FFT&NTT&多项式相关

打了FFT
感觉以后多项式不虚了 _滑稽
PS

关于详见没写完....

code

#include<cmath>
#include<cstdio>
#include<cstring>
#include<algorithm>
// pi = 3.14;
inline int read() {
    int x = 0,f = 1 ;
    char c = getchar();
    while(c < '0' || c > '9') {  if(c == '-') f = -1; c = getchar() ;}
    while(c <= '9' && c >= '0') x = x * 10 + c- '0' ,c = getchar ();
    return x *f;
}
const int maxn  = 5000007;
const double pi = acos(-1.0);
struct Complex {
    double x,y;
    Complex (double xx = 0,double yy = 0) { x = xx,y = yy; }
}f[maxn],g[maxn];
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); }
int n,m ;
int l,r[maxn];
int limit = 1;
void FFT (Complex * A,int type) {
    for(int i = 0;i < limit;i ++ )
        if(i < r[i]) std::swap(A[i],A[r[i]]); //get _迭代序列
    for(int mid = 1;mid < limit; mid <<= 1) {
        Complex wn  (cos(pi / mid) , type * sin(pi / mid));
        for(int R = mid << 1 ,j = 0;j < limit ; j += R) {
            Complex w(1,0);
            for(int k = 0;k < mid;k ++ ,w = w * wn) {
                Complex x = A[j + k] , y = w *A[j + mid + k]; 
                A[j + k] = x + y;
                A[j + mid + k] = x - y;
            }
        }
    }
}
int main() {
    n = read(), m = read();
    for(int i = 0;i <= n; ++ i) f[i] = read();
    for(int i = 0;i <= m; ++ i) g[i] = read();
    while(limit <= n + m) limit <<= 1,l ++;
    for(int i = 0;i < limit;i ++) 
        r[i] = (r[i >> 1] >> 1) | ((i & 1) << (l-1));
    FFT(f,1);FFT(g,1);
    for(int i = 0;i <= limit;++ i) f[i] = f[i] * g[i];
    FFT(f,- 1);
    for(int i = 0;i <= n + m;++ i) 
        printf("%d ",(int) (f[i].x / limit + 0.5));
    return 0; 
}