dfs find the multiple

Given a positive integer n, write a program to find out a nonzero multiple m of n whose decimal representation contains only the digits 0 and 1. You may assume that n is not greater than 200 and there is a corresponding m containing no more than 100 decimal digits. 

Input
The input file may contain multiple test cases. Each line contains a value of n (1 <= n <= 200). A line containing a zero terminates the input.
Output
For each value of n in the input print a line containing the corresponding value of m. The decimal representation of m must not contain more than 100 digits. If there are multiple solutions for a given value of n, any one of them is acceptable.
Sample Input

2
6
19
0

Sample Output

10
100100100100100100
111111111111111111
题目大致意思就是找到只含有0和1的数同时能被n整除，dfs暴力就能过。

#include<iostream>
using namespace std;
bool flag;
unsigned long long ans;
void	dfs(unsigned long long m,int n,int k)
{
   
	
	if(flag)
	{
   
		return ;
	}
	if(k>19)
	{
   
		return ;
	}
	if(m%n==0)
	{
   
		flag=1;
		ans=m;
		return ;
	}
	dfs(m*10,n,k+1);
	dfs(m*10+1,n,k+1);
}
int main()
{
   
	int n;
	while(cin>>n,n)
	{
   
		flag=0;
		dfs(1,n,1);//~~我蒟蒻~~ 
		cout<<ans<<endl;
	}
 }