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 Python (3.4.3) (judge verdict: Accepted)

#!/usr/bin/env python3
import sys
input = sys.stdin.readline

n = int(input())
s = str(n)
K = int(input())
l = len(str(n))

# dp[loose/tight][count][length]
dp = [[[0] * l for _ in range(K + 1)] for _ in range(2)]
for i, item in enumerate(s):
    val = int(item)
    # Initialize
    if i == 0:
        dp[0][0][i] = 1
        for j in range(1, 10):
            if j < val:
                dp[0][1][i] += 1
            if j == val:
                dp[1][1][i] += 1
        continue
    # Use 0
    if val == 0:
        for k in range(K+1):
            dp[0][k][i] += dp[0][k][i-1]
            dp[1][k][i] += dp[1][k][i-1]
    else:
        for k in range(K+1):
            dp[0][k][i] += dp[0][k][i-1] + dp[1][k][i-1]
    for j in range(1, 10):
        if val == j:
            for k in range(K):
                dp[0][k+1][i] += dp[0][k][i-1]
                dp[1][k+1][i] += dp[1][k][i-1]
        elif j < val:
            for k in range(K):
                dp[0][k+1][i] += dp[0][k][i-1] + dp[1][k][i-1]
        else:
            for k in range(K):
                dp[0][k+1][i] += dp[0][k][i-1]

print(dp[0][-1][-1] + dp[1][-1][-1])