## Intuition The problem requires us to find the minimum absolute difference between a given number `a` and another number `b` that can be formed using only two distinct digits. The key insight is to generate numbers `b` using these digits and compare them to `a` to find the smallest difference. Since `n=2`, we can generate numbers using combinations of these two digits and check which one is closest to `a`. ## Approach 1. **Generate Possible Numbers**: Since we only have two digits, we can generate numbers using these digits. We can form numbers by considering all possible combinations of these two digits up to a certain length. 2. **Compare with `a`**: For each generated number `b`, compute the absolute difference `|a - b|`. Keep track of the minimum difference encountered. 3. **Iterate Over Lengths**: The length of the number `b` can vary. We start from 1-digit numbers and go up to a reasonable length. Given that `a` can be as large as `10^{17}`, we consider numbers up to a length of 18 digits. 4. **Brute Force Approach**: Since `n=2`, the number of possible combinations is manageable. We can use a brute force approach to generate all possible numbers `b` and find the one that minimizes the difference. ## Complexity - **Time Complexity**: O(2^L * L * T), where `L` is the maximum length of numbers we consider (up to 18), and `T` is the number of test cases. This is feasible given the constraints. - **Space Complexity**: O(1), as we only store a few variables for calculations. ## Solution (C++) ```cpp #include using namespace std; void solve() { long long a; int n; cin >> a >> n; vector d(n); for (int i = 0; i < n; ++i) { cin >> d[i]; } long long min_diff = LLONG_MAX; // Generate numbers using digits d[0] and d[1] for (int len = 1; len <= 18; ++len) { // Generate all numbers of length `len` using the digits in `d` for (int mask = 0; mask < (1 << len); ++mask) { long long b = 0; for (int i = 0; i < len; ++i) { int digit = (mask & (1 << i)) ? d[1] : d[0]; b = b * 10 + digit; } min_diff = min(min_diff, abs(a - b)); } } cout << min_diff << '\n'; } int main() { ios::sync_with_stdio(false); cin.tie(nullptr); int t; cin >> t; while (t--) { solve(); } return 0; } ``` This solution reads multiple test cases, generates possible numbers `b` using the two given digits, and computes the minimum absolute difference with `a`. It efficiently handles the constraints by limiting the length of numbers considered and using bit manipulation to generate combinations.

Cirno and Number (Easy Version)

Problem Details