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
题意理解： 给定一个n的值，找到 一个n的倍数m，并且m的十进制只有0和1 （0 1 10 11 100 101 110 111 ………………等）任意一个m都可。
#include <iostream>
using namespace std;
int flag; //标记一下是否找打了需要的m
void dfs(long long int x, int n, int k) // x代表的找的m,n代表给定的n,k代表递归的次数
{
if(flag || k==19) return ; //因为x 定义的longlong，顶多到19位，递归x不难看出，一次增加1位，所以最多递归19次
if(x%n==0)
{
flag=1; //找到了满足条件的m，将标记变量变为1
cout<<x<<endl;
return;
}
dfs(x*10,n,k+1); //深度遍历 0 10 100 110 ……
dfs(x*10+1,n,k+1);//深度遍历 1 11 101 111 ……
}
int main()
{
int n;
while(cin >>n,n)
{
flag=0;
dfs(1,n,0);
}
return 0;
}
