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：给出一个数n，然后求得一个数字k，数字满足：能被n整除，每一位只有0,1。这样的数字k会有很多个，然以输出一个就可以。
思路：n的最大值为200，用dfs从数字k的个位开始往高位搜索，每一位只有0或1。找到能被n整除的时候输出就可以了。
ps:(一开始不知道到底应该搜到多少位时可以return，后来看到网上说最多到19位就可以)
代码如下：
#include <iostream>
using namespace std;
int n,flag;
void dfs(int k,long long cur){ //k:位数 cur:当前数字
if(k == 19 || flag) return ;
if(cur%n == 0){
cout<<cur<<endl;
flag = 1;
return ;
}
dfs(k+1,cur*10); //高位为 0
dfs(k+1,cur*10+1); //高位为 1
}
int main()
{
while(cin>>n,n != 0){
flag = 0;
dfs(0,1);
}
return 0;
}