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>
#include <algorithm>
using namespace std;

long long comb(int n, int r) {
  vector<vector<long long>> v(n + 1,vector<long long>(n + 1, 0));
  for (int i = 0; i < v.size(); i++) {
    v[i][0] = 1;
    v[i][i] = 1;
  }
  for (int j = 1; j < v.size(); j++) {
    for (int k = 1; k < j; k++) {
      v[j][k] = (v[j - 1][k - 1] + v[j - 1][k]);
    }
  }
  return v[n][r];
}

int main(void){
    // Your code here!
    int N,K;
    cin >> N;
    cin >> K;
    
    unsigned digit=0;
    while(N!=0){
        N /= 10;
        digit++;
    }

    int count = 0;
    for(int i=K;i<digit+1;i++){
        if (i==digit){
            
        }else{
            count += 9*comb(i-1,K-1);
        }
    }
    cout << count << endl;
}