题目描述：设$A,B$为$n-1$次多项式，求$A*B^C$在系数模$n+1$，长度为$n$的循环卷积。
数据范围：$n\leq 5*10^5,C\leq 10^9$，且$n$的质因子不超过7，$n+1$为质数。
这就是一个循环卷积，在$n=2^k$的情况下可以直接使用FFT/NTT，但是这里不行。
由于$n$的质因子很小，设$n=2^{k_1}*3^{k_2}*5^{k_3}*7^{k_4}$，我们考虑将$n$分治为$p$份（NTT就是$p=2$的情况，这里$p=2,3,5,7$）
$$F(\omega_n^i)=\sum_{i=0}^{p-1}\omega_n^iF_i(\omega_{\frac{n}{p}}^i)$$
设$a_i=F(x)[x^i]$
$$F_r(x)=\sum_{i}a_{ip+r}x^i$$
而$rev$数组其实就是把模$p$同余的放在一起。
NTT的逆变换就是把次数上的$i$改成$-i$，不过代码实现里面我直接写了对$A[1:n-1]$进行reverse（这里说的就是反序）
1 #include<bits/stdc++.h>
2 #define Rint register int
3 using namespace std;
4 typedef long long LL;
5 const int N = 500003;
6 int n, C, mod, pri[N], tot, Wn[N];
7 inline void add(int &a, int b){a += b; if(a >= mod) a -= mod;}
8 inline int kasumi(int a, int b){
9 int res = 1;
10 while(b){
11 if(b & 1) res = (LL) res * a % mod;
12 a = (LL) a * a % mod;
13 b >>= 1;
14 }
15 return res;
16 }
17 inline void factor(int n){
18 for(Rint i = 2;i * i <= n;i ++)
19 if(!(n % i)) pri[++ tot] = i, n /= i, -- i;
20 if(n > 1) pri[++ tot] = n;
21 }
22 inline int primitive(){
23 for(Rint i = 2;;i ++){
24 bool flag = true;
25 for(Rint j = 1;j <= tot && flag;j ++)
26 if(kasumi(i, n / pri[j]) == 1) flag = false;
27 if(flag) return i;
28 }
29 }
30 int a[N], b[N], tmp[N];
31 inline void Rev(int *A){
32 for(Rint i = tot, block = n;i;block /= pri[i], i --){
33 for(Rint num = 0, j = 0;j < n;j += block)
34 for(Rint k = 0;k < pri[i];k ++)
35 for(Rint l = 0;l < block;l += pri[i])
36 tmp[num ++] = A[j + k + l];
37 for(Rint i = 0;i < n;i ++) A[i] = tmp[i];
38 }
39 }
40 inline void NTT(int *A, int type){
41 Rev(A);
42 for(Rint i = 1, block = 1;i <= tot;i ++){
43 int mid = block, wi = Wn[n / (block *= pri[i])];
44 for(Rint j = 0;j < n;j ++) tmp[j] = 0;
45 for(Rint j = 0;j < n;j += block){
46 int wk = 1;
47 for(Rint k = 0;k < block;k ++){
48 for(Rint l = k % mid, w = 1;l < block;l += mid, w = (LL) w * wk % mod)
49 add(tmp[j + k], (LL) w * A[j + l] % mod);
50 wk = (LL) wk * wi % mod;
51 }
52 }
53 for(Rint j = 0;j < n;j ++) A[j] = tmp[j];
54 }
55 if(type == -1){
56 std :: reverse(A + 1, A + n);
57 for(Rint i = 0;i < n;i ++)
58 A[i] = (LL) A[i] * n % mod;
59 }
60 }
61 int main(){
62 scanf("%d%d", &n, &C); mod = n + 1;
63 for(Rint i = 0;i < n;i ++) scanf("%d", a + i);
64 for(Rint i = 0;i < n;i ++) scanf("%d", b + i);
65 factor(n);
66 Wn[0] = 1; Wn[1] = primitive();
67 for(Rint i = 2;i <= n;i ++) Wn[i] = (LL) Wn[i - 1] * Wn[1] % mod;
68 NTT(a, 1); NTT(b, 1);
69 for(Rint i = 0;i < n;i ++)
70 a[i] = (LL) a[i] * kasumi(b[i], C) % mod;
71 NTT(a, -1);
72 for(Rint i = 0;i < n;i ++)
73 printf("%d\n", a[i]);
74 }
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