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 i,k,m,c,cf=0,cc=1,cg=0,ch=0;
  char n[105];
  scanf("%s",&n);
  scanf("%ld",&k);
  for(c=0;n[c]!=0;c++){
    n[c]-='0';
    if(c>0){
      if(cf==0){
        cf=n[c];
      }else{
        cf+=9;
    	if(c>1){
          ch+=n[c];
          if(cg==0){
        	cg=n[c];
          }else{
        	cg+=9;
        	cc++;
          }
    	}
      }
    }
  }
  if(k==1){
    m=(c-1)*9;
    m+=n[0];
  }else if(k==2){
    m=(c-1)*(c-2)/2*81;
    m+=cf+(n[0]-1)*(c-1)*9;
  }else if(k==3){
    m=(c-1)*(c-2)*(c-3)/2*243;
    m+=(n[0]-1)*(c-1)*(c-2)/2*81;
    m+=(cc-1)*(cc-2)/2*81;
    m+=cg*(ch>1)+(n[c-cc]-1)*(cc-1)*9;
  }
  printf("%ld",m);
  return 0;
}