hdu 6395 Sequence (简单矩乘)

P/n大多数情况是不变的, 取值只有$O(\sqrt{P})$种, 可以用$p/(p/i)$跳过重复的值, 复杂度$O(logn\sqrt{P})$

要注意

• P跟模数P有冲突
• 要特判p/i==0和p/(p/i)>n的情况
• 题目给的$CF_{n-2}+DF_{n-1}$, 写矩阵要D在前C在后

//fn   =  D   C   x   fn-1
//fn-1    1   0   0   fn-2
//1       0   0   1   1

struct Mat {
    ll v[4][4];
    Mat() {memset(v, 0, sizeof v);}
    Mat operator * (const Mat& b) const {
        Mat c;
        REP(k,1,3)REP(i,1,3)REP(j,1,3) {
            c.v[i][j] = (v[i][k]*b.v[k][j]+c.v[i][j])%P;
        }
        return c;
    }
    Mat operator ^ (ll nn) {
        Mat b, a=*this;
        REP(i,1,3) b.v[i][i]=1;
        while(nn) {
            if(nn&1LL) b=b*a;
            nn>>=1LL,a=a*a;
        }
        return b;
    }
};

void work() {
	int A, B, C, D, p, n;
	scanf("%d%d%d%d%d%d", &A, &B, &C, &D, &p, &n);
	if (n==1) return printf("%d\n",A),void();
	if (n==2) return printf("%d\n",B),void();
	Mat r;
	r.v[1][1]=r.v[2][2]=r.v[3][3]=1;
	REP(i,3,n) {
		Mat g;
		g.v[1][1]=D,g.v[1][2]=C,g.v[2][1]=g.v[3][3]=1,g.v[1][3]=p/i;
		if (p<i) {
			r = (g^(n-i+1))*r;
			break;
		}
		int k = p/(p/i);
		if (k<=n) r = (g^(k-i+1))*r;
		else r = (g^(n-i+1))*r;
		i = k;
	}
	ll ans = r.v[1][1]*B%P+r.v[1][2]*A%P+r.v[1][3];
	printf("%lld\n", ans%P);
}

int main() {
	int t;
	scanf("%d", &t);
	while (t--) work();
}