FFT与NTT的模板

网上相关博客不少，这里给自己留个带点注释的模板，以后要是忘了作提醒用。

以洛谷3803多项式乘法裸题为例。

FFT：

 #include <cstdio>
 #include <cmath>
 #include <cctype>
 #include <algorithm>
 #define ri readint()
 #define gc getchar()

 int readint() {
     int x = , s = , c = gc;
     while (c <= )    c = gc;
     if (c == '-')    s = -, c = gc;
     for (; isdigit(c); c = gc)    x = x *  + c - ;
     return x * s;
 }

 const int maxn =  * 1e6 + ;
 const double PI = acos(-1.0);

 struct Complex {
     double x, y;
     Complex(double a = , double b = ):x(a), y(b){}
 };
 Complex operator + (Complex A, Complex B) { return Complex(A.x + B.x, A.y + B.y); }
 Complex operator - (Complex A, Complex B) { return Complex(A.x - B.x, A.y - B.y); }
 Complex operator * (Complex A, Complex B) { return Complex(A.x * B.x - A.y * B.y, A.x * B.y + A.y * B.x); }

 Complex a[maxn], b[maxn];
 int n, m;
 int r[maxn], l, limit = ;

 void fft(Complex *A, int type) {
     for (int i = ; i < limit; i++)
         if (i < r[i])
             std::swap(A[i], A[r[i]]);
     //迭代方式模拟递归写法，需要理解递归是怎么做的才能看懂这个
     for (int mid = ; mid < limit; mid <<= ) {
         //本来单位根是2*PI/len，这里len替换成2*mid，2就约掉了
         Complex Wn(cos(PI / mid), type * sin(PI / mid));
         for (int R = mid << , j = ; j < limit; j += R) {
             Complex w(, );//单位根的k次幂
             for (int k = ; k < mid; k++, w = w * Wn) {
                 //蝴蝶变换
                 Complex x = A[j+k], y = w * A[j+k+mid];
                 A[j+k] = x + y;
                 A[j+k+mid] = x - y;
             }
         }
     }
 }

 int main() {
     n = ri, m = ri;
     for (int i = ; i <= n; i++)
         a[i].x = ri;
     for (int i = ; i <= m; i++)
         b[i].x = ri;

     while (limit <= n + m) {//长度变为2^l
         limit <<= ;
         l++;
     }
     for (int i = ; i < limit; i++)//二进制镜像
         r[i] = (r[i>>] >> ) | ((i&) << (l-));
     fft(a, );
     fft(b, );
     for (int i = ; i < limit; i++)
         a[i] = a[i] * b[i];
     fft(a, -);
     for (int i = ; i <= n + m; i++)
         printf("%d ", (int)(a[i].x / limit + 0.5));
     return ;
 }

　NTT是用模域取代了复数域，性质相同只是换了单位根，所以板子基本相同。我这两个相比NTT确实比FFT快一点的：

 #include <bits/stdc++.h>
 #define ll long long
 #define ri readll()
 #define gc getchar()
 #define rep(i, a, b) for (int i = a; i <= b; i++)
 using namespace std;

 const int P = , G = , Gi = , maxn =  * 1e6 + ;
 //P的原根为3,3%P的逆元为332748118
 //原根意味着：3^(P-1) % P = 1,其中P-1是3%P的阶，本应是φ(P)，这里恰好为大素数
 ll n, m;
 ll a[maxn], b[maxn];
 int limit = , l, r[maxn];

 ll readll() {
     ll x = 0ll, s = 1ll;
     char c = gc;
     while (c <= )    c = gc;
     if (c == '-')    s = -1ll, c = gc;
     for (; isdigit(c); c = gc)    x = x *  + c - ;
     return x * s;
 }

 ll ksm(ll a, ll b, int mod) {
     ll res = 1ll;
     for (; b; b >>= ) {
         if (b & )    res = res * a % mod;
         a = a * a % mod;
     }
     return res;
 }

 void NTT(ll *A, int flag) {
     rep(i, , limit)
     if (i < r[i])
         swap(A[i], A[r[i]]);

     for (int mid = ; mid < limit; mid <<= ) {
         //如果是变换则单位根为3^[(P-1)/(len)] % P,逆变换则用逆元
         ll Wn = ksm(flag ? G : Gi, (P-) / (mid*), P);
         for (int R = mid << , j = ; j < limit; j += R) {
             ll w = 1ll;
             for (int k = ; k < mid; k++, w = w * Wn % P) {
                 ll x = A[j+k], y = A[j+k+mid] * w % P;
                 A[j+k] = (x + y) % P;
                 A[j+k+mid] = (x - y + P) % P;
             }
         }
     }
 }

 int main() {
     n = ri, m = ri;
     rep(i, , n)    a[i] = (ri + P) % P;
     rep(i, , m)    b[i] = (ri + P) % P;

     while (limit < n + m + ) {
         limit <<= ;
         l++;
     }
     rep(i, , limit)    r[i] = (r[i>>] >> ) | ((i & ) << (l - ));
     NTT(a, );    NTT(b, );
     rep(i, , limit)    a[i] = a[i] * b[i] % P;
     NTT(a, );

     ll inv = ksm(limit, P - , P);//最后变换回来要乘长度的逆元
     rep(i, , n + m)    printf("%lld ", a[i] * inv % P);

     return ;
 }