loj 3090 「BJOI2019」勘破神機 - 數學

題目傳送門

　　傳送門

題目大意

　　設$F_{n}$表示用$1\times 2$的骨牌填$2\times n$的網格的方案數，設$G_{n}$$表示用$1\times 2$的骨牌填$3\times n$的網格的方案數.

• 給定$l, r, k$，求$\frac{1}{r - l + 1}\sum_{i = l}^{r} \binom{F_{i}}{k}$.
• 給定$l, r, k$，求$\frac{1}{r - l + 1}\sum_{i = l}^{r} \binom{G_{i}}{k}$.

　　之前好像在loj / uoj羣上問過求Fibonacci求立方和。sad。。。

　　校內考的時候忘了求數列通項，sad。。。

　　首先用Stirling數把組合數拆成通常冪。

　　對於第一部分，直接用$Fibonacci$通項，由於$5$不是模$998244353$的二次剩餘，所以把答案在運算中表示成$a + b\sqrt{5}$的形式。

　　第$n$項的$k$次冪大概是$(a^n + b^n)^{K}$，這個方便直接求和。用二項式定理展開就行了。

　　對於第二部分。考察選手是否上過高考數學課（bushi

　　顯然當$2 \nmid n$時$G_{n} = 0$，現在考慮$G'_{n} = G_{2n}$。

　　注意到最後只有這幾種填法：

　　所以有

$$
\begin{align}
G'_{n} &= 3G_{n - 1} + 2\sum_{i = 0}^{n - 2}G_{i} \\
&= 4G_{n - 1} - 3G_{n - 2} - 2\sum_{i = 0}^{n - 3} G_{i - 1} + 2\sum_{i =0}^{n - 2}G_i \\
&= 4G_{n - 1} - G_{n - 2}
\end{align}
$$

　　用特徵根法推通項（不會的話可以回教室了）。然後做完了。

　　（我被Jode錘爆了。瑟瑟發抖.jpg）

Code

#include <bits/stdc++.h>
using namespace std;
typedef bool boolean;

#define ll long long

void exgcd(int a, int b, int& x, int& y) {
	if (!b) {
		x = 1, y = 0;
	} else {
		exgcd(b, a % b, y, x);
		y -= (a / b) * x;
	}
}

int inv(int a, int n) {
	int x, y;
	exgcd(a, n, x, y);
	return (x < 0) ? (x + n) : (x);
}

const int Mod = 998244353;

template <const int Mod = :: Mod>
class Z {
	public:
		int v;

		Z() : v(0) {	}
		Z(int x) : v(x){	}
		Z(ll x) : v(x % Mod) {	}

		friend Z operator + (const Z& a, const Z& b) {
			int x;
			return Z(((x = a.v + b.v) >= Mod) ? (x - Mod) : (x));
		}
		friend Z operator - (const Z& a, const Z& b) {
			int x;
			return Z(((x = a.v - b.v) < 0) ? (x + Mod) : (x));
		}
		friend Z operator * (const Z& a, const Z& b) {
			return Z(a.v * 1ll * b.v);
		}
		friend Z operator ~(const Z& a) {
			return inv(a.v, Mod);
		}
		friend Z operator - (const Z& a) {
			return Z(0) - a;
		}
		Z& operator += (Z b) {
			return *this = *this + b;
		}
		Z& operator -= (Z b) {
			return *this = *this - b;
		}
		Z& operator *= (Z b) {
			return *this = *this * b;
		}
		friend boolean operator == (const Z& a, const Z& b) {
			return a.v == b.v;
		} 
};

Z<> qpow(Z<> a, int p) {
	Z<> rt = Z<>(1), pa = a;
	for ( ; p; p >>= 1, pa = pa * pa) {
		if (p & 1) {
			rt = rt * pa;
		}
	}
	return rt;
}

typedef Z<> Zi;

template <const int I>
class ComplexTemp {
	public:
		Zi r, v;

