poj1426 宽搜

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.

考试的时候一看200位直接就被吓到了，直接跳过，后来听说答案开long long 可过，就是普通宽搜，感觉亏了一个亿，既然每一位不是0就是1，首先第一位一定是1，然后每次这个数乘10，或者乘10加1，就是可以产生每一位都是1或者0的数字了。

代码：

#include<iostream>
#include<cstdio>
#include<queue>
#define ll long long
using namespace std;
int n;
queue<long long>q;
ll ans;
int main()
{
    while(1)
    {
        cin>>n;
        if(!n)break;
        while(!q.empty())q.pop();
        q.push(1);
        while(!q.empty())
        {
            ll k=q.front();q.pop();
            if(k%n==0)
            {
                ans=k;break;
            }
            q.push(k*10),q.push(k*10+1);
        }
        printf("%lld
",ans);
    }

    return 0;
}