第一道BFS

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

BFS对于树来说，是按照层来搜索的
推广到图也类似
这道题是典型的树
题意：
给你一个数n；
让你找到一个数是它的倍数，并且只由0和1组成。
一个数怎么才能只有0和一呢？
1->1*10=10->10+1=11>110>111>1110>>1111
除了这一类数没有其他的啦= =
这一点是关键
那么这个是怎么变成的呢？
*10 or *10+1;

代码
我们只需要按层改变就好啦

#include <iostream>#include<queue>using namespace std;queue<int>qint bfs(){   while(q.empty()) q.pop();   q.push(1);   while (true){      int p= q.front();      q.pop;      if(p%n==0) return p;      q.push(p*10);      q.push(p*10+1);      }      return 0;}int main(){    int n;    while (cin>>n&&n)    cout<<bfs()<<endl;}