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传送门
题目翻译：给定两个 \(n\) 次多项式 \(A,B\) 和一个整数 \(C\)，求 \(A\times B^C\) 在模 \(x^n\) 意义下的卷积
显然就是个循环卷积，所以只要代入 \(\omega_n^{k}\) 进去求出点值，然后插值就好了
？？？\(n\) 不是 \(2^k\) 的形式，不能直接 \(NTT\)
怎么办呢？
根据题目性质，可以把 \(n\) 拆成 \(2^{a_1}3^{a_2}5^{a_3}7^{a_4}\) 的形式
这启示我们每次不是每次分成两半而是拆分成 \(3/5/7\) 次，然后再合并点值
设 \(F(x)=\sum a_ix^i，F_r(x)=\sum a_{ip+r}x^i\)
那么 \(F(x)=\sum x^rF(x^p)\)
根据单位复数的性质(消去引理和折半引理)那么
\[F(\omega_n^{an+b})=\sum \omega_{np}^{(an+b)r}F_r(w_n^b)\]
那么只需要写一个每次分 \(p\) 份的 \(FFT\) 就好了

# include <bits/stdc++.h>
using namespace std;
typedef long long ll;const int maxn(5e5 + 5);int n, c, a[maxn], b[maxn], tmp[maxn], g, pri[233333], tot, pw[2][maxn], mod, r[maxn];inline int Pow(ll x, int y) {register ll ret = 1;for (; y; y >>= 1, x = x * x % mod)if (y & 1) ret = ret * x % mod;return ret;
}inline void Inc(int &x, int y) {x = x + y >= mod ? x + y - mod : x + y;
}int Dfs(int s, int p, int cur, int blk) {if (cur == tot + 1) return s + p;register int nxt;nxt = blk / pri[cur];return Dfs(s + nxt * (p % pri[cur]), (p - p % pri[cur]) / pri[cur], cur + 1, nxt);
}inline void DFT(int *p, int opt) {register int i, j, k, l, q, t, cur;for (i = 0; i < n; ++i) tmp[r[i]] = p[i];for (i = 0; i < n; ++i) p[i] = tmp[i], tmp[i] = 0;for (i = 1, cur = tot; i < n; i *= pri[cur], --cur) {for (t = i * pri[cur], j = 0; j < n; j += t)for (k = 0; k < t; k += i)for (l = 0; l < i; ++l)for (q = 0; q < pri[cur]; ++q)Inc(tmp[j + k + l], (ll)pw[opt == -1][n / t * (k + l) * q % n] * p[j + i * q + l] % mod);for (j = 0; j < n; ++j) p[j] = tmp[j], tmp[j] = 0;}if (opt == -1) for (c = Pow(n, mod - 2), i = 0; i < n; ++i) p[i] = (ll)p[i] * c % mod;
}int main() {register int i, j, x;scanf("%d%d", &n, &c), mod = n + 1;for (x = n, i = 2; i * i <= x; ++i)while (x % i == 0) pri[++tot] = i, x /= i;if (x > 1) pri[++tot] = x;for (i = 2; ; ++i) {for (g = i, j = 1; g && j <= tot; ++j)if (Pow(g, n / pri[j]) == 1) g = 0;if (g) break;}for (i = 0; i < n; ++i) scanf("%d", &a[i]);for (i = 0; i < n; ++i) scanf("%d", &b[i]);pw[0][0] = pw[1][0] = 1, pw[0][1] = g, pw[1][1] = Pow(g, mod - 2);for (i = 2; i < n; ++i) pw[0][i] = (ll)pw[0][i - 1] * g % mod, pw[1][i] = (ll)pw[1][i - 1] * pw[1][1] % mod;for (i = 0; i < n; ++i) r[i] = Dfs(0, i, 1, n);DFT(a, 1), DFT(b, 1);for (i = 0; i < n; ++i) a[i] = (ll)a[i] * Pow(b[i], c) % mod;DFT(a, -1);for (i = 0; i < n; ++i) printf("%d\n", a[i]);return 0;
}

转载于:https://www.cnblogs.com/cjoieryl/p/10211854.html