POJ 1426 Find The Multiple （dfs？？！！）

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

给你一个n，让你找一个n的倍数，它的每个位不是0就是1。
这次题解用咆哮体写题解！这题是搜索你敢信？这题答案不超过unsigned __int64 你敢信？啊？ 啊！ 啊？
网络上还有说什么考到同余定理的人？你不知道答案不超unsigned __int64，你敢写？
代码如下：

 #include<iostream>
 #include<cstdlib>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<cmath>
 #include<queue>
 using namespace std;
 bool found;
 int n;
 void dfs(unsigned __int64 x,int step)
 {
     if(found)
     return ;
     if(x%n==)
     {
         printf("%I64u\n",x);
         found=true;
         return ;
     }
     if(k==)
         return ;
     dfs(x*,step+);
     dfs(x*+,step+);
 }
 int main()
 {
     while(cin>>n,n)
     {
         found=false;
         dfs(,);
     }
     return ;
 }

最后提醒下自己写prinf 是老老实实写%I64u...