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 <iostream>
#include <cmath>

int d(int,int);
int BC(int,int);

int main(void){
  int N,K;
  std::cin >> N;
  std::cin >> K;
  int ans = d(N,K);
  std::cout << ans << std::endl; 
  return 0;
}

int d(int N,int K){
  int num = 1; 
  if(K!=0){
    // 桁数を求める
    int n = (int)std::log10(N);
    // n+1が桁数 n+1 < Kなら終了
    if(n+1 < K){
      return 0;
    }
    // 最上位の数
    int Xn = (int)(N/std::pow(10,n));
    // Xnが0なら0にセット
    if(Xn == 0){
      num = 0;
    }
    // あまりの数
    int remain = N - Xn * std::pow(10,n);
    return num*(Xn-1)*BC(n,K-1)*std::pow(9,K-1) + BC(n,K)*std::pow(9,K) + d(remain,K-1);
  }else{
    return 0;
  }
  return 0;
}

int BC(int n, int k) {
  if(n < k)
    return 0;
   if (k == 0 || k == n){
   	  return 1;
   }
   return BC(n - 1, k - 1) + BC(n - 1, k);
}