Luogu4191 [CTSC2010]性能优化【多项式，循环卷积】

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：$nleq 5*10^5,Cleq 10^9$，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）

#include<bits/stdc++.h>
#define Rint register int
using namespace std;
typedef long long LL;
const int N = 500003;
int n, C, mod, pri[N], tot, Wn[N];
inline void add(int &a, int b){a += b; if(a >= mod) a -= mod;}
inline int kasumi(int a, int b){
    int res = 1;
    while(b){
        if(b & 1) res = (LL) res * a % mod;
        a = (LL) a * a % mod;
        b >>= 1;
    }
    return res;
}
inline void factor(int n){
    for(Rint i = 2;i * i <= n;i ++)
        if(!(n % i)) pri[++ tot] = i, n /= i, -- i;
    if(n > 1) pri[++ tot] = n;
}
inline int primitive(){
    for(Rint i = 2;;i ++){
        bool flag = true;
        for(Rint j = 1;j <= tot && flag;j ++)
            if(kasumi(i, n / pri[j]) == 1) flag = false;
        if(flag) return i;
    }
}
int a[N], b[N], tmp[N];
inline void Rev(int *A){
    for(Rint i = tot, block = n;i;block /= pri[i], i --){
        for(Rint num = 0, j = 0;j < n;j += block)
            for(Rint k = 0;k < pri[i];k ++)
                for(Rint l = 0;l < block;l += pri[i])
                    tmp[num ++] = A[j + k + l];
        for(Rint i = 0;i < n;i ++) A[i] = tmp[i];
    }
}
inline void NTT(int *A, int type){
    Rev(A);
    for(Rint i = 1, block = 1;i <= tot;i ++){
        int mid = block, wi = Wn[n / (block *= pri[i])];
        for(Rint j = 0;j < n;j ++) tmp[j] = 0;
        for(Rint j = 0;j < n;j += block){
            int wk = 1;
            for(Rint k = 0;k < block;k ++){
                for(Rint l = k % mid, w = 1;l < block;l += mid, w = (LL) w * wk % mod)
                    add(tmp[j + k], (LL) w * A[j + l] % mod);
                wk = (LL) wk * wi % mod;
            }
        }
        for(Rint j = 0;j < n;j ++) A[j] = tmp[j];
    }
    if(type == -1){
        std :: reverse(A + 1, A + n);
        for(Rint i = 0;i < n;i ++)
            A[i] = (LL) A[i] * n % mod;
    }
}
int main(){
    scanf("%d%d", &n, &C); mod = n + 1;
    for(Rint i = 0;i < n;i ++) scanf("%d", a + i);
    for(Rint i = 0;i < n;i ++) scanf("%d", b + i);
    factor(n);
    Wn[0] = 1; Wn[1] = primitive();
    for(Rint i = 2;i <= n;i ++) Wn[i] = (LL) Wn[i - 1] * Wn[1] % mod;
    NTT(a, 1); NTT(b, 1);
    for(Rint i = 0;i < n;i ++)
        a[i] = (LL) a[i] * kasumi(b[i], C) % mod;
    NTT(a, -1);
    for(Rint i = 0;i < n;i ++)
        printf("%d
", a[i]);
}