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;
  int T, N;
  bool ex[((int)1e6 + 10)];
  int tv[10];
  int v[10];
  int reg[((int)1e6 + 10)];
  int reginx = 0;
  int main() {
    scanf("%d", &T);
    while (T--) {
      scanf("%d", &N);
      bool dup = false;
      for (int i = 0; i < N; i++) {
        scanf("%d", v + i);
        if (v[i] == 0) dup = true;
        if (v[i] < 0) v[i] = -v[i];
      }
      sort(v, v + N);
      for (int i = 1; i < N && !dup; i++)
        if (v[i] == v[i - 1]) dup = true;
      if (dup) {
        printf("YES\n");
        continue;
      }
      bool ok = false;
      fill(ex, ex + ((int)1e6 + 10), false);
      reginx = 0;
      reg[reginx++] = v[0];
      ex[v[0]] = true;
      for (int i = 1; i < N && !ok; i++) {
        int val = v[i];
        int rsize = reginx;
        for (int k = 0; k < rsize; k++) {
          int newval = val + reg[k];
          if (ex[newval]) {
            ok = true;
            break;
          } else {
            reg[reginx++] = newval;
            ex[newval] = true;
          }
        }
        if (ex[val]) {
          ok = true;
          break;
        } else {
          reg[reginx++] = val;
          ex[val] = true;
        }
      }
      if (ok)
        printf("YES\n");
      else
        printf("NO\n");
    }
    return 0;
  }
  