由题

余数为零 则该串必定为1开头的01串

由BFS特性可得首解最优

因此队头1

向后添加0 或 1

进行取模搜索

当前取模值单一后推运算即可

否则展开规律相同 无意义

#pragma GCC optimize(2)
#include <cstdio>
#include <iostream>
#include <cstdlib>
#include <cmath>
#include <cctype>
#include <string>
#include <cstring>
#include <algorithm>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <ctime>
#include <vector>
#include <fstream>
#include <list>
#include <iomanip>
#include <numeric>
using namespace std;
typedef unsigned long long ll;

const int MAXN = 1e6 + 10;

ll exz[MAXN] = {0};

/*map <int, bool> exz;*/

ll n;

struct bfsnode
{
    string ss;
    
    ll num;
    
    bfsnode(string ts, ll x)
    {
        ss = ts;
        num = x;
    }
};

void searh()
{
    bfsnode a("1", 1 % n);

    queue <bfsnode> Q;

    Q.push(a);

    while(!Q.empty())
    {
        bfsnode b = Q.front();
        
        Q.pop();
        
        if(b.num == 0)
        {
            cout<<b.ss<<endl;
            return ;
        }

        else
        {
            bfsnode t(b.ss + "0", b.num * 10 % n);
            
            if(exz[b.num] == 0)
            {
                exz[b.num] = 1;
                Q.push(t);
            }
            

            bfsnode k(b.ss + "1", (b.num * 10 + 1) % n);
            
            if(exz[k.num] == 0)
            {
                exz[b.num] = 1;
                Q.push(k);
            }
        }
    }   
}

int main()
{
    //ios::sync_with_stdio(false);

    //cin.tie(0);     cout.tie(0);

    cin>>n;

    searh();

    return 0;
}