多项式乘法 FFT模板

题目传送门

在下只是来存个板子，，（板子还是洛谷找的2333）

证明的话，太（wo）难（bu）写（hui），就先留个坑吧，，，

NOIP后，如果没退役，我会回来填坑的，，

#include<bits/stdc++.h>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<queue>
#include<deque>
#include<list>
#include<set>
#include<vector>
#include<iostream>
#define ll long long
#define re register
#define inf 0x7f7f7f7f
#define inl inline
#define sqr(x) (x*x)
//#define eps 1e-8
#define debug printf("debug
");
//#pragma comment(linker, "/STACK:1024000000,1024000000")
//#pragma GCC optimize (2)
//#pragma G++ optimize (2)
using namespace std;
//const ll mod;
const double Pi=acos(-1.0);
const ll MAXN=1e7+10;
inl ll read() {
    re ll x = 0; re int f = 1;
    char ch = getchar();
    while(ch<'0'||ch>'9') { if(ch== '-' ) f = -1; ch = getchar(); }
    while(ch>='0'&&ch<='9') {x=(x<<1)+(x<<3)+(ch^48);ch=getchar();}
    return x * f;
}
inl char readc() {
    char ch=getchar();
    while(('z'<ch||ch<'a')&&('Z'<ch||ch<'A')) ch=getchar();
    return ch;
}
inl void write(re ll x){
    if(x>=10)write(x/10);
    putchar(x%10+'0');
}
inl void writeln(re ll x){
    if(x<0) {x=-x;putchar('-');}
    write(x); puts("");
}
inl ll gcd(re ll x,re ll y){while(y^=x^=y^=x%=y);return x;}
inl ll Lcm(re ll a,re ll b) {return a/gcd(a,b)*b;}
inl void FR() {
    freopen(".in","r",stdin);
    freopen(".out","w",stdout);
}
inl void FC() {
    fclose(stdin);
    fclose(stdout);
}
struct cmplex{
    double x,y;
    cmplex (double xx=0,double yy=0) {x=xx;y=yy;}
    cmplex operator + (const cmplex &ch) const {return cmplex(x+ch.x,y+ch.y);} 
    cmplex operator - (const cmplex &ch) const {return cmplex(x-ch.x,y-ch.y);} 
    cmplex operator * (const cmplex &ch) const {return cmplex(x*ch.x-y*ch.y,x*ch.y+y*ch.x);} 
}a[MAXN],b[MAXN];
ll n,m,l,r[MAXN],limit=1;
inl void fft(cmplex *a,ll type) {
    for(re ll i=0;i<limit;i++) {if(i<r[i]) swap(a[i],a[r[i]]);}//要迭代序列 
    for(re ll mid=1;mid<limit;mid<<=1) {
        cmplex Wn(cos(Pi/mid),type*sin(Pi/mid));//单位根 
        for(re ll R=mid<<1,j=0;j<limit;j+=R) {//R区间长，枚举到j 
            cmplex w(1,0);//幂 
            for(re ll k=0;k<mid;k++,w=w*Wn) {//左半部分 
                cmplex x=a[j+k],y=w*a[j+mid+k];
                a[j+k]=x+y;a[j+mid+k]=x-y;
            }
        }
    }
}
int main() {
//  FR();
    n=read(),m=read();
    for(re ll i=0;i<=n;i++) a[i].x=read();
    for(re ll i=0;i<=m;i++) b[i].x=read();
    while(limit<=n+m) limit<<=1,l++;
    for(re ll i=0;i<limit;i++) {r[i]=(r[i>>1]>>1)|((i&1)<<(l-1));}//特殊处理奇数 
    fft(a,1);fft(b,1);
    for(re ll i=0;i<=limit;i++) a[i]=a[i]*b[i];
    fft(a,-1);
    for(re ll i=0;i<n+m;i++) {write((ll)(a[i].x/limit+0.5));putchar(' ');}
    writeln((ll)(a[n+m].x/limit+0.5));
//  FC();
    return 0;
}