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 <algorithm>
#include <array>
#include <cstddef>
#include <cstdint>
#include <iomanip>
#include <iostream>
#include <map>
#include <set>
#include <vector>

using namespace std;

int main()
{
    string n;
    cin >> n;
    int k;
    cin >> k;

    reverse(n.begin(), n.end());

    if (k == 1)
    {
        int ans = 0;
        for (int i = 0; i < n.size(); i++)
        {
            if (i == n.size() - 1)
            {
                ans += n[i] - '1' + 1;
            }
            else
            {
                ans += 9;
            }
        }
        cout << ans << endl;
    }
    else if (k == 2)
    {
        int ans = 0;
        for (int i = 0; i < n.size(); i++)
        {
            for (int j = i + 1; j < n.size(); j++)
            {
                if (j == n.size() - 1)
                {
                    ans += (n[j] - '1') * 9;
                    if (i == n.size() - 2)
                    {
                        ans += (n[i] - '1' + 1);
                    }
                    else
                    {
                        ans += 9;
                    }
                }
                else
                {
                    ans += 9 * 9;
                }
            }
        }
        cout << ans << endl;
    }
    else
    {
        int64_t ans = 0;
        for (int i = 0; i < n.size(); i++)
        {
            for (int j = i + 1; j < n.size(); j++)
            {
                for (int l = j + 1; l < n.size(); l++)
                {
                    if (l == n.size() - 1)
                    {
                        ans += (n[l] - '1') * 81;

                        if (j == n.size() - 2)
                        {
                            ans += (n[j] - '1') * 9;

                            if (i == n.size() - 3)
                            {
                                ans += (n[i] - '1' + 1);
                            }
                            else
                            {
                                ans += 9;
                            }
                        }
                        else
                        {
                            ans += 9 * 9;
                        }
                    }
                    else
                    {
                        ans += 9 * 9 * 9;
                    }
                }
            }
        }
        cout << ans << endl;
    }
}