洛谷P3803 【模板】多项式乘法（FFT）

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FFT我啥都不会，先坑着

//minamoto
#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;
#define getc() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++)
char buf[1<<21],*p1=buf,*p2=buf;
inline int read(){
    #define num ch-'0'
    char ch;bool flag=0;int res;
    while(!isdigit(ch=getc()))
    (ch=='-')&&(flag=true);
    for(res=num;isdigit(ch=getc());res=res*10+num);
    (flag)&&(res=-res);
    #undef num
    return res;
}
char sr[1<<21],z[20];int C=-1,Z;
inline void Ot(){fwrite(sr,1,C+1,stdout),C=-1;}
inline void print(int x){
    if(C>1<<20)Ot();if(x<0)sr[++C]=45,x=-x;
    while(z[++Z]=x%10+48,x/=10);
    while(sr[++C]=z[Z],--Z);sr[++C]=' ';
}
const int N=1e7+5;const double Pi=acos(-1.0);
struct complex{
    double x,y;
    complex(double xx=0,double yy=0){x=xx,y=yy;}
    inline complex operator +(complex b){return complex(x+b.x,y+b.y);}
    inline complex operator -(complex b){return complex(x-b.x,y-b.y);}
    inline complex operator *(complex b){return complex(x*b.x-y*b.y,x*b.y+y*b.x);}
}a[N],b[N];
int n,m,l,r[N],limit=1;
void FFT(complex *A,int type){
    for(int i=0;i<limit;++i)
    if(i<r[i]) swap(A[i],A[r[i]]);
    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(){
//    freopen("testdata.in","r",stdin);
    n=read(),m=read();
    for(int i=0;i<=n;++i) a[i].x=read();
    for(int i=0;i<=m;++i) b[i].x=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(a,1),FFT(b,1);
    for(int i=0;i<=limit;++i) a[i]=a[i]*b[i];
    FFT(a,-1);
    for(int i=0;i<=n+m;++i)
    print((int)(a[i].x/limit+0.5));
    Ot();
    return 0;
}