Problem:
Score : 700 points

Problem Statement

Find the smallest possible sum of the digits in the decimal notation of a positive multiple of K.

Constraints

2 \leq K \leq 10^5

K is an integer.

Input

Input is given from Standard Input in the following format:

K

Output

Print the smallest possible sum of the digits in the decimal notation of a positive multiple of K.

Sample Input 1

6

Sample Output 1

3

12=6×2 yields the smallest sum.

Sample Input 2

41

Sample Output 2

5

11111=41×271 yields the smallest sum.

Sample Input 3

79992

Sample Output 3

36

Solution in C (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <float.h>
#include <inttypes.h>
#include <limits.h>
#include <signal.h>
#include <stdbool.h>
#include <stdint.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <time.h>
#include <unistd.h>
#include <math.h>
#ifdef __cplusplus
#include <bits/stdc++.h>
#endif
//#pragma GCC target("avx2")
//#pragma GCC optimize("O3,unroll-loops")

#define getchar getchar_unlocked
#define putchar putchar_unlocked
int32_t nextint(void){ char c=getchar(); while(c!='-'&&(c<'0'||'9'<c)) c=getchar(); bool s=false; if(c=='-'){s=true;c=getchar();} uint32_t x=0; while('0'<=c && c<='9'){ x=x*10+c-'0'; c=getchar(); } return s?-x:x; }
int64_t nextlong(void){ char c=getchar(); while(c!='-'&&(c<'0'||'9'<c)) c=getchar(); int s=0; if(c=='-'){s=1;c=getchar();} uint64_t x=0; while('0'<=c && c<='9'){ x=x*10+c-'0'; c=getchar(); } return s?-x:x; }
uint32_t nextstr(char *s){ char c=getchar(); while(c==' '||c=='\n') c=getchar(); uint32_t len=0; while(c!=' '&&c!='\n'){ *s++=c; len++; c=getchar(); } *s='\0'; return len; }

int loop=19;
void stop(int sig){
	loop--;
}

int f(long x){
	int ans=0;
	while(x>0){
		ans+=x%10;
		x=x/10;
	}
	return ans;
}

int main(void){
	signal(SIGALRM, stop);
	ualarm(100000, 100000);
	int n=nextint();
	int ans=INT_MAX;
	for(long i=n; loop; i+=n){
		printf("", loop);
		int now=f(i);
		if(now<ans) ans=now;
	}
	printf("%d\n", ans);
}