https://www.oj.swust.edu.cn/problem/show/2781
式子很好推
很明显卷积即可
可以使用
NTT或者拆系数FFT
这里使用的是NTT
#include <algorithm>
#include <cstdio>
#include <cstring>
#include<iostream>
using namespace std;
int mod=1e9+7;
namespace Math
{
inline int pw(int base, int p, const int mod)
{
static int res;
for (res = 1; p; p >>= 1, base = static_cast<long long> (base) * base % mod) if (p & 1) res = static_cast<long long> (res) * base % mod;
return res;
}
inline int inv(int x, const int mod)
{
return pw(x, mod - 2, mod);
}
}
using namespace Math;
const int mod1 = 998244353, mod2 = 1004535809, mod3 = 469762049, G = 3;
const long long mod_1_2 = static_cast<long long> (mod1) * mod2;
const int inv_1 = Math::inv(mod1, mod2), inv_2 = Math::inv(mod_1_2 % mod3, mod3);
struct Int
{
int A, B, C;
explicit inline Int() { }
explicit inline Int(int __num) : A(__num), B(__num), C(__num) { }
explicit inline Int(int __A, int __B, int __C) : A(__A), B(__B), C(__C) { }
static inline Int reduce(const Int &x)
{
return Int(x.A + (x.A >> 31 & mod1), x.B + (x.B >> 31 & mod2), x.C + (x.C >> 31 & mod3));
}
inline friend Int operator + (const Int &lhs, const Int &rhs)
{
return reduce(Int(lhs.A + rhs.A - mod1, lhs.B + rhs.B - mod2, lhs.C + rhs.C - mod3));
}
inline friend Int operator - (const Int &lhs, const Int &rhs)
{
return reduce(Int(lhs.A - rhs.A, lhs.B - rhs.B, lhs.C - rhs.C));
}
inline friend Int operator * (const Int &lhs, const Int &rhs)
{
return Int(static_cast<long long> (lhs.A) * rhs.A % mod1, static_cast<long long> (lhs.B) * rhs.B % mod2, static_cast<long long> (lhs.C) * rhs.C % mod3);
}
inline int get()
{
long long x = static_cast<long long> (B - A + mod2) % mod2 * inv_1 % mod2 * mod1 + A;
return (static_cast<long long> (C - x % mod3 + mod3) % mod3 * inv_2 % mod3 * (mod_1_2 % mod) % mod + x) % mod;
}
} ;
#define maxn 131072
namespace Poly
{
#define N (maxn << 1)
int lim, s, rev[N];
Int Wn[N | 1];
inline void init(int n)
{
s = -1, lim = 1;
while (lim < n) lim <<= 1, ++s;
for (register int i = 1; i < lim; ++i) rev[i] = rev[i >> 1] >> 1 | (i & 1) << s;
const Int t(Math::pw(G, (mod1 - 1) / lim, mod1), Math::pw(G, (mod2 - 1) / lim, mod2), Math::pw(G, (mod3 - 1) / lim, mod3));
*Wn = Int(1);
for (register Int *i = Wn; i != Wn + lim; ++i) *(i + 1) = *i * t;
}
inline void NTT(Int *A, const int op = 1)
{
for (register int i = 1; i < lim; ++i) if (i < rev[i]) std::swap(A[i], A[rev[i]]);
for (register int mid = 1; mid < lim; mid <<= 1)
{
const int t = lim / mid >> 1;
for (register int i = 0; i < lim; i += mid << 1)
{
for (register int j = 0; j < mid; ++j)
{
const Int W = op ? Wn[t * j] : Wn[lim - t * j];
const Int X = A[i + j], Y = A[i + j + mid] * W;
A[i + j] = X + Y, A[i + j + mid] = X - Y;
}
}
}
if (!op)
{
const Int ilim(Math::inv(lim, mod1), Math::inv(lim, mod2), Math::inv(lim, mod3));
for (register Int *i = A; i != A + lim; ++i) *i = (*i) * ilim;
}
}
#undef N
}
using namespace Poly;
int n, k;
Int A[maxn << 1], B[maxn << 1];
int N=1e5+10;
long long fac[maxn];
long long ner[maxn];
int main()
{
fac[0]=1;
scanf("%d%d", &n, &k);
for(int i=1; i<=N; i++)
fac[i]=fac[i-1]*i%mod;
n+=1;
ner[N]=pw(fac[N],mod-2,mod);
for(int i=N-1; i>=0; i--)
ner[i]=(ner[i+1]*(i+1))%mod;//阶乘逆元
for (int i = 0, x; i < n; ++i)
{
A[i] =Int((int)(ner[i]*pw(i,k,mod)%mod));
}
for (int i = 0, x; i < n; ++i)
{
B[i] =Int((int)ner[i]);
}
init(n+n);
NTT(A), NTT(B);
for (int i = 0; i < lim; ++i) A[i] = A[i] * B[i];
NTT(A, 0);
for (int i = 1; i < n; ++i)
{
printf("%lld", (long long)(A[i].get())%mod*fac[i]%mod);
if(i!=n-1)
printf(" ");
}
return 0;
}