		ComplexTemp() : r(0), v(0) {	}
		ComplexTemp(Zi r) : r(r), v(0) {	}
		ComplexTemp(Zi r, Zi v) : r(r), v(v) {	}
		
		friend ComplexTemp operator + (const ComplexTemp& a, const ComplexTemp& b) {
			return ComplexTemp(a.r + b.r, a.v + b.v);
		}
		friend ComplexTemp operator - (const ComplexTemp& a, const ComplexTemp& b) {
			return ComplexTemp(a.r - b.r, a.v - b.v);
		}
		friend ComplexTemp operator - (const ComplexTemp& a, const int& b) {
			return ComplexTemp(a.r - b, a.v);			
		} 
		friend ComplexTemp operator * (const ComplexTemp& a, const ComplexTemp& b) {
			return ComplexTemp(a.r * b.r + a.v * b.v * I, a.r * b.v + a.v * b.r);
		}
		friend ComplexTemp operator * (const ComplexTemp& a, const Zi& x) {
			return ComplexTemp(a.r * x, a.v * x);
		}
		friend ComplexTemp operator / (const ComplexTemp& a, const ComplexTemp& b) {
			ComplexTemp c = b.conj();
			return a * c * ~((b * c).r);
		}
		ComplexTemp conj() const {
			return ComplexTemp(r, -v);
		}
		boolean operator == (ComplexTemp b) {
			return r == b.r && v == b.v;
		}
};

const int Kmx = 510;

Zi s1[Kmx][Kmx];
Zi comb[Kmx][Kmx];
Zi fac[Kmx], _fac[Kmx];
void prepare(int n) {
	fac[0] = 1;
	for (int i = 1; i <= n; i++) {
		fac[i] = fac[i - 1] * i;
	}
	_fac[n] = ~fac[n];
	for (int i = n; i; i--) {
		_fac[i - 1] = _fac[i] * i;
	}

	s1[0][0] = 1;
	for (int i = 1; i <= n; i++) {
		s1[i][0] = 0, s1[i][i] = 1;
		for (int j = 1; j < i; j++) {
			s1[i][j] = s1[i - 1][j - 1] + s1[i - 1][j] * (i - 1);
		}
	}

	comb[0][0] = 1;
	for (int i = 1; i <= n; i++) {
		comb[i][0] = comb[i][i] = 1;
		for (int j = 1; j < i; j++) {
			comb[i][j] = comb[i - 1][j - 1] + comb[i - 1][j];
		}
	}
}

template <typename T>
T qpow(T a, ll p) {
	T rt = Zi(1), pa = a;
	for ( ; p; p >>= 1, pa = pa * pa) {
		if (p & 1) {
			rt = rt * pa;
		}
	}
	return rt;
}

namespace subtask1 {

typedef ComplexTemp<5> Complex;

const Zi inv2 ((Mod + 1) >> 1);
const Complex q1 (inv2, inv2), q2 (inv2, -inv2);
Complex pwq1[Kmx], pwq2[Kmx], pwqq[Kmx][Kmx];

Complex get_sum(int k1, int k2, ll n) {
	Complex x = pwq1[k1] * pwq2[k2];
	if (x == Zi(1))
		return Zi(n);
	return (pwqq[k1][k2] - 1) / (x - 1);
}

inline void init() {
	pwq1[0] = pwq2[0] = Zi(1);
	for (int i = 1; i < Kmx; i++) {
		pwq1[i] = pwq1[i - 1] * q1;
		pwq2[i] = pwq2[i - 1] * q2;
	}
}

Zi work(ll n, int K) {
	Complex coef = qpow(Complex(0, ~Zi(5)), K);
	Complex ret (0, 0);
	for (int k = 0; k <= K; k++) {
//		Complex tmp = get_sum(pwq1[k] * pwq2[K - k], n) * comb[K][k];
		Complex tmp = get_sum(k, K - k, n) * comb[K][k];
		if ((K - k) & 1) {
			ret = ret - tmp;
		} else {
			ret = ret + tmp;
		}
	}
	ret = ret * coef;
	assert(ret.v.v == 0);
//	cerr << n << " " << K << " " << ret.r.v << '\n';
	return ret.r;
}

