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: Runtime Error)

#include <iostream>
#include <vector>
using namespace std;
int main() {
  string s;
  int k;
  cin >> s >> k;
  int l = s.size();
  vector<int> n(l);
  for (int i = 0; i < l; ++i) n.at(i) = s.at(i) - '0';

  vector<vector<int>> dp0(l, vector<int>(k, 0)), dp1(l, vector<int>(k, 0));

  for (int i = 0; i < k; ++i) {
    dp1.at(i).at(i) = 1;
  }

  for (int i = 0; i < l; ++i) {
    for (int j = 0; j < k; ++j) {
      if (j > i) {
        continue;
      }
      if (i == 0 & j == 0) {
        dp0.at(i).at(j) = n.at(i) - 1;
      } else if (j == 0) {
        dp0.at(i).at(j) = dp0.at(i-1).at(j) + dp1.at(i-1).at(j) + 9;
      } else {
        dp0.at(i).at(j) = dp0.at(i-1).at(j-1) * 9 + dp1.at(i-1).at(j-1) * (n.at(i) - 1) + dp1.at(i-1).at(j) + dp0.at(i-1).at(j);
      }
    }
  }
  cout << dp0.at(l-1).at(k-1) + dp1.at(l-1).at(k-1) << endl;
  return 0;
}