模数循环节——cf547A

朱雀 2023-10-12 08:22 10阅读 0赞

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;
    }

转载于:https://www.cnblogs.com/zsben991126/p/11440655.html