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>
#include<stdlib.h>
#include<stdint.h>
#include<inttypes.h>

typedef int32_t i32;
typedef uint64_t u64;

void run (void) {
  i32 n;
  scanf ("%" SCNi32, &n);
  i32 *a = (i32 *) calloc (n, sizeof (i32));
  i32 sum = 0;
  for (i32 i = 0; i < n; ++i) {
    scanf ("%" SCNi32, a + i);
    sum += a[i];
  }
  const i32 len = (sum + sizeof (u64) - 1) / sizeof (u64);
  u64 *dp = (u64 *) calloc (len, sizeof (u64));
  dp[0] = 1;
  for (i32 i = 0; i < n; ++i) {
    i32 v = a[i];
    i32 p = v / sizeof (u64);
    i32 q = v % sizeof (u64);
    if (q == 0) {
      for (i32 j = len - 1; j - p >= 0; --j) {
        dp[j] |= dp[j - p];
      }
    } else {
      for (i32 j = len - 1; j - p - 1 >= 0; --j) {
        u64 y = dp[j - p];
        u64 z = dp[j - p - 1];
	dp[j] |= z >> (sizeof (u64) - q);
	dp[j] |= y << q;
      }
      dp[p] |= dp[0] << q;
    }
  }
  i32 m = (sum + 1)/ 2;
  while (((dp[m / sizeof (u64)] >> (m % sizeof (u64))) & 1) == 0) m++;
  printf ("%" PRIi32 "\n", m);
}

int main (void) {
  run ();
  return 0;
}