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>
#include<stdlib.h>
#define delMinus(x) ((x)<0 ? 0 : (x))
int getDp(int ***dp,int i,int j,int k){
	if(i==-1&&j==0&&k==1)return 1;
	if(i<0||j<0)return 0;
	return dp[i][j][k];
}
int getDigit(char *s){
	int i=0;
	while(s[i]!='\0'){
		i++;
	}
	return i;
}
int main(void){
	char s[126];
	scanf("%s",s);
	int K,N=getDigit(s);
	scanf("%d",&K);
	int ***dp=malloc(sizeof(int**)*N);
	for(int i=0;i<N;i++){
		dp[i]=malloc(sizeof(int*)*K+1);
		int cs=s[i]-'0';
		for(int j=0;j<K+1;j++){
			dp[i][j]=malloc(sizeof(int)*2);
			dp[i][j][0]=getDp(dp,i-1,j-1,0)*9+getDp(dp,i-1,j,0)+delMinus((cs-1)*getDp(dp,i-1,j-1,1))+getDp(dp,i-1,j,1);
			dp[i][j][1]=(cs==0 ? getDp(dp,i-1,j,1) : getDp(dp,i-1,j-1,1));
			//printf("%2d(%2d ",dp[i][j][0],dp[i][j][1]);
		}
		//putchar('\n');
	}
	printf("%d\n",dp[N-1][K][0]+dp[N-1][K][1]);
	return 0;
}