hdu4686Arc of Dream 矩阵快速幂

//a[n] = ax*a[n-1] + ay
//b[n] = bx*b[n-1] + by
//给出n求segma(a[i]*b[i])(0<=i<=n-1)
//f[n] = a[n]*b[n] = ax*bx*f[n-1] + ax*by*a[n-1] + ay*bx*b[n-1] + ay*by ;
//s[n] = segma(a[i]*b[i]) = s[n-1] + f[n]
//可以构造矩阵
// |f[n]  | |ax*bx ax*by ay*by ay*bx  0| |f[n-1]|
// |a[n]  | |  1     0     0     0    0| |a[n-1]|
// |b[n]  | |  0    ax     0     ay   0| |b[n-1]|
// |1     | |  0     0     0     1    0| | 1    |
// |s[n]  | |  1     0     0     0    1| |s[n-1]|
//然后矩阵快速幂就可求解
#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std ;
typedef long long ll ;
const ll mod = 1e9+7 ;
const int maxn = 10 ;
struct node
{
    ll p[maxn][maxn] ;
};
void debug(node a)
{
    for(int i = 1;i <= 5;i++)
      for(int j = 1;j <= 5;j++)
      printf("%lld%c" , a.p[i][j] , j ==  5 ?'\n':' ') ;
    puts("") ;
}
node mul(node a , node b)
{
    node c ;
    memset(c.p , 0 , sizeof(c.p)) ;
    for(int i = 1;i <= 5;i++)
      for(int j = 1;j <= 5;j++)
        for(int k = 1;k <= 5;k++)
        c.p[i][j] = (c.p[i][j] + a.p[i][k]*b.p[k][j])%mod ;
    return c ;
}
node pow(node a , ll k)
{
    node c ;
    memset(c.p , 0 , sizeof(c.p)) ;
    for(int i = 1;i <= 5;i++)
    c.p[i][i] = 1 ;
    while(k)
    {
        if(k&1)
          c = mul(c , a) ;
        a = mul(a , a) ;
        k >>= 1 ;
    }
    return c ;
}
int main()
{
    //freopen("in.txt" , "r" , stdin) ;
    ll ax , ay , a0 ;
    ll bx , by , b0 ;
    ll n ;
    while(~scanf("%lld" , &n))
    {
        scanf("%lld%lld%lld" , &a0 , &ax , &ay ) ;
        scanf("%lld%lld%lld" , &b0 , &bx , &by ) ;
        node a ;
        memset(a.p , 0 , sizeof(a.p)) ;
        a.p[1][1] = (ax*bx)%mod ;
        a.p[1][2] = (ax*by)%mod ;
        a.p[1][3] = (ay*bx)%mod ;
        a.p[1][4] = (ay*by)%mod ;
        a.p[2][2] = ax%mod ;
        a.p[2][4] = ay%mod ;
        a.p[3][3] = bx%mod ;
        a.p[3][4] = by%mod ;
        a.p[4][4] = 1 ;
        a.p[5][1] = 1 ;
        a.p[5][5] = 1 ;
        if(n == 0)
        {
            puts("0") ;
            continue ;
        }
        node c = pow(a , n-1) ;
        ll a1 =  (a0*ax%mod + ay)%mod ;
        ll b1 =  (b0*bx%mod + by)%mod ;
        ll s1 =  a0*b0%mod ;
        ll f1 =  a1*b1%mod ;
        node x ;
        memset(x.p , 0 , sizeof(x.p)) ;
        x.p[1][1] = f1 ;
        x.p[2][1] = a1 ;
        x.p[3][1] = b1 ;
        x.p[4][1] = 1 ;
        x.p[5][1] = s1 ;
        node ans = mul(c , x) ;
        printf("%lld\n" , ans.p[5][1]%mod) ;
    }
}