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>

using namespace std;
using ll = long long;


int main()
{
	string s; int p;
	cin >> s >> p;
	for(auto &i : s)i -= '0';
	int n = s.size();
	vector<vector<vector<ll>>> dp(n + 1, vector<vector<ll>>(110, vector<ll>(2))); //i桁目まで見て0がjこでフラグがk;
	dp[0][0][0] = 1;

	for(int i = 0; i < n; i++) {
		for(int j = 0; j < 105; j++) {
			for(int k = 0; k < 2; k++) {

				for(int d = 0; d <= (k ? 9 : s[i]); d++) {
					dp[i + 1][j + (d != 0)][k || d < s[i]] += dp[i][j][k];
				}
			}
		}
	}
	cout << dp[n][p][0] + dp[n][p][1] << endl;
}