用下降幂搞搞事情
[ frac{m(m-1) + k(k+1) -(m-k)(m-k-1)}{2} \
=frac{m^2 - m + k^2 + k - m^2 - k^2 + 2mk - k + m}{2} \
= mk \
y_{m} = sum_{k=0}^n A_{k}omega^{mk}\=omega^{frac{m(m-1)}{2}}sum_{k=0}^n A_komega^{frac{k(k+1)}{2}}omega^{frac{-(m-k)(m-k-1)}{2}}
]
后面是一个卷积形式，可以快速实现求 (f(x),f(x^1), f(x^2) .. f(x^n)) 以及循环卷积意义下的多项式快速幂。
例题 UOJ500 任意基DFT
观察发现
[q_m = q_0x^m+sum_{i=0}^{m-1} xy^i =(frac{y}{x-1}+q_0)x^m-frac{y}{x-1}\
]
设 (c = frac{y}{x-1}) ，那么有
[f(q_m) = sum_{i=0}^n a_i((c+q_0)x^m-c)^i \
=sum_{i=0}^n a_isum_{j=0}^i{ichoose j}(c+q_0)^jx^{mj}(-c)^{i-j} \
=sum_{i=0}^n(c+q_0)^ix^{mi}sum_{j=i}^n{i choose j}(-c)^{i-j}a_i
]
令 (b_j = (c+q_0)^i sum_{j=i}^n{i choose j}(-c)^{i-j}a_i) ，这是一个卷积形式，可以用一遍 ( ext{FFT}) 求出所有的 (b) ，那么
[f(q_m) = sum_{i=0}^nb_i x^{mi}
]
再跑一遍 bluestein's 对 (1-m) 求解即可，注意用线性递推求幂次来优化常数。
code
/*program by mangoyang*/
#pragma GCC optimize("Ofast", "inline")
#include<bits/stdc++.h>
#define inf (0x7f7f7f7f)
#define Max(a, b) ((a) > (b) ? (a) : (b))
#define Min(a, b) ((a) < (b) ? (a) : (b))
typedef long long ll;
using namespace std;
template <class T>
inline void read(T &x){
int ch = 0, f = 0; x = 0;
for(; !isdigit(ch); ch = getchar()) if(ch == '-') f = 1;
for(; isdigit(ch); ch = getchar()) x = x * 10 + ch - 48;
if(f) x = -x;
}
const int N = 1 << 21, mod = 998244353, G = 3;
int a[N], b[N], A[N], B[N], C[N], d[N], e[N], js[N], inv[N], n, q, q0, qx, qy;
inline void up(int &x, int y){
x = x + y >= mod ? x + y - mod : x + y;
}
inline int Pow(int a, int b){
int ans = 1;
for(; b; b >>= 1, a = 1ll * a * a % mod)
if(b & 1) ans = 1ll * ans * a % mod;
return ans;
}
namespace poly{
int rev[N], len, lg;
inline void init(int lenth){
for(len = 1, lg = 0; len <= lenth; len <<= 1, lg++);
for(int i = 0; i < len; i++)
rev[i] = (rev[i>>1] >> 1) | ((i & 1) << (lg - 1));
}
inline void dft(int *a, int sgn){
for(int i = 0; i < len; i++)
if(i < rev[i]) swap(a[i], a[rev[i]]);
for(int k = 2; k <= len; k <<= 1){
int w = Pow(G, (mod - 1) / k);
if(sgn == -1) w = Pow(w, mod - 2);
for(int i = 0, now; i < len; i += k){
now = 1;
for(int j = i; j < i + (k >> 1); j++){
int x = a[j], y = 1ll * a[j+(k>>1)] * now % mod;
a[j] = x + y >= mod ? x + y - mod : x + y;
a[j+(k>>1)] = x - y < 0 ? x - y + mod : x - y;
now = 1ll * now * w % mod;
}
}
}
if(sgn == -1){
int x = Pow(len, mod - 2);
for(int i = 0; i < len; i++) a[i] = 1ll * a[i] * x % mod;
}
}
}
int main(){
read(n), read(q), js[0] = 1, inv[0] = 1;
for(int i = 1; i <= n; i++){
js[i] = 1ll * js[i-1] * i % mod;
inv[i] = Pow(js[i], mod - 2);
}
for(int i = 0; i <= n; i++) read(a[i]);
read(q0), read(qx), read(qy);
int c = 1ll * qy * Pow(qx - 1, mod - 2) % mod;
for(int i = 0, ss = 1; i <= n; i++){
A[n-i] = 1ll * js[i] * a[i] % mod;
B[i] = 1ll * inv[i] * ss % mod;
ss = 1ll * ss * (mod - c) % mod;
}
poly::init(n + n + 1);
poly::dft(A, 1), poly::dft(B, 1);
for(int i = 0; i < poly::len; i++)
C[i] = 1ll * A[i] * B[i] % mod;
poly::dft(C, -1);
for(int i = 0; i <= n; i++){
b[i] = 1ll * C[n-i] * inv[i] % mod;
b[i] = 1ll * b[i] * Pow(c + q0, i) % mod;
}
memset(A, 0, sizeof(A));
memset(B, 0, sizeof(B));
memset(C, 0, sizeof(C));
d[0] = 1, e[0] = 1;
int ix = Pow(qx, mod - 2);
for(int i = 1, ss = 1, tt = 1; i <= n + q + 1; i++){
ss = 1ll * qx * ss % mod;
tt = 1ll * ix * tt % mod;
d[i] = 1ll * d[i-1] * ss % mod;
e[i] = 1ll * e[i-1] * tt % mod;
}
for(int i = 0; i <= n; i++)
A[i] = 1ll * b[i] * d[i] % mod;
for(int i = -n + 1; i <= q; i++)
B[i+n] = i <= 0 ? e[abs(i)] : e[i-1];
poly::init(n + n + q);
poly::dft(A, 1), poly::dft(B, 1);
for(int i = 0; i < poly::len; i++)
C[i] = 1ll * A[i] * B[i] % mod;
poly::dft(C, -1);
int ans = 0;
for(int i = 1; i <= q; i++)
ans ^= 1ll * C[i+n] * d[i-1] % mod;
cout << ans << endl;
return 0;
}