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: Runtime Error)

from math import log10
n=input()
k=int(input())

#ないときも後で
l=len(n)
ans1=0
if l>k:
    if k==1:
        ans1=l-1
    elif k==2:
        ans1=(l-1)*(l-2)//2
    else:
        ans1=(l-1)*(l-2)*(l-3)//6
ans1*=9**k
if l>=k:
    if k==1:
        ans=0
    elif k==2:
        ans=(l-1)
    else:
        ans=(l-1)*(l-2)//2
ans*=(int(n[0])-1)*9**(k-1)

dp=[[0]*3 for _ in range(l-1)]
dp[-1][0]=1
n=list(n)
n.reverse()
for i,c in enumerate(n[:-1]):
    c=int(c)
    dp[i][0]=1
    dp[i][1]=dp[i-1][1]+dp[i-1][0]*c
    dp[i][2]=dp[i-1][2]+dp[i-1][1]*c
ans+=dp[-1][k-1]
print(ans+ans1)