题目传送门

　　传送门

题目大意

　　设$F_{n}$表示用$1\times 2$的骨牌填$2\times n$的网格的方案数，设$G_{n}$$表示用$1\times 2$的骨牌填$3\times n$的网格的方案数.

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

#include <bits/stdc++.h>
using namespace std;
typedef bool boolean;#define ll long longvoid 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;
}