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 <string.h>
#include <stdlib.h>
#include <math.h>

int calc_sum(int digit) {
	int sum = 0;
	for (int i = 1; i < digit+1; ++i) sum += i;
	return sum;
}

int main(void) {

	char N[101];
	int K;
	int count = 0;
	int flag = 1;
	scanf("%s",N);
	scanf("%d",&K);
	int digit = strlen(N)-1;
	int* dig_num;
	dig_num = (int*)calloc(sizeof(int),digit+1);
	for (int i = 0; i < strlen(N); ++i) {
		dig_num[i] = (int)(N[i] - '0');
	}

	if (K == 1) {
		count += dig_num[0];
		count += 9*digit;
	}
	else if (K == 2) {
		count += 9*9*calc_sum(digit-1);
		for (int i = 0; i < digit; ++i) {
			if (dig_num[i] == 0) continue;
			count += 9*(digit-i)*(dig_num[i]-1);
			if (i != 0) {
				flag = 0;
				count += dig_num[i];
				break;
			}
		}
		if (flag) {
			count += dig_num[digit];
		}
		else ;
	}
	else if (K == 3) {
		count += 9*9*9*calc_sum(digit-2);
		for (int i = 0; i < digit; ++i) {
			if (dig_num[i] == 0) continue;
			count += 9*9*(digit-i-1)*(dig_num[i]-1);
			if (i != 0) {
				flag = 0;
				count += dig_num[i];
				break;
			}
		}
		if (flag) {
			count += dig_num[digit];
		}
		else ;
	}

	printf("%d\n",count);
	free(dig_num);

	return 0;
}