Zi solve(ll n, int K) {
	Zi ans = 0;
	Complex q1n = qpow(q1, n + 1);
	Complex q2n = qpow(q2, n + 1);
	pwqq[0][0] = Zi(1);
	for (int i = 0; i <= K; i++) {
		for (int j = (i == 0); i + j <= K; j++) {
			pwqq[i][j] = ((!i) ? (pwqq[i][j - 1] * q2n) : (pwqq[i - 1][j] * q1n));
		}
	}
	for (int i = 1; i <= K; i++) {
		Zi tmp = work(n, i) * s1[K][i] * _fac[K];
		if ((K - i) & 1) {
			ans -= tmp;
		} else {
			ans += tmp;
		}
	}
	return ans;
}

void __main__(ll l, ll r, int K) {
	++l, ++r;
	Zi ans = solve(r, K) - solve(l - 1, K);
	ans = ans * ~Zi(r - l + 1);
	printf("%d\n", ans.v);
}

}

namespace subtask2 {

typedef ComplexTemp<3> Complex;

const Zi inv2 ((Mod + 1) >> 1);
const Zi inv3 ((Mod + 1) / 3);
const Zi inv6 = inv2 * inv3;
const Complex c1 (inv2, -inv6), c2 (inv2, inv6);
const Complex q1 (2, Mod - 1), q2 (2, 1);
Complex pwq1[Kmx], pwq2[Kmx], pwc1[Kmx], pwc2[Kmx];
Complex pwqq[Kmx][Kmx];

Complex get_sum(int k1, int k2, ll n) {
	Complex x = pwq1[k1] * pwq2[k2];
	if (x == Zi(1))
		return Zi(n);
	return (pwqq[k1][k2] - 1) / (x - 1);
}

inline void init() {
	pwq1[0] = pwq2[0] = pwc1[0] = pwc2[0] = Zi(1);
	for (int k = 1; k < Kmx; k++) {
		pwq1[k] = pwq1[k - 1] * q1;
		pwq2[k] = pwq2[k - 1] * q2;
		pwc1[k] = pwc1[k - 1] * c1;
		pwc2[k] = pwc2[k - 1] * c2;
	}
}

Zi work(ll n, int K) {
	Complex ret (0, 0);
	for (int k = 0; k <= K; k++) {
		Complex tmp = get_sum(k, K - k, n) * comb[K][k];
		Complex b = pwc1[k] * pwc2[K - k];
		tmp = tmp * b;
		ret = ret + tmp;
	}
	assert(ret.v.v == 0);
//	cerr << n << " " << K << " " << ret.r.v << '\n';
	return ret.r;
}

Zi solve(ll n, int K) {
	n >>= 1;
	Zi ans = 0;
	Complex q1n = qpow(q1, n + 1);
	Complex q2n = qpow(q2, n + 1);
	pwqq[0][0] = Zi(1);
	for (int i = 0; i <= K; i++) {
		for (int j = (i == 0); i + j <= K; j++) {
			pwqq[i][j] = ((!i) ? (pwqq[i][j - 1] * q2n) : (pwqq[i - 1][j] * q1n));
		}
	}
	for (int i = 1; i <= K; i++) {
		Zi tmp = work(n, i) * s1[K][i] * _fac[K];
		if ((K - i) & 1) {
			ans -= tmp;
		} else {
			ans += tmp;
		}
	}
	return ans;
}

void __main__(ll l, ll r, int K) {
	Zi ans = solve(r, K) - solve(l - 1, K);
	ans = ans * ~Zi(r - l + 1);
	printf("%d\n", ans.v);
}

}

int Case, ___;
ll l, r;
int K;
int main() {
	prepare(502);
	scanf("%d%d", &Case, &___);
	if (___ == 3) {
		subtask2::init();
	} else {
		subtask1::init();
	}
	while (Case--) {
		scanf("%lld%lld%d", &l, &r, &K);
		if (___ == 2) {
			subtask1::__main__(l, r, K);
		} else {
			subtask2::__main__(l, r, K);
		}
	}
	return 0;
}