Arc of Dream HDU - 4686

Arc of Dream HDU - 4686
思路：

#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
typedef long long ll;
const ll mod=1000000007;
struct  matrix{
    ll x[5][5];
};
matrix multi(matrix a,matrix b){
    matrix temp;
    memset(temp.x,0,sizeof(temp.x));
    for(int i=0;i<5;i++)
        for(int j=0;j<5;j++)
            for(int k=0;k<5;k++)
            {
                temp.x[i][j]+=a.x[i][k]*b.x[k][j];
                temp.x[i][j]%=mod;
            }
    return temp;
}
matrix quick_multi(matrix a,ll n)//矩阵快速幂
{
    matrix temp=a;
    n--;
    while(n){
        if(n&1)
            temp=multi(temp,a);
        a=multi(a,a);
        n>>=1;
    }
    return temp;
}
/*void p(matrix a)
{
    for(int i=0;i<5;i++)
    {
        for(int j=0;j<5;j++)
            printf("%lld ",a.x[i][j]);
        printf("\n");
    }
}*/
int main()
{
    ll n;
    ll a0,ax,ay,b0,bx,by;
    while(scanf("%lld",&n)!=EOF)
    {
        matrix A;
        scanf("%lld%lld%lld",&a0,&ax,&ay);
        scanf("%lld%lld%lld",&b0,&bx,&by);
        memset(A.x,0,sizeof(A.x));
        if(n<=1) n==1?printf("%lld\n",(a0*b0)%mod):printf("0\n");
        else
        {
            A.x[0][0]=1;A.x[0][1]=(ax*bx)%mod;A.x[0][2]=(ax*by)%mod;A.x[0][3]=(bx*ay)%mod;A.x[0][4]=(ay*by)%mod;
            A.x[1][1]=(ax*bx)%mod;A.x[1][2]=(ax*by)%mod;A.x[1][3]=(bx*ay)%mod;A.x[1][4]=(ay*by)%mod;
            A.x[2][2]=ax%mod;A.x[2][4]=ay%mod;
            A.x[3][3]=bx%mod;A.x[3][4]=by%mod;
            A.x[4][4]=1;
            matrix B;
            memset(B.x,0,sizeof(B.x));
            B.x[0][0]=(a0*b0)%mod;B.x[1][0]=(a0*b0)%mod;B.x[2][0]=a0%mod;B.x[3][0]=b0%mod;B.x[4][0]=1;
            A=quick_multi(A,n-1);
            B=multi(A,B);
            printf("%lld\n",B.x[0][0]%mod);
        }
    }
    return 0;
}