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;

signed main() {
	string S;
	int K;
	cin >> S >> K;
	int N = (int)S.size();
	// Ai以下を見て, Sと等しいか, 以下か
	vector<vector<ll>> dp_equal(N+1, vector<ll>(4, 0));
	vector<vector<ll>> dp_less(N+1, vector<ll>(4, 0));
	// 0個見て, 0でない数字が0個 つまり0
	dp_equal[0][0] = 1;

	for(int i=1;i<=N;i++) {
		int now = (int)(S[i-1] - '0');
		// i桁目まで全部0
		dp_less[i][0] = 1;
		for(int j=1;j<=K;j++) {
			// equal -> equalについて
			if(now > 0) dp_equal[i][j] = dp_equal[i-1][j-1];
			else dp_equal[i][j] = dp_equal[i-1][j];

			// less -> lessについて
			// 1~9のどれかを置ける
			dp_less[i][j] += dp_less[i-1][j-1] * 9;
			// 0を置く
			dp_less[i][j] += dp_less[i-1][j];

			// now == 0のときはequal -> lessにならない
			if(now > 0) {
				// 0を置く
				dp_less[i][j] += dp_equal[i-1][j];
				// equalからlessになる, 1以上now-1以下
				dp_less[i][j] += dp_equal[i-1][j-1] * (now - 1);
			}
		}
	}
	cout << dp_equal[N][K] + dp_less[N][K] << endl;
}