模数循环节——cf547A

campjls讲过模数循环节的问题，今天做cf才做到这类题

h1->a1的长度为len1，a1->a1的长度为cir1

h2->a2的长度为len2，a2->a2的长度为cir2

要注意特判，再用exgcd求

len1+cir1*t1 = len2+cir2*t2的一组整数解，把t1回代就是答案

//#pragma comment(linker,"/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include<vector>
#include<cmath>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<algorithm>
#include <stack>
#include <list>
using namespace std;
const int SZ=,INF=0x7FFFFFFF;
typedef long long lon;
 
lon exgcd(lon a,lon b,lon &x,lon &y,lon &d)
  {
       if(b==)
      {
          x=;
         y=;
         d=a;
         return a;
     }
     lon gcd=exgcd(b,a%b,x,y,d);
    lon x2=x,y2=y;
   x=y2;
   y=x2-(a/b)*y2;
    return gcd;
}
 
int main()
{
    std::ios::sync_with_stdio();
    //freopen("d:\\1.txt","r",stdin);
    lon m;
    cin>>m;
    lon h1,a1,x1,y1,h2,a2,x2,y2;
    cin>>h1>>a1>>x1>>y1>>h2>>a2>>x2>>y2;
    set<pair<lon,lon>> st;
    st.insert(make_pair(h1,h2));
    lon cnt1=,cnt2=,pri1=,pri2=,t1=,t2=;
    for(lon time=;;++time)
    {
        lon newa1=(x1*h1+y1)%m,newa2=(x2*h2+y2)%m;
        if(newa1==a1)
        {
            if(cnt1==)pri1=time;
            else if(cnt1==) t1=time-pri1;
            ++cnt1;
        }
        if(newa2==a2)
        {
            if(cnt2==)pri2=time;
            else if(cnt2==)t2=time-pri2;
            ++cnt2;
        }
        if(pri1&&pri2&&t1&&t2)break;
 
        if(newa1==a1&&newa2==a2)
        {
            cout<<time<<endl;
            return ;
        }
        if(time>1e7)
        {
            cout<<-<<endl;
            return ;
        }
        h1=newa1,h2=newa2;
    }
    lon x,y,d;
    exgcd(abs(t1),abs(t2),x,y,d);
    if(abs(pri1-pri2)%d==)
    {
        x*=(pri2-pri1)/d;
        y*=(pri2-pri1)/d;
        y*=-;
        if(x<)
        {
            int beishu=abs((x)/(t2/d));
            x=x+beishu*abs(t2/d);
            y=y+beishu*abs(t1/d);
        }
        if(y<)
        {
            int num=;
            num=abs((y)/(t1/d));
            y=y+num*abs(t1/d);
            if(y<)
            {
                ++num;
                y+=abs(t1/d);
            }
            x+=num*abs(t2/d);
        }
        if(x<)x+=abs(t2/d);
        if(x>&&y>)
        {
            int num1=abs((x)/(t2/d));
            int num2=abs((y)/(t1/d));
            int miner=min(num1,num2);
            x-=miner*abs(t2/d);
            y-=miner*abs(t1/d);
        }
        cout<<pri1+x*t1<<endl;
    }
    else cout<<-<<endl;
    return ;
}