
#define fr first
#define sc second
#define ll  long long
#define mp make_pair
#define mod 1000000007

using namespace std;

ll dp[2001*2001+100];
ll x[2002*2002];
ll x1[2002*2002];
vector<ll>res;
int en;
void f(int i,ll s){
    if(i == en){
        if(s==0)return;
        res.push_back(s);
        return;
    }
    f(i+1,s+x1[i]);
    f(i+1,s);
}

int main()  {
    int n,k=0;
    cin>>n;
    for(int i=0 ;i <n; i ++){
        scanf("%I64d",&x[i]);
    }

    sort(x,x+n);
    int n1 = n/2;
    int st = n1-10;
    en = n1+10;
    st = max(st,0);
    en = min(en,n);
    for(int i=st ;i <en ; i++){
        x1[k++] = x[i];
    }
    en-= st;
    f(0,0);
    sort(res.begin(),res.end());
    cout<<res[res.size()/2]<<endl;
    return 0;
}