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 (GCC 9.2.1) (judge verdict: Accepted)

#include <stdio.h>

const int pow[4] = {1, 9, 81, 729};

int combination(int n, int k)
{
	if (n < k) return 0;
	else if (k == 0 || n == k) return 1;
	
	int i, x;
	for (i = n, x = 1; i >= n - k + 1; i--) x *= i;
	for (i = 2; i <= k; i++) x /= i;
	return x;
}

long long recursion(char N[], int l, int K)
{
	if (K == 0) return 1;
	else if (l == 0) return 0;
	else if (N[0] == '0') return recursion(&(N[1]), l - 1, K);
	else return recursion(&(N[1]), l - 1, K - 1) + combination(l - 1, K - 1) * pow[K-1] * (N[0] - '1') + combination(l - 1, K) * pow[K];
}

int main()
{
	int l, K;
	char N[102];
	scanf("%s", N);
	scanf("%d", &K);
	for (l = 0; N[l] != 0; l++);
	printf("%lld\n", recursion(N, l, K));
	fflush(stdout);
	return 0;
}