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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#define ll long long
#define ull unsigned ll
#define rep(i,l,r)for(ll i=(l);i<(r);i++)
#define min(p,q)((p)<(q)?(p):(q))
#define bit(n,m)(((n)>>(m))&1)

ll a[2010];
ll s[40010];
ll sum;
int main(){
	int n;
	scanf("%d",&n);
	rep(i,0,n){
		scanf("%lld",a+i);
		sum+=a[i];
	}
	//sum/2以上で最も小さいもの⇔sum-(sum/2以下で最も大きいもの)

	//iが作れる⇔bit(s[i/64],i%64)が1
	s[0]=1;
	rep(i,0,n){
		for(int j=34000-a[i]/64;j>=0;j--){
			if(a[i]%64)s[j+a[i]/64+1]|=s[j]>>(64-a[i]%64);
			s[j+a[i]/64]|=s[j]<<(a[i]%64);
		}
	}
	for(ll ans=sum/2;;ans--)if(bit(s[ans/64],ans%64)){
		printf("%lld",sum-ans);
		return 0;
	}
}