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 PyPy3 (2.4.0) (judge verdict: Wrong Answer)

# ABC154e


def main():
    import sys
    import math
    input = sys.stdin.readline
    sys.setrecursionlimit(10 ** 6)
    # 再帰関数を使わない限りPypyで出すこと

    n = int(input())
    ns = str(n)
    k = int(input())
    keta = len(ns)
    # dp[i+1][smaller][j]=i桁目までで、未満フラグをsmaller、0の個数j
    dp = [[[0]*(keta+1) for i in range(2)]
          for j in range(keta + 1)]

    dp[0][0][0] = 1

    for i in range(keta):
        for smaller in range(2):
            for j in range(i+1):
                for x in range(10 if smaller == 1 else int(ns[i])+1):
                    # if i == 0 and x == 0:
                    #    continue
                    dp[i + 1][1 if smaller == 1 or x < int(ns[i])
                              else 0][j if x != 0 else j+1] += dp[i][smaller][j]
    # print(dp)
    # print(dp[keta][0][keta-k])
    #print(dp[keta][1][keta - k])
    print(dp[keta][0][keta-k]+dp[keta][1][keta - k])


if __name__ == '__main__':
    main()