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 <algorithm>
#include <cstdio>
#include <cstring>
#include <queue>
#include <utility>
#include <vector>
using namespace std;
typedef long long ll;

int main() {
  char s[110];
  int k, n, ans = 0;
  scanf("%s", s);
  scanf("%d", &k);
  n = strlen(s);
  int dp0[110][5] = {}, dp1[110][5] = {};
  for (int i = 1; i <= 9; i++) {
    if (i < s[0] - '0') {
      dp0[1][1]++;
    } else if (i == s[0] - '0') {
      dp1[1][1]++;
    }
  }
  dp0[1][0] = 1;
  for (int i = 2; i <= n; i++) {
    for (int j = 0; j < 4; j++) {
      dp0[i][j] += dp0[i - 1][j];
      dp0[i][j + 1] += 9 * dp0[i - 1][j];
      if (s[i - 1] - '0') {
        dp0[i][j + 1] += (s[i - 1] - '0' - 1) * dp1[i - 1][j];
        dp1[i][j + 1] += dp1[i - 1][j];
        dp0[i][j] += dp1[i - 1][j];
      } else {
        dp1[i][j] += dp1[i - 1][j];
      }
    }
  }
  printf("%d\n", dp0[n][k] + dp1[n][k]);
}