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)

#154_E
n = input()
s = len(n)
x = int(input())

#dp[i][flg][cnt] ... i桁目まで、未満フラグ、0以外の数
dp = [[[0 for _ in range(4)] for _ in range(2)] for _ in range(s+1)]
dp[0][0][0] = 1

for i in range(1, s+1):
    for k in [0, 1, 2, 3]:
        if n[i-1] == '0':
            dp[i][0][k] += dp[i-1][0][k]
        else:
            dp[i][1][k] += dp[i-1][0][k]
            if k > 0:
                dp[i][0][k] += dp[i-1][0][k-1]
                dp[i][1][k] += dp[i-1][0][k-1] * (int(n[i-1]) - 1)
        
        dp[i][1][k] += dp[i-1][1][k]
        if k > 0:
            dp[i][1][k] += dp[i-1][1][k-1] * 9
   
print(dp[s][0][x] + dp[s][1][x])