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)

// Copyright 2020 yaito3014
#include <bits/stdc++.h>
using namespace std;

int main(){
	string N;
	int K;
	cin >> N >> K;
	int len = N.length();
	// dp[i][j][smaller]
	vector<vector<array<int,2>>> dp(len+1,vector<array<int,2>>(K+1));
	for(auto&v:dp)for(auto&w:v)w.fill(0);
	dp[0][0][0] = 1;
	for(int i = 0; i < len; ++i){
		for(int j = 0; j <= K; ++j){
			for(auto smaller : { true, false }){
				int k_n = smaller?9:N[i]-'0';
				for(int k = 0; k <= k_n; ++k){
					if(k!=0&&j==K)continue;
					dp[i+1][j+(k==0?0:1)][smaller||k<N[i]-'0']
						+= dp[i][j][smaller];
				}
			}
		}
	}
	cout << dp[len][K][0] + dp[len][K][1] << endl;
}