## Intuition The problem requires transforming an array of positive integers into an array of zeros using a series of operations. Each operation allows us to choose a subset of the array and replace it with a new subset, provided the sum of the new subset equals the sum of the chosen subset. The key insight is that we can use a greedy approach to iteratively reduce the array elements by choosing strategic subsets. ## Approach 1. **Understanding the Operation**: The operation allows us to select a subset of the array and replace it with another subset of the same sum. This means we can gradually reduce the values in the array by choosing subsets that allow us to set elements to zero. 2. **Greedy Reduction**: Since the problem guarantees that it is possible to reduce the array to zero in at most 17 operations, we can use a greedy approach: - At each step, try to zero out a significant portion of the array. - Use the largest possible power of two to reduce the elements, as this allows us to handle large numbers efficiently. 3. **Implementation Strategy**: - For each test case, start with the original array. - Iteratively apply operations to reduce the array: - Choose a subset of the array. - Replace it with a subset that zeros out as many elements as possible. - Continue this process until all elements are zero. 4. **Output**: For each test case, output the number of operations and the operations themselves. ## Complexity - **Time Complexity**: O(n) per test case, where n is the number of elements in the array. This is because each operation can be done in linear time. - **Space Complexity**: O(n) for storing the operations. ## Solution (C++) ```cpp #include using namespace std; int main() { ios::sync_with_stdio(false); cin.tie(nullptr); int t; cin >> t; while (t--) { int n; cin >> n; vector a(n); for (int i = 0; i < n; ++i) { cin >> a[i]; } // We will use a greedy approach to zero out the array vector> operations; while (true) { bool all_zero = true; for (int i = 0; i < n; ++i) { if (a[i] != 0) { all_zero = false; break; } } if (all_zero) break; // Find the largest power of two less than or equal to the largest element long long max_elem = *max_element(a.begin(), a.end()); long long power_of_two = 1; while (power_of_two <= max_elem) { power_of_two *= 2; } power_of_two /= 2; // Create an operation vector b(n, 0); for (int i = 0; i < n; ++i) { if (a[i] >= power_of_two) { b[i] = power_of_two; a[i] -= power_of_two; } } operations.push_back(b); } cout << operations.size() << "\n"; for (const auto& op : operations) { for (long long x : op) { cout << x << " "; } cout << "\n"; } } return 0; } ``` This solution uses a greedy approach to iteratively reduce the array to zero by applying operations that zero out elements using powers of two. The solution is efficient and adheres to the problem's constraints.

Make it Zero

Problem Details