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;
  int L, m;
  char a[1000];
  int pan() {
    int j, i;
    set<char> s;
    for (i = 0; i < L; i++) {
      s.insert(a[i]);
    }
    if (s.size() > m)
      return 1;
    else
      return 0;
  }
  int main() {
    int j, i, w;
    scanf("%d", &w);
    while (w--) {
      scanf("%s %d", a, &m);
      L = strlen(a);
      i = 0;
      while (pan()) {
        set<char> s;
        for (i = 0; i < L; i++) {
          s.insert(a[i]);
          if (s.size() > m) {
            while (a[i] == '9') i--;
            a[i]++;
            for (j = i + 1; j < L; j++) a[j] = '0';
            break;
          }
        }
      }
      printf("%s\n", a);
    }
    return 0;
  }
  