Problem:
Score : 500 points

Problem Statement

Find the number of integers between 1 and N (inclusive) that contains exactly K non-zero digits when written in base ten.

Constraints

1 \leq N < 10^{100}

1 \leq K \leq 3

Input

Input is given from Standard Input in the following format:

N
K

Output

Print the count.

Sample Input 1

100
1

Sample Output 1

19

The following 19 integers satisfy the condition:

1,2,3,4,5,6,7,8,9,10,20,30,40,50,60,70,80,90,100

Sample Input 2

25
2

Sample Output 2

14

The following 14 integers satisfy the condition:

11,12,13,14,15,16,17,18,19,21,22,23,24,25

Sample Input 3

314159
2

Sample Output 3

937

Sample Input 4

9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999
3

Sample Output 4

117879300

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <iostream>
#include <string>
#include <vector>

using namespace std;
using ll = long long int;

int main() {
  string S;
  int K;
  cin >> S >> K;

  vector<int> N(S.size());
  for (size_t i = 0; i < S.size(); i++) {
    N[i] = S[i] - '0';
  }

  ll dp[200][4][2] = {};
  dp[0][0][0] = 1;

  for (size_t i = 0; i < N.size(); i++) {
    for (size_t d = 0; d < 10; d++) {
      for (size_t k = 0; k < 4; k++) {
        for (size_t l = 0; l <= 1; l++) {
          int nk = k;
          if (d != 0) {
            nk++;
          }
          if ((l == 0 && N[i] < d) || nk > 3) {
            continue;
          }
          int nl = l;
          if (N[i] > d) {
            nl = 1;
          }
          dp[i + 1][nk][nl] += dp[i][k][l];
        }
      }
    }
  }

  cout << (dp[N.size()][K][1] + dp[N.size()][K][0]) << endl;
}