Find The Multiple POJ

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

一开始没读懂题(=_=)，就是找到n的一个倍数m，只由0，1组成的十进制数，这道题居然能用BFS做，，而且还很简单，，直接上代码吧，BFS大法好

#include <iostream>  #include <algorithm>  #include <cstdio>  #include <queue>    using namespace std;  long long n;long long BFS()  {      queue<long long> q;      q.push(1);      while(!q.empty())      {          long long t = q.front();          q.pop();          if(t % n == 0)              return t;          q.push(10*t);          q.push(10*t + 1);      }      return -1; }  int main()  {      while(~scanf("%lld", &n) && n)    {        printf("%lld\n", BFS());    }        return 0;  }