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 <iostream>
#include <string.h>
#include <algorithm>
#include <vector>
#include <queue>

using namespace std;
typedef long long ll;
const ll INF=10000000000000000LL;
int n,a[2001];
priority_queue<int,vector<int>,greater<int> >que;
ll cheak(bool c[2001]){
	ll t=0;
	for(int i=0;i<n;++i){
		if(c[i])t+=a[i];
	}
//	cout<<t<<endl;
	return t;
}
void reg(int n1,bool d[2001]){
	if(n1==n)que.push(cheak(d));
	else{
		reg(n1+1,d);
		d[n1]=true;
		reg(n1+1,d);
		d[n1]=false;
	}
}
int main() {
	cin>>n;
	bool b[2001]={};
	ll s=1;
	for(int i=0;i<n;++i){
		cin>>a[i];
		s*=2;
	}
	s/=2;
	reg(0,b);
	ll l;
	for(int i=0;i<s+1;++i){
		l=que.top();que.pop();
//		cout<<l<<endl;
	}
	cout<<l<<endl;
	return 0;
}