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>
#include <string.h>

int main(){

	int K,keta;
	char N[110];
	scanf("%s",N);
	scanf("%d",&K);
	keta = strlen(N);
	
	int dp[keta+1][K+1][2];
	
	for(int i=0;i<keta+1;i++){
		for(int j=0;j<K+1;j++){
			dp[i][j][0] = 0;
			dp[i][j][1] = 0;
		}
	}
	dp[0][0][0] = 1;
	
	for(int i=0;i<keta;i++){
		int nd = N[i]-'0';
		for(int j=0;j<K+1;j++){
			for(int k=0;k<2;k++){
				for(int d=0;d<10;d++){
					int ni=i+1,nj=j,nk=k;
					if( d!=0 ) nj++;
					if( nj>K ) continue;
					if( k==0 ){
						if( d>nd ) continue;
						if( d<nd ) nk=1;
					}
					dp[ni][nj][nk] += dp[i][j][k];
				}
			}
		}
	}
	
	printf("%d",dp[keta][K][1]+dp[keta][K][0]);
	return 0;
}