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: Accepted)

#include<stdio.h>
int main(){
  char s[110];
  int i,j,m,d[110][5][2]={};
  scanf("%s %d",s,&m);
  d[0][0][1]=1;
  for(i=0;s[i];i++){
    for(j=0;j<4;j++){//printf("%d %d:",d[i][j][0],d[i][j][1]);
      d[i+1][j  ][0]+=d[i][j][0];//mini+0
      d[i+1][j+1][0]+=d[i][j][0]*9;//mini+other
      if(s[i]=='0')d[i+1][j][1]+=d[i][j][1];
      else{
	d[i+1][j  ][0]+=d[i][j][1];
	d[i+1][j+1][1]+=d[i][j][1];
	d[i+1][j+1][0]+=d[i][j][1]*(s[i]-'0'-1);
      }
      //d[i+1][j+1][0]+=d[i][j][1]*(s[i]-'0');
    }//printf("\n");
  }
  //for(j=0;j<4;j++)printf("%d %d:",d[i][j][0],d[i][j][1]);printf("\n");
  printf("%d\n",d[i][m][0]+d[i][m][1]);
  return 0;
}