题目大意：给出an，bn的递推，求ai*bi（i=0，1，……n-1）的和（an=a(n-1)*Ax+Ay， bn=b(n-1)*Bx+By， a0=A0， b0=B0， Ax,Bx,Ay,By已知）。
题解：矩阵快速幂不用说了，式子也好推，开了ll基本没有坑点，奇葩的是这题是做连通分量做到的，有丶意思。
#include <iostream>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <algorithm>
using namespace std;
typedef long long ll;
const int N = 6;
const int MOD = 1e9 + 7;
ll A0, B0, Ax, Bx, Ay, By;
ll p, q, r, s;
struct M
{
ll mat[N];
M(){}
M( int _ )
{
mat[1] = A0*B0%MOD;
mat[2] = A0; mat[3] = B0;
mat[4] = 1;
mat[5] = A0*B0%MOD;
}
};
struct _M
{
ll mat[N][N];
_M(){}
_M( int _ )
{
mat[1][1] = p; mat[1][2] = q; mat[1][3] = r; mat[1][4] = s; mat[1][5] = 0;
mat[2][1] = mat[2][3] = mat[2][5] = 0; mat[2][2] = Ax%MOD; mat[2][4] = Ay%MOD;
mat[3][1] = mat[3][2] = mat[3][5] = 0; mat[3][3] = Bx%MOD; mat[3][4] = By%MOD;
mat[4][1] = mat[4][2] = mat[4][3] = mat[4][5] = 0; mat[4][4] = 1;
mat[5][1] = p; mat[5][2] = q; mat[5][3] = r; mat[5][4] = s; mat[5][5] = 1;
}
};
void init()
{
p = Ax*Bx%MOD;
q = Ax*By%MOD;
r = Bx*Ay%MOD;
s = Ay*By%MOD;
}
M M12( M a, _M b )
{
M c;
memset(c.mat,0,sizeof(c.mat));
for ( int i = 1; i < N; i ++ )
{
ll cur = 0;
for ( int j = 1; j < N; j ++ )
cur += (a.mat[j]%MOD*b.mat[i][j]%MOD)%MOD;
c.mat[i] = cur % MOD;
}
return c;
}
_M M22( _M a, _M b )
{
_M c;
memset(c.mat,0,sizeof(c.mat));
for ( int i = 1; i < N; i ++ )
{
for ( int j = 1; j < N; j ++ )
{
ll cur = 0;
for ( int k = 1; k < N; k ++ )
cur += (a.mat[i][k]%MOD*b.mat[k][j]%MOD)%MOD;
c.mat[i][j] = cur % MOD;
}
}
return c;
}
M qp( M a, _M b, ll c )
{
while ( c )
{
if ( c & 1 ) a = M12( a, b );
c >>= 1;
b = M22(b,b);
}
return a;
}
int main()
{
ll m;
while ( ~scanf("%lld", &m) )
{
scanf("%lld%lld%lld", &A0, &Ax, &Ay);
scanf("%lld%lld%lld", &B0, &Bx, &By);
if ( !m ) puts("0");
else if ( m == 1 )
{
ll now = A0%MOD*B0%MOD%MOD;
printf("%lld\n", now);
}
else
{
init();
M ans(1);
_M base(1);
/* for ( int i = 1; i < N; i ++)
cout << ans.mat[i] << endl;
for ( int i = 1; i < N; i ++ )
{
for ( int j = 1; j < N; j ++ )
cout << base.mat[i][j] << " ";
cout << endl;
}
*/
ans = qp( ans, base, m-1);
printf("%lld\n", (ans.mat[5]%MOD+MOD)%MOD);
}
}
return 0;
}