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 C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>

int main(){
   long ans;
   int i,j,n[100],k,nn;
   char s[101];
 
   scanf("%s",s);
   scanf("%d",&k);
   for (i=0;s[i]!='\0';i++) n[i]=s[i]-'0';
   nn=i;
   if (nn<k) puts("0");
   else if (k==1) {
      printf("%d\n",n[0]+(nn-1)*9);
   } else if (k==2) {
      ans=n[1];
      ans+=9*(nn-2);
      ans+=(n[0]-1)*9*(nn-1);
      for (i=1;i<nn-1;i++) ans+=9*9*i;
      printf("%ld\n",ans);
   } else {
      ans=n[1]*n[2];
      ans+=n[1]*9*(nn-3);
      ans+=9*9*(nn-2)*(nn-3)/2;
      ans+=(n[0]-1)*9*9*(nn-1)*(nn-2)/2;
      ans+=9*9*9*(nn-1)*(nn-2)*(nn-3)/6;
      printf("%ld\n",ans);
   }
   return 0;
}