模数循环节——cf547A

Posted 2022-09-16 zsben991126

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=1000010,INF=0x7FFFFFFF;
typedef long long lon;
 
lon exgcd(lon a,lon b,lon &x,lon &y,lon &d)
  
       if(b==0)
      
          x=1;
         y=0;
         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(0);
    //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=0,cnt2=0,pri1=0,pri2=0,t1=0,t2=0;
    for(lon time=1;;++time)
    
        lon newa1=(x1*h1+y1)%m,newa2=(x2*h2+y2)%m;
        if(newa1==a1)
        
            if(cnt1==0)pri1=time;
            else if(cnt1==1) t1=time-pri1;
            ++cnt1;
        
        if(newa2==a2)
        
            if(cnt2==0)pri2=time;
            else if(cnt2==1)t2=time-pri2;
            ++cnt2;
        
        if(pri1&&pri2&&t1&&t2)break;
        
        if(newa1==a1&&newa2==a2)
        
            cout<<time<<endl;
            return 0;
        
        if(time>1e7)
        
            cout<<-1<<endl;
            return 0;
        
        h1=newa1,h2=newa2;
    
    lon x,y,d;
    exgcd(abs(t1),abs(t2),x,y,d);
    if(abs(pri1-pri2)%d==0)
    
        x*=(pri2-pri1)/d;
        y*=(pri2-pri1)/d;
        y*=-1;
        if(x<0)
        
            int beishu=abs((x)/(t2/d));
            x=x+beishu*abs(t2/d);
            y=y+beishu*abs(t1/d);
         
        if(y<0)
        
            int num=0;
            num=abs((y)/(t1/d));
            y=y+num*abs(t1/d);
            if(y<0)
            
                ++num;
                y+=abs(t1/d);
            
            x+=num*abs(t2/d);
        
        if(x<0)x+=abs(t2/d);
        if(x>0&&y>0)
        
            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<<-1<<endl;
    return 0;