HDU 4686 Arc of Dream 矩阵快速幂

若能根据关系 推出转移矩阵 就是简单题

然而这需要多做题的经验

 

 

ll Ax, Ay, A0, B0 ,Bx,By;

struct Mat {
    ll m[5][5];
};

Mat E;

Mat P;

void init() {
    for (int i = 0; i < 5; i++) E.m[i][i] = 1;
    P.m[0][0] = 1ll;
    Ax %= MOD;
    Bx %= MOD;
    A0 %= MOD;
    B0 %= MOD;
    P.m[0][1] = (Ax * Bx)%MOD;
    P.m[0][2] = (By * Ax)%MOD;
    P.m[0][3] = (Ay * Bx)%MOD;
    P.m[0][4] = (Ay * By)%MOD;
    P.m[1][1] = (Ax * Bx)%MOD;
    P.m[1][2] = (By * Ax)%MOD;
    P.m[1][3] = (Ay * Bx)%MOD;
    P.m[1][4] = (Ay * By)%MOD;
    P.m[2][2] = Ax;
    P.m[2][4] = Ay;
    P.m[3][3] = Bx;
    P.m[3][4] = By;
    P.m[4][4] = 1ll;
}

Mat mul(const Mat& a, const Mat& b) {
    Mat c;
    for (int i = 0; i < 5; i++) {
        for (int j = 0; j < 5; j++) {
            c.m[i][j] = 0;
            for (int k = 0; k < 5; k++) c.m[i][j] = (c.m[i][j] + (a.m[i][k] * b.m[k][j])%MOD)%MOD;
        }
    }
    return c;
}

Mat quickPower(Mat a, ll b) {
    Mat ans = E;
    Mat base = a;
    while (b) {
        if (b & 1) ans = mul(ans, base);
        base = mul(base, base);
        b >>= 1;
    }
    return ans;
}

int main() {
    ll n;
    while (~scanf("%lld%lld%lld%lld%lld%lld%lld", &n, &A0, &Ax, &Ay, &B0, &Bx, &By)) {
        init();
        ll res = (A0 * B0) % MOD;
        if (n == 0) {
            puts("0");
            continue;
        }
        if (n == 1) {
            printf("%lld\n", res);
            continue;
        }
        Mat ans = quickPower(P, n - 1);
        res = (((ans.m[0][0] * res) % MOD + (ans.m[0][1] * res) % MOD)%MOD + (((A0 * ans.m[0][2]) % MOD + B0 * ans.m[0][3] % MOD)%MOD + 1ll * ans.m[0][4] % MOD)%MOD)%MOD;
        printf("%lld\n", res);
    }
}