传送门

题意：

初始有一个数$1$，你每次可以将其$\ast 10$或者$+(x-1)$，现在给你$x$，问最少经过多少步能到达$n$。
$1\le n\le 1e6,1\le m\le 1e9$

思路：

同余最短路，定义$dis[i]$代表通过两个操作能到的数$\mathrm{m}\mathrm{o}\mathrm{d}n$后为$i$的最少操作次数，那么答案就是$dis[0]$了，直接扩展就好啦。
最后注意一下$n=1$的情况，别被坑了。

// Problem: Yuki with emofunc and playf
// Contest: NowCoder
// URL: https://ac.nowcoder.com/acm/contest/18196/B
// Memory Limit: 524288 MB
// Time Limit: 2000 ms
// 
// Powered by CP Editor (https://cpeditor.org)//#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4.1,sse4.2,avx,avx2,popcnt,tune=native")
//#pragma GCC optimize(2)
#include<cstdio>
#include<iostream>
#include<string>
#include<cstring>
#include<map>
#include<cmath>
#include<cctype>
#include<vector>
#include<set>
#include<queue>
#include<algorithm>
#include<sstream>
#include<ctime>
#include<cstdlib>
#include<random>
#include<cassert>
#define X first
#define Y second
#define L (u<<1)
#define R (u<<1|1)
#define pb push_back
#define mk make_pair
#define Mid ((tr[u].l+tr[u].r)>>1)
#define Len(u) (tr[u].r-tr[u].l+1)
#define random(a,b) ((a)+rand()%((b)-(a)+1))
#define db puts("---")
using namespace std;//void rd_cre() { freopen("d://dp//data.txt","w",stdout); srand(time(NULL)); }
//void rd_ac() { freopen("d://dp//data.txt","r",stdin); freopen("d://dp//AC.txt","w",stdout); }
//void rd_wa() { freopen("d://dp//data.txt","r",stdin); freopen("d://dp//WA.txt","w",stdout); }typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> PII;const int N=1000010,mod=1e9+7,INF=0x3f3f3f3f;
const double eps=1e-6;int n,x;
int dis[N];
bool vis[N];int main()
{
//	ios::sync_with_stdio(false);
//	cin.tie(0);cin>>n>>x;if(n==1) {puts("0");return 0;}x--; queue<int>q; q.push(1);dis[1]=0; vis[1]=true;x%=n;while(q.size()) {int u=q.front(); q.pop();int now=1ll*u*10%n;if(!vis[now]) {vis[now]=1; q.push(now);dis[now]=dis[u]+1;}now=(1ll*u+x)%n;if(!vis[now]) {vis[now]=1; q.push(now);dis[now]=dis[u]+1;}}if(!vis[0]) puts("-1");else printf("%d\n",dis[0]);return 0;
}
/**/