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: Accepted)

#include <bits/stdc++.h>
#include <algorithm>
using namespace std;
 
#define rep(i, a, b) for (int i = a; i < b; i++)
#define print(a) cout << (a) << endl; 

int dp[101][4][2];

int main() {
    string s; 
    int k;
    cin >> s;
    cin >> k;
    int n = s.size();
    int ret = 0;
    int i0 = s[0] - '0';
    dp[0][1][1] = 1;
    dp[0][1][0] = i0 - 1;
    dp[0][0][1] = 0;
    dp[0][0][0] = 1;

    rep(i, 1, n) {
        int is = s[i] - '0';
        dp[i][0][1] = 0;
        dp[i][0][0] = 1; 
        if (is > 0) {
            rep(k, 1, 4) {
                dp[i][k][0] = dp[i-1][k-1][0] * 9 
                    + dp[i-1][k][0] + dp[i-1][k-1][1] * (is-1) + dp[i-1][k][1]; 
                dp[i][k][1] = dp[i-1][k-1][1]; 
            }
        } else {
            rep(k, 1, 4) {
                dp[i][k][0] = dp[i-1][k][0] + dp[i-1][k-1][0]*9; 
                dp[i][k][1] = dp[i-1][k][1]; 
            } 
        }
    }
    cout << dp[n-1][k][0] + dp[n-1][k][1] << endl;

    return 0;
}