hdu 4686 (矩阵快速幂)

Posted 2020-09-25 simpleknight

题意：

          

         a ₀ = A0 
         a _i = a _i-1*AX+AY 
         b ₀ = B0 
         b _i = b _i-1*BX+BY 

         求AoD(N) mod 1000000007

思路：

         做这道题时思路没有打开，一直纠结于如何把公式加一起.....

         正确是做法是把AoD这个和加到你的构造矩阵里来递推计算

         a_ib_i=(A_x*B_x)*a_i-1*b_i-1 + (A_x*B_y)*a_i-1 + (A_y*B_x)*b_i-1 + A_y*B_y

        

 

代码：

#include <cstdio>
#include <cstring>
#include <iostream>

using namespace std;

typedef long long ll;
const int N=5,M=5,P=5;
const ll MOD=1000000007;
struct  Matrix
{
    ll m[N][N];
};

Matrix  A;

Matrix  I={1,0,0,0,0,
           0,1,0,0,0,
           0,0,1,0,0,
           0,0,0,1,0,
           0,0,0,0,1};

Matrix multi(Matrix a,Matrix b)
{
    Matrix ans;
    for(int i=0;i<N;i++)
    {
       for(int j=0;j<M;j++)
       {
          ans.m[i][j]=0;
          for(int k=0;k<P;k++)
          {
             ans.m[i][j]+=a.m[i][k]*b.m[k][j]%MOD;
          }
          ans.m[i][j]%=MOD;
       }
    }
    return ans;
}

Matrix power(Matrix a,ll k)
{
    Matrix ans=I,p=a;
    while(k)
    {
      if(k&1)
      {
        ans=multi(ans,p);
      }
      k>>=1;
      p=multi(p,p);
    }
    return ans;
}

int main(int argc, char const *argv[])
{
    ll n;
    ll A0,Ax,Ay,B0,Bx,By;
    while(cin>>n)
    {
        scanf("%lld %lld %lld",&A0,&Ax,&Ay);
        scanf("%lld %lld %lld",&B0,&Bx,&By);
        if(n==0)
        {
            cout<<"0"<<endl;
            continue;
        }
        A0%=MOD;
        Ax%=MOD;
        Ay%=MOD;
        B0%=MOD;
        Bx%=MOD;
        By%=MOD;
        A.m[0][0]=1;
        A.m[0][1]=(Ax*Bx)%MOD;
        A.m[0][2]=(Ax*By)%MOD;
        A.m[0][3]=(Ay*Bx)%MOD;
        A.m[0][4]=(Ay*By)%MOD;
        A.m[1][0]=0;
        A.m[1][1]=(Ax*Bx)%MOD;
        A.m[1][2]=(Ax*By)%MOD;
        A.m[1][3]=(Ay*Bx)%MOD;
        A.m[1][4]=(Ay*By)%MOD;
        A.m[2][0]=0;
        A.m[2][1]=0;
        A.m[2][2]=(Ax)%MOD;
        A.m[2][3]=0;
        A.m[2][4]=(Ay)%MOD;
        A.m[3][0]=0;
        A.m[3][1]=0;
        A.m[3][2]=0;
        A.m[3][3]=(Bx)%MOD;
        A.m[3][4]=(By)%MOD;
        A.m[4][0]=0;
        A.m[4][1]=0;
        A.m[4][2]=0;
        A.m[4][3]=0;
        A.m[4][4]=1;
        Matrix ans=power(A,(n-1));
        ll num=0;
        ll a=(A0*B0)%MOD;
        ll b=(A0*B0)%MOD;
        ll c=A0%MOD;
        ll d=B0%MOD;
        ll e=1;
        num=(num+(ans.m[0][0]*a)%MOD)%MOD;
        num=(num+(ans.m[0][1]*b)%MOD)%MOD;
        num=(num+(ans.m[0][2]*c)%MOD)%MOD;
        num=(num+(ans.m[0][3]*d)%MOD)%MOD;
        num=(num+(ans.m[0][4]*e)%MOD)%MOD;
        printf("%lld\\n",num);
    }
    return 0;
}