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{egin{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}
ight.
]
(m = 3)时
[left{egin{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}
ight.
]
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}inom{k}{j}c_{1}^{j}x_{1}^{ij}c_{2}^{k - j}x_{2}^{i(k - j)})
(sum_{j = 0}^{k}inom{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('
')
#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 ivorysi
freopen("f1.in","r",stdin);
#endif
int 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();
}
}