description: You are given a sequence of n integers a_1,   a_2,   ...,   a_n.
  
  Does there exist a sequence of n integers b_1,   b_2,   ...,   b_n such that the following property holds?
  
    * For each 1 ≤ i ≤ n, there exist two (not necessarily distinct) indices j and k (1 ≤ j,   k ≤ n) such that a_i = b_j - b_k. 
  
  Input
  
  The first line contains a single integer t (1 ≤ t ≤ 20) — the number of test cases. Then t test cases follow.
  
  The first line of each test case contains one integer n (1 ≤ n ≤ 10).
  
  The second line of each test case contains the n integers a_1,   ...,   a_n (-10^5 ≤ a_i ≤ 10^5).
  
  Output
  
  For each test case, output a line containing YES if a sequence b_1,   ...,   b_n satisfying the required property exists, and NO otherwise.
  
  Example
  
  Input
  
  
  5
  5
  4 -7 -1 5 10
  1
  0
  3
  1 10 100
  4
  -3 2 10 2
  9
  25 -171 250 174 152 242 100 -205 -258
  
  
  Output
  
  
  YES
  YES
  NO
  YES
  YES
  
  Note
  
  In the first test case, the sequence b = [-9,   2,   1,   3,   -2] satisfies the property. Indeed, the following holds: 
  
    * a_1 = 4 = 2 - (-2) = b_2 - b_5; 
    * a_2 = -7 = -9 - (-2) = b_1 - b_5; 
    * a_3 = -1 = 1 - 2 = b_3 - b_2; 
    * a_4 = 5 = 3 - (-2) = b_4 - b_5; 
    * a_5 = 10 = 1 - (-9) = b_3 - b_1. 
  
  
  
  In the second test case, it is sufficient to choose b = [0], since a_1 = 0 = 0 - 0 = b_1 - b_1.
  
  In the third test case, it is possible to show that no sequence b of length 3 satisfies the property.
solution: #include <bits/stdc++.h>
  using namespace std;
  set<int> s;
  int arr[15], x[100005];
  int get_sum(int mask, int n) {
    int ret = 0;
    for (int i = 0; i < n; i++) {
      if (mask & (1 << i)) ret += arr[i];
    }
    return ret;
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int t;
    cin >> t;
    for (int i = 0; i < t; i++) {
      int n;
      cin >> n;
      s.clear();
      for (int j = 0; j < 1e5 + 5; j++) {
        x[j] = 0;
      }
      for (int j = 0; j < n; j++) {
        cin >> arr[j];
        if (arr[j] < 0) arr[j] = -arr[j];
        x[arr[j]] = j + 1;
        s.insert(arr[j]);
      }
      if (s.size() < n) {
        cout << "YES\n";
        continue;
      }
      int b1 = 0;
      for (int mask1 = 0; mask1 < (1 << n); mask1++) {
        for (int mask2 = 0; mask2 < (1 << n); mask2++) {
          if (mask1 & mask2) continue;
          int a = get_sum(mask1, n) - get_sum(mask2, n);
          if (a > 1e5 || a < 0) continue;
          if (!x[a]) continue;
          if (!(mask1 & (1 << (x[a] - 1))) && !(mask2 & (1 << (x[a] - 1)))) {
            b1 = 1;
          }
        }
      }
      if (b1)
        cout << "YES\n";
      else
        cout << "NO\n";
    }
    return 0;
  }
  