LOJ#3090. 「BJOI2019」勘破神机

为了这题我去学习了一下BM算法。。

很容易发现这2的地方是\(F_{1} = 1,F_{2} = 2\)的斐波那契数列

3的地方是\(G_{1} = 3,G_{2} = 11\)其中下标表示长度的\(\frac{1}{2}\)，可以得到\(G_{3} = 4G_{2} - G_{1}\)

然后我们列一波特征根方程，可以得到

\(m = 2\)时
$$
\left{\begin{matrix}
x_{1} = \frac{1 + \sqrt{5}}{2}
\ x_{2} = \frac{1 - \sqrt{5}}{2}
\ c_{1} = \frac{5 + \sqrt{5}}{10}
\ c_{2} = \frac{5 - \sqrt{5}}{10}

\end{matrix}\right.
$$

\(m = 3\)时
$$
\left{\begin{matrix}
x_{1} = 2 + \sqrt{3}
\ x_{2} = 2 - \sqrt{3}
\ c_{1} = \frac{3 + \sqrt{3}}{6}
\ c_{2} = \frac{3 - \sqrt{3}}{10}

\end{matrix}\right.
$$
然后我们会可以把阶乘\(n(n - 1)(n - 2)...(n - k + 1)\)当成一个k次多项式

然后暴力算出每一项的系数，然后我们要算的就是

\(\sum_{i = l}^{r} f_{i}^{k}\)

然后把通项带进去

\(\sum_{i = l}^{r} (c_{1}x_{1}^{i} + c_{2}x_{2}^{i})^{k}\)

\(\sum_{i = l}^{r}\sum_{j = 0}^{k}\binom{k}{j}c_{1}^{j}x_{1}^{ij}c_{2}^{k - j}x_{2}^{i(k - j)}\)

\(\sum_{j = 0}^{k}\binom{k}{j}c_{1}^{j}c_{2}^{k - j}\sum_{i = l}^{r}(x_{1}^{j}x_{2}^{k - j})^{i}\)

后面是等比数列求和，算一下就好了

复杂度是\(k^{2}\log r\)的

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define eps 1e-10
#define MAXN 2005
#define ba 47
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template<class T>
void read(T &res) {res = 0;T f = 1;char c = getchar();while(c < '0' || c > '9') {if(c == '-') f = -1;c = getchar();}while(c >= '0' && c <= '9') {res = res * 10 +c - '0';c = getchar();}res *= f;
}
template<class T>
void out(T x) {if(x < 0) {x = -x;putchar('-');}if(x >= 10) {out(x / 10);}putchar('0' + x % 10);
}
const int MOD = 998244353;
int inc(int a,int b) {return a + b >= MOD ? a + b - MOD : a + b;
}
int mul(int a,int b) {return 1LL * a * b % MOD;
}
void update(int &x,int y) {x = inc(x,y);
}
int fpow(int x,int c) {int res = 1,t = x;while(c) {if(c & 1) res = mul(res,t);t = mul(t,t);c >>= 1;}return res;
}
template<int T>
struct Complex {int r,i;Complex(int _r = 0,int _i = 0) {r = _r;i = _i;}friend Complex operator + (const Complex &a,const Complex &b) {return Complex(inc(a.r,b.r),inc(a.i,b.i));}friend Complex operator - (const Complex &a,const Complex &b) {return Complex(inc(a.r,MOD - b.r),inc(a.i,MOD - b.i));}friend Complex operator * (const Complex &a,const Complex &b) {return Complex(inc(mul(a.r,b.r),mul(T,mul(a.i,b.i))),inc(mul(a.r,b.i),mul(b.r,a.i)));}friend Complex fpow(Complex x,int64 c) {Complex res(1,0),t = x;while(c) {if(c & 1) res = res * t;t = t * t;c >>= 1;}return res;}friend Complex Query(Complex x,int64 n) {if(x.r == 1 && !x.i) return (n + 1)% MOD;Complex u = 1 - fpow(x,n + 1),d = 1 - x,t;t = d;t.i = MOD - t.i;int iv = fpow((t * d).r,MOD - 2);u = u * t * iv;return u;}
};
int m,k,f[1005],fac[1005],invfac[1005];
int64 l,r;
int C(int n,int m) {if(n < m) return 0;return mul(fac[n],mul(invfac[m],invfac[n - m]));
}
namespace mequal2 {Complex<5> x1,x2,c1,c2;void Main() {int inv2 = fpow(2,MOD - 2),inv10 = fpow(10,MOD - 2);x1 = Complex<5>(inv2,inv2);x2 = Complex<5>(inv2,MOD - inv2);c1 = Complex<5>(mul(5,inv10),inv10);c2 = Complex<5>(mul(5,inv10),MOD - inv10);int ans = mul(invfac[k],fpow((r - l + 1) % MOD,MOD - 2));Complex<5> res;for(int j = 1 ; j <= k ; ++j) {Complex<5> s;for(int h = 0 ; h <= j ; ++h) {Complex<5> t = C(j,h) * fpow(c1,h) * fpow(c2,j - h),q = fpow(x1,h) * fpow(x2,j - h),p;p = Query(q,r) - Query(q,l - 1);s = s + t * p;}res = res + s * f[j];}ans = mul(ans,res.r);out(ans);enter;}
}
namespace mequal3 {Complex<3> x1,x2,c1,c2;void Main() {int inv6 = fpow(6,MOD - 2);x1 = Complex<3>(2,1);x2 = Complex<3>(2,MOD - 1);c1 = Complex<3>(mul(3,inv6),inv6);c2 = Complex<3>(mul(3,inv6),MOD - inv6);int ans = mul(invfac[k],fpow((r - l + 1) % MOD,MOD - 2));r = r / 2;if(l & 1) l = (l + 1) / 2;else l = l / 2;Complex<3> res;if(l <= r) {for(int j = 1 ; j <= k ; ++j) {Complex<3> s;for(int h = 0 ; h <= j ; ++h) {Complex<3> t = C(j,h) * fpow(c1,h) * fpow(c2,j - h),q = fpow(x1,h) * fpow(x2,j - h),p;p = Query(q,r) - Query(q,l - 1);s = s + t * p;}res = res + s * f[j];}}ans = mul(ans,res.r);out(ans);enter;}
}
void Solve() {read(l);read(r);read(k);memset(f,0,sizeof(f));f[1] = 1;for(int i = 1 ; i < k ; ++i) {for(int j = i + 1 ; j >= 1 ; --j) {f[j] = inc(mul(MOD - i,f[j]),f[j - 1]);}}if(m == 2) mequal2::Main();else mequal3::Main();
}
int main() {
#ifdef ivorysifreopen("f1.in","r",stdin);
#endifint T;read(T);read(m);fac[0] = 1;for(int i = 1 ; i <= 1000 ; ++i) fac[i] = mul(fac[i - 1],i);invfac[1000] = fpow(fac[1000],MOD - 2);for(int i = 999 ; i >= 0 ; --i) invfac[i] = mul(invfac[i + 1],i + 1);for(int i = 1 ; i <= T ; ++i) {Solve();}
}