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: Wrong Answer)

#500
N = int(input())
K = int(input())
string = str(N)
digit = len(string)
dp = [[[0,0] for _ in range(K+1)] for _ in range(digit+1)]

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

for i in range(digit):
    for j in range(K):
        dp[i+1][0][1] = 1
        dp[i+1][j+1][0] += dp[i][j][0]
        if string[i] == "0":
            dp[i+1][j+1][0] += dp[i][j+1][0]
        dp[i+1][j+1][1] += 9*dp[i][j][1] + dp[i][j+1][1]
        if string[i] != "0":
            dp[i+1][j+1][1] += dp[i][j+1][0] + (int(string[i])-1)*dp[i][j][0]
            
if N - 10*(digit-1) and K >= 2:
    n = dp[digit][K][1]
else:
    n = dp[digit][K][0] + dp[digit][K][1]
print(n)