description: It is a complicated version of problem F1. The difference between them is the constraints (F1: k ≤ 2, F2: k ≤ 10).
  
  You are given an integer n. Find the minimum integer x such that x ≥ n and the number x is k-beautiful.
  
  A number is called k-beautiful if its decimal representation having no leading zeroes contains no more than k different digits. E.g. if k = 2, the numbers 3434443, 55550, 777 and 21 are k-beautiful whereas the numbers 120, 445435 and 998244353 are not.
  
  Input
  
  The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t test cases follow.
  
  Each test case consists of one line containing two integers n and k (1 ≤ n ≤ 10^9, 1 ≤ k ≤ 10).
  
  Output
  
  For each test case output on a separate line x — the minimum k-beautiful integer such that x ≥ n.
  
  Example
  
  Input
  
  
  6
  2021 3
  177890 2
  34512 3
  724533 4
  998244353 1
  12345678 10
  
  
  Output
  
  
  2021
  181111
  34533
  724542
  999999999
  12345678
solution: #include <bits/stdc++.h>
  using namespace std;
  const int N = 100005;
  int t, n, k, st[30], vis[32];
  int main() {
    scanf("%lld", &t);
    while (t--) {
      scanf("%d%d", &n, &k);
      st[0] = 0;
      while (n) st[++st[0]] = n % 10, n /= 10;
      memset(vis, 0, sizeof(vis));
      while (1) {
        int status = 0, num = 0;
        for (register int i = 1; i <= st[0]; ++i)
          if (st[i] >= 10) st[i] = 0, st[i + 1]++;
        for (register int i = st[0]; i >= 1; --i) {
          num += !(status >> st[i] & 1);
          status |= 1 << st[i];
          if (num > k) {
            ++st[i];
            for (register int j = 1; j <= i - 1; ++j) st[j] = 0;
            break;
          }
        }
        if (num <= k) {
          for (register int i = st[0]; i >= 1; --i) printf("%d", st[i]);
          puts("");
          break;
        }
      }
    }
    return 0;
  }
  