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 (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#include <string.h>
#include <math.h>

long Combination(long n, long r) {
  long answer = 1;
  if (n-r < r) {
    r = n-r;
  }
  for (long i = 1; i <= r; i++) {
    answer *= n+1-i;
    answer /= i;
  }
  return answer;
}

int main(void) {

  char n[200];
  scanf("%s", n);
  int length = strlen(n);
  int k;
  scanf("%d", &k);
  long count;
  long sum = 0;
  int flag;
  for (int digits = k; digits < length; digits++) {
    count = pow(9,k);
    count *= Combination(digits-1,k-1);
    sum += count;
  }
  if (k == 1) {
    sum += n[0]-'0';
  } else if (k == 2) {
    sum += (n[0]-'0'-1)*9*(length-1);
    flag = 0;
    for (int i = 1; i < length; i++) {
      if (flag == 0) {
        sum += n[i]-'0';
      } else {
        sum += 9;
      }
      if (n[i] != '0') {
        flag = 1;
      }
    }
  } else {
    sum += (n[0]-'0'-1)*pow(9,2)*Combination(length-1,2);
    for (int second = 1; second < length-1; second++) {
      if (n[second] != '0') {
        sum += (n[second]-'0'-1)*9*(length-second-1);
      }
    }
    for (int second = 1; second < length-1; second++) {
      if (n[second] == '0') {
        continue;
      }
      for (int third = second+1; third < length; third++) {
        flag = 0;
        for (int i = 1; i < third; i++) {
          if (i == second) {
            continue;
          }
          if (n[i] != '0') {
            flag = 1;
            break;
          }
        }
        if (flag == 0) {
          sum += n[third]-'0';
        } else {
          sum += 9;
        }
      }
    }
    flag = 0;
    for (int second = 1; second < length-1; second++) {
      if (flag == 1) {
        sum += (9-n[second]+'0')*9*(length-second-1);
      }
      if (n[second] != '0') {
        flag = 1;
      }
    }
  }
  printf("%ld\n", sum);

  return 0;
}