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)

N = int(input())
K = int(input())
ans = 0
while N!=0:
    n = str(N)
    if len(n)<K:break

    if n[1:].count("9")==len(n)-1:
        if K==1:ans+=int(n[0])
        elif K==2:
            for i in range(1,len(n)):
                ans+=int(n[0])*9
        else:
            for i in range(1,len(n)-1):
                for j in range(i+1,len(n)):
                    ans+=int(n[0])*9*9
    else:
        if K==1:ans+=1
        elif K==2:
            k = 0
            for i in range(1,len(n)):
                if k==0:
                    ans += int(n[i])
                    if n[i]!="0":k=1
                else:
                    ans+=9
        else:
            k = 0
            for i in range(1,len(n)-1):
                for j in range(2,len(n)):
                    if k==0:
                        ans += int(n[i])*9
                    else:
                        ans += 9*9
                if n[i]!="0":k=1
    if len(n)==1:N = 0
    elif n[1:].count("9")==len(n)-1:N = 10**(len(n)-1)-1
    else:N = int(n[0])*(10**(len(n)-1))-1
print(ans)