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 getDigit(char *input){
	int i=0;
	while(input[i]!='\0'){
		i++;
	}
	return i;
}
int main(void){
	char s[126];
	scanf("%s",s);
	int K;
	int n=getDigit(s);
	scanf("%d",&K);
	int dp[126][4][2]={};
	dp[0][0][0]=1;
	for(int i=0;i<n;i++){
		for(int j=0;j<4;j++){
			for(int k=0;k<2;k++){
				int nd=s[i]-'0';//n��i�Ԗڂ̐���
				for(int d=0;d<10;d++){
					int ni=i+1;//�s��
					int nj=j;//
					int 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];
				}
			}
		}
		putchar('\n');
	}
	for(int i=0;i<=n;i++){
		for(int j=0;j<=K;j++){
			printf("%d(%d) ",dp[i][j][0],dp[i][j][1]);
		}
		putchar('\n');
	}
	int answer=dp[n][K][0]+dp[n][K][1];
	printf("%d\n",answer);
	return 0;
}