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<bits/stdc++.h>
using namespace std;

int main() {
	
	string n;
	int k;
	cin >> n >> k;
	int m = n.size();
	vector<vector<vector<int64_t>>> dp(m,vector<vector<int64_t>>(2,vector<int64_t>(k+1)));
	dp.at(0).at(1).at(1) = n.at(0)-1-'0';
	dp.at(0).at(1).at(0) = 1;
	dp.at(0).at(0).at(1) = 1;
	for(int i=1; i<m; i++){
		for(int j=0; j<k+1; j++){
			if(j)dp.at(i).at(1).at(j) += 9*dp.at(i-1).at(1).at(j-1);
			dp.at(i).at(1).at(j) += dp.at(i-1).at(1).at(j);
			if(n.at(i)>0){
				if(j)dp.at(i).at(1).at(j) += ((n.at(i))-1-'0') * dp.at(i-1).at(0).at(j-1);
				dp.at(i).at(1).at(j) += dp.at(i-1).at(0).at(j);
			}
			if(n.at(i)>0){
				if(j)dp.at(i).at(0).at(j) += dp.at(i-1).at(0).at(j-1);
			}
			else{
				dp.at(i).at(0).at(j) += dp.at(j-1).at(0).at(j);
			}
		}
	}
	cout << dp.at(m-1).at(1).at(k)+dp.at(m-1).at(0).at(k) << endl;
	
}