D - Find The Multiple(2)

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

#include<iostream>
#include<cstdio>
using namespace std;
typedef long long ll;
int flag;//标志变量
int n;//数字位数
void dfs(int dig,ll m)
{
    if(dig>19||flag)//在n<=200的范围内，n的倍数位数最大是19
        return;
    if(m%n==0)//可以整除
    {
        flag=1;//找到解
        cout<<m<<endl;
        return;
    }
    else//没找到
    {
        dig++;//对数据位数进行增加，再递归
        dfs(dig,m*10);//分别遍历两种不一样的尾数的情况
        dfs(dig,m*10+1);
    }
}
int main()
{
    int dig;
    while(~scanf("%d",&n)&&n!=0)
    {
        dig=1;//位数最少为1
        flag=0;//重置变量
        dfs(dig,1);
    }
    return 0;
}