Find The Multiple

                                                                                  ​​POJ - 1426 ​​

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

 

题目的意思是输入一个 n , 然后找一个由0 和 1组成的十进制数是 n 的 整倍数  , 

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <queue>
#define Swap(a,b)  a ^= b ^= a ^= b
using namespace std ;
const int MAX = 1e7+10;
const int inf = 0xffffff;
typedef long long LL;
void bfs(int n )
{
  //  bfs 超时了
  queue<LL> q ;
  q.push(1) ;
  while(!q.empty())
  {
    int i ;
    LL x;
    x  = q.front() ;
    q.pop() ;
    if(x %n ==0)
    {
      printf("%lld\n",x);
      return ;
    }
    q.push(x*10) ;
    q.push(x*10+1) ;
  }
}
bool flag ;
int n ;
void DFS(int cur , LL num )
{
  if(cur ==19 || flag)
  {
    return ;
  }
  if(num %n == 0)
  {
    cout<<num <<endl;
    flag = 1 ; // 找到做标记
    return  ;
  }
  DFS(cur+1, num*10) ;
  DFS(cur+1, num*10+1) ;
  return ;
}
int main()
{


  while(scanf("%d",&n)&&n)
  {
    flag = 0 ;
    DFS(0,1);
  }

  return 0 ;
}