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 <iostream>
#include <cstdio>
#include <map>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
#include <set>
#include <bitset>
const int N = 100005;
using namespace std;
typedef long long ll;
int a[N];
int n;
bitset<4000010> f;
int main()
{
    int i,j,k=0;
    scanf("%d",&n);
    f[0]=1;
    for(i=1;i<=n;i++)  //得到n个数构成的集合f,有则为1,无则为0
      {
       scanf("%d",&j);
       f=f|(f<<j);
       k+=j;
       //printf("%d\n",f);
      }
      /*for(int i=1;i<=k;i++)
      {
          if(f[i]==0)
            printf("0\n");
          else
            printf("1\n");
      }*/
    for(i=(k+1)/2;!f[i];i++);//从总值的一半开始找
    printf("%d\n",i);
    return 0;
}