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: Wrong Answer)

#include <bits/stdc++.h>
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
using namespace std;
using ll = long long;

ll chose(int n, int r) {
  ll a = 1, b = 1;
  rep(i, r) {
    a *= n-i;
    b *= r-i;
  }
  return a / b;
}

int main() {
  string N;
  int K;
  cin >> N >> K;
  int n = N.size();
  ll ans = 0;
  ans = chose(n-1, K)*pow(9, K);
  int s = 0;
  int e = 0;
  rep(i, K) {
    s = pow(10, i);
    e = 10*e + (N[i] - '0');
  }
  for (int i = s; i <= e; i++) {
    int a = i;
    int c = 0;
    while(a > 0) {
      int b = a%10;
      a /= 10;
      if (b != 0) c++;
    }
    if (i == e && c != K) {
      int d = 0;
      int x; int y;
      for (int j = 0; j < n; j++) {
        if (N[j] != '0') d++;
        if (d == K) {
          x = N[j] - '0';
          y = j;
          break;
        }
      }
      ans += x;
      ans += 9*(n-y-1);
    }
    else {
      ans += chose(n-K, K-c)*pow(9, K-c);
    }
  }
  cout << ans << endl;
}