Poj 1426--Find The Multiple（bfs或dfs）

Find The Multiple

Time Limit: 1000MS		Memory Limit: 10000K

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：输入一个数n，然后输出m，其中m是n的倍数，且m中的每位必须是1或者0；

开始的时候读错题意了，这个地方可以思考一下，开始的时候我从1开始入栈，然后把1的10倍入栈，然后把1的10倍+1的数入栈，以后以次类推。

这样就可以把所有的m满足各位为0或1的数，存到栈里，每次入栈时，只要再满足此时的m是n的倍数，输出m，程序就可以结束了。

因为这个题说了，答案不唯一，只要输出满足要求的就可以了，输出满足条件的最小的那个m；

#include<iostream>
#include<queue>
using namespace std;
int n;
long long BFS()
{
    long long t;
    queue<long long>q;
    while(!q.empty())
        q.pop();
    q.push(1);
    while(1)
    {
        t = q.front();
        if(t%n==0)
            return t;
        q.pop();
        q.push(t*10);
        q.push(t*10+1);
    }
}
int main()
{
    while(cin>>n, n)
    {
        long long res = BFS();
         cout<<res<<endl;
    }
    return 0;
}