POJ 1426 (BFS)

思路：用队列存储“打的表”，即从1开始，1的10倍满足条件，1的10倍+1也满足条件，判断队头是否%n==0就行。

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<string>
#include<vector>
#include<stack>
#include<bitset>
#include<cstdlib>
#include<cmath>
#include<set>
#include<list>
#include<deque>
#include<map>
#include<queue>
using namespace std;
#define inf 0x3f3f3f3f
#define ll long long
const int maxn=200005;
const double eps=1e-8;
const double PI = acos(-1.0);
void bfs(ll n)
{
    queue<ll> q;
    q.push(1LL);
    while(!q.empty())
    {
        ll t=q.front();
        if(t%n==0)
        {
            cout<<t<<endl;
            break;
        }
        q.pop();
        q.push(t*10);
        q.push(t*10+1);
    }
}
int main()
{
    std::ios::sync_with_stdio(false);
    std::cin.tie(0);
    std::cout.tie(0);
    ll n;
    while(cin>>n&&n)
    {
        bfs(n);
    }
    return 0;
}