description: In Berland, n different types of banknotes are used. Banknotes of the i-th type have denomination 10^{a_i} burles (burles are the currency used in Berland); the denomination of banknotes of the first type is exactly 1.
  
  Let's denote f(s) as the minimum number of banknotes required to represent exactly s burles. For example, if the denominations of banknotes used in Berland are 1, 10 and 100, then f(59) = 14: 9 banknotes with denomination of 1 burle and 5 banknotes with denomination of 10 burles can be used to represent exactly 9 ⋅ 1 + 5 ⋅ 10 = 59 burles, and there's no way to do it with fewer banknotes.
  
  For a given integer k, find the minimum positive number of burles s that cannot be represented with k or fewer banknotes (that is, f(s) > k).
  
  Input
  
  The first line contains a single integer t (1 ≤ t ≤ 10^4) — number of test cases.
  
  The first line of each test case contains two integers n and k (1 ≤ n ≤ 10; 1 ≤ k ≤ 10^9).
  
  The next line contains n integers a_1, a_2, ..., a_n (0 = a_1 < a_2 < ... < a_n ≤ 9).
  
  Output
  
  For each test case, print one integer — the minimum positive number of burles s that cannot be represented with k or fewer banknotes.
  
  Example
  
  Input
  
  
  4
  3 13
  0 1 2
  2 777
  0 4
  3 255
  0 1 3
  10 1000000000
  0 1 2 3 4 5 6 7 8 9
  
  
  Output
  
  
  59
  778
  148999
  999999920999999999
solution: #include <bits/stdc++.h>
  using namespace std;
  const long long inf = 0x3f3f3f3f3f3f3f3f;
  template <typename T>
  inline void read(T& X) {
    X = 0;
    int w = 0;
    char ch = 0;
    while (!isdigit(ch)) {
      w |= ch == '-';
      ch = getchar();
    }
    while (isdigit(ch)) X = (X << 3) + (X << 1) + (ch ^ 48), ch = getchar();
    if (w) X = -X;
  }
  long long a[20], p[20];
  int main() {
    int t;
    for (int i = 0; i <= 9; i++) p[i] = pow(10, i);
    scanf("%d", &t);
    while (t--) {
      long long k, n;
      long long ans = 0;
      scanf("%lld %lld", &n, &k);
      for (int i = 1; i <= n; i++) {
        scanf("%lld", &a[i]);
        a[i] = p[a[i]];
      }
      for (int i = 1; i <= n; i++) {
        if (i == n) {
          ans = ans + (k + 1) * a[i];
          break;
        }
        long long now = (a[i + 1] - a[i]) / a[i];
        if (now <= k) {
          ans = ans + a[i + 1] - a[i];
          k = k - now;
        } else {
          ans = ans + a[i] * (k + 1);
          break;
        }
      }
      printf("%lld\n", ans);
    }
    return 0;
  }
  