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。

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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
      
        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 }