Find The Multiple
Time Limit: 1000MS   Memory Limit: 10000K
Total Submissions: 38700   Accepted: 16149   Special Judge

Description

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

题意 ： 输出任意一个由 0 1组成的十进制数字且该数可以整除 n  ，通过dfs来写. 两组条件是，dfs(ans*10） ， dfs(ans * 10 + 1).

#include <cstdio>
#include <iostream> 
#include <cstring>
#include <cstdlib>
#define ull unsigned long long int
using namespace std;
int flag = 0;
void dfs(ull step , ull n)
{

	
		if(flag == 1)
		{
		//	flag = 0 ;
			return ;
		}
	if(step % n == 0)
	{
		flag = 1;
		cout << step << endl;
		return ;
	}
	if(step > 1000000000000000000)
	{
		return;
	}
	
		dfs(step*10 , n);
		dfs(step*10+1 , n);

}
int main()
{
	ull n;
	while(cin >> n)
	{
		flag = 0;
		if(n == 0)break;
		dfs(1 , n);
	}
	
	return 0;
}

要注意的一个点： 
if(step > 1000000000000000000)
	{
		return;
	}

该式要放在

if(step % n == 0)
	{
		flag = 1;
		cout << step << endl;
		return ;
	}
之下。