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;
  long long t, n, b[100010], lcm, a[100010], ans[100010];
  long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
  int main() {
    scanf("%lld", &t);
    while (t--) {
      scanf("%lld", &n);
      for (long long i = 1; i <= n; ++i) {
        scanf("%lld", &b[i]);
        a[i] = abs(b[i]);
      }
      if (n & 1) {
        for (long long i = 1; i <= n - 3; i += 2) {
          ans[i] = -b[i + 1];
          ans[i + 1] = b[i];
        }
        lcm = a[n - 1] * a[n - 2] / gcd(a[n - 1], a[n - 2]);
        lcm = lcm * a[n] / gcd(lcm, a[n]);
        ans[n] = -2 * (lcm / b[n]);
        ans[n - 1] = lcm / b[n - 1];
        ans[n - 2] = lcm / b[n - 2];
      } else
        for (long long i = 1; i <= n; i += 2) {
          ans[i] = -b[i + 1];
          ans[i + 1] = b[i];
        }
      for (long long i = 1; i <= n; ++i) printf("%lld ", ans[i]);
      printf("\n");
    }
    return 0;
  }
  