[矩阵乘法][费马小定理]Product Oriented Recurrence (Codeforces Round #566 Div. 2 E)

题目描述

    f(x)=c^(2x-6)*f(x-1)*f(x-2)*f(x-3),给定n, f(1), f(2), f(3), c,求f(n) mod(1e9+7) 

      4<=n<=1e18 ,  1<=f(1),f(2),f(3),c<=1e9 

 

题解

    g(x)=c^x*f(x)   g(x)=g(x-1)*g(x-2)*g(x-3)

    令g(x)=g(1)^t1*g(2)^t2*g(3)^t3

   分别求t1 , t2 , t3 是 三元斐波那契数列

   [x2  x3  x4]=[x1  x2  x3]*[0  0  1]

                                          [1  0  1]

                                          [0  1  1]

  因为是指数 所以由费马小定理对(1e+7 -1)取模

 

  g(n)=c^n*f(n)  费马小定理求逆元  完了。。

 

#include<cstdio>
#include<cstring>
using namespace std;
const long long mod=1000000000+7;
long long n,f1,f2,f3,c;
struct hehe{long long m[4][4];}t1,t2,t3,a,tur,he;
hehe operator * (hehe &a,hehe &b)
  {hehe re;  memset(re.m,0,sizeof(re.m));
   
   for(int i=1;i<=3;i++)
    for(int j=1;j<=3;j++)
     for(int k=1;k<=3;k++)
       {re.m[i][j]=re.m[i][j]+a.m[i][k]%(mod-1)*b.m[k][j]%(mod-1);
       }
   return re;      
  }
inline long long ksm(long long a,long long p)
 {long long re=1;
  while(p)
   {if(p&1) re=re*a%mod;
    p>>=1;
    a=a*a%mod%mod;
   }
  return re;     
 }  
int main()
 {       
 scanf("%lld %lld %lld %lld %lld",&n,&f1,&f2,&f3,&c);    
 
 f1=f1*c%mod; f2=f2*c%mod*c%mod; f3=f3*c%mod*c%mod*c%mod;   
 he.m[1][1]=he.m[2][2]=he.m[3][3]=1;   
 tur.m[1][3]=tur.m[2][1]=tur.m[2][3]=tur.m[3][2]=tur.m[3][3]=1;   
  
 t1.m[1][1]=1; t2.m[1][2]=1; t3.m[1][3]=1;
    
 //tur^n-1    *a
    long long k=n-1;
 while(k)   
   {if(k&1) {he=he*tur;} 
       k>>=1;
       tur=tur*tur;
   } 
 t1=t1*he; t2=t2*he; t3=t3*he; 
  
   
 long long ans=ksm(f1,t1.m[1][1])*ksm(f2,t2.m[1][1])%mod*ksm(f3,t3.m[1][1])%mod,inv=ksm(c,mod-2);   // ans=c^n*f[n]
 printf("%lld",ans*ksm(inv,n)%mod);
return 0;
 }