description: Vupsen and Pupsen were gifted an integer array. Since Vupsen doesn't like the number 0, he threw away all numbers equal to 0 from the array. As a result, he got an array a of length n.
  
  Pupsen, on the contrary, likes the number 0 and he got upset when he saw the array without zeroes. To cheer Pupsen up, Vupsen decided to come up with another array b of length n such that ∑_{i=1}^{n}a_i ⋅ b_i=0. Since Vupsen doesn't like number 0, the array b must not contain numbers equal to 0. Also, the numbers in that array must not be huge, so the sum of their absolute values cannot exceed 10^9. Please help Vupsen to find any such array b!
  
  Input
  
  The first line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases. The next 2 ⋅ t lines contain the description of test cases. The description of each test case consists of two lines.
  
  The first line of each test case contains a single integer n (2 ≤ n ≤ 10^5) — the length of the array.
  
  The second line contains n integers a_1, a_2, …, a_n (-10^4 ≤ a_i ≤ 10^4, a_i ≠ 0) — the elements of the array a.
  
  It is guaranteed that the sum of n over all test cases does not exceed 2 ⋅ 10^5.
  
  Output
  
  For each test case print n integers b_1, b_2, …, b_n — elements of the array b (|b_1|+|b_2|+… +|b_n| ≤ 10^9, b_i ≠ 0, ∑_{i=1}^{n}a_i ⋅ b_i=0).
  
  It can be shown that the answer always exists.
  
  Example
  
  Input
  
  
  3
  2
  5 5
  5
  5 -2 10 -9 4
  7
  1 2 3 4 5 6 7
  
  
  Output
  
  
  1 -1
  -1 5 1 -1 -1
  -10 2 2 -3 5 -1 -1
  
  Note
  
  In the first test case, 5 ⋅ 1 + 5 ⋅ (-1)=5-5=0. You could also print 3 -3, for example, since 5 ⋅ 3 + 5 ⋅ (-3)=15-15=0
  
  In the second test case, 5 ⋅ (-1) + (-2) ⋅ 5 + 10 ⋅ 1 + (-9) ⋅ (-1) + 4 ⋅ (-1)=-5-10+10+9-4=0.
solution: #include <bits/stdc++.h>
  using namespace std;
  int a[100010];
  int b[100010];
  bool IN(int x) { return x >= -10000 && x <= 10000; }
  int main() {
    int T;
    scanf("%d", &T);
    while (T--) {
      int n;
      scanf("%d", &n);
      for (int i = 1; i <= n; i++) {
        scanf("%d", &a[i]);
      }
      if (n % 2 == 0) {
        for (int i = 1; i <= n; i += 2) {
          b[i] = a[i + 1], b[i + 1] = -a[i];
        }
      } else {
        for (int i = 4; i <= n; i += 2) {
          b[i] = a[i + 1], b[i + 1] = -a[i];
        }
        int t1 = a[1] < 0, t2 = a[2] < 0, t3 = a[3] < 0;
        a[1] = abs(a[1]), a[2] = abs(a[2]), a[3] = abs(a[3]);
        if (a[1] == a[2] && a[2] == a[3])
          b[1] = 1, b[2] = 1, b[3] = -2;
        else {
          int flag = 0;
          if (a[1] == a[2]) flag = 1, swap(a[2], a[3]);
          assert(a[1] != a[2]);
          if (IN(a[1] + a[2]))
            b[1] = 1, b[2] = 1;
          else
            assert(IN(a[1] - a[2])), b[1] = 1, b[2] = -1;
          int sum = a[1] * b[1] + a[2] * b[2];
          b[3] = -sum, b[1] *= a[3], b[2] *= a[3];
          if (flag) swap(a[2], a[3]), swap(b[2], b[3]);
        }
        if (t1) b[1] *= -1;
        if (t2) b[2] *= -1;
        if (t3) b[3] *= -1;
      }
      for (int i = 1; i <= n; i++) {
        printf("%d ", b[i]);
      }
      printf("\n");
    }
    return 0;
  }
  