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 (0) 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
大致题意：
给个n,找n的十进制倍数，仅有 0 和 1 组成，找一个就行，随便哪一个。
解题思路：
观察以下0，1串：
1
10 11
100 101 110 111
从n=1开始， 重复 n*10 与 n*10+1 ，由此遍历所有01串。
BFS，DFS皆可，以下采用DFS。
1 #include <iostream>
2 #include <queue>
3 #include <cstdio>
4 using namespace std;
5 queue<unsigned long long> s;
6 int n;//输入
7 unsigned long long temp,p;
8 void bfs()
9 {
10 while(!s.empty())
11 s.pop();
12 temp=1;
13 s.push(temp);
14 while(!s.empty())
15 {
16 p=s.front();
17 s.pop();
18 if(p%n==0)
19 {
20 printf("%lld\n",p);//lld！
21 break;
22 }
23 s.push(p*10);
24 s.push(p*10+1);
25 }
26 }
27 int main(){
28 while(scanf("%d",&n),n)
29 {
30 bfs();
31 }
32 return 0;
33 }