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 <bits/stdc++.h>
#define PB push_back
#define MP make_pair
#define F first
#define S second
#define SZ(x) ((int)(x).size())
#define ALL(x) (x).begin(),(x).end()
#ifdef _DEBUG_
	#define debug(...) printf(__VA_ARGS__)
#else
	#define debug(...) (void)0
#endif
using namespace std;
typedef long long ll;
typedef pair<int,int> PII;
typedef vector<int> VI;
int a[2010];
bitset<4000100> dp;
int main() {
	int n;
	int sum = 0;
	cin>>n;
	for(int i = 1;i<=n;i++){
		cin>>a[i];
		sum+=a[i];
	}
	dp.set(0);
	for(int i = 1; i <= n; i++){
		dp |= dp<<a[i];
	}
	if(sum%2) sum = sum/2+1;
	else sum /= 2; 
	for(int i = sum; i<=4000000; i++){
		if(dp[i]) {
			printf("%d\n",i);
			return 0;
		}
	}
	return 0;
}