BZOJ 4827 [Hnoi2017]礼物 ——FFT

题目上要求一个循环卷积的最小值，直接破环成链然后FFT就可以了。

然后考虑计算的式子，可以分成两个部分分开计算。

前半部分FFT，后半部分扫一遍。

#include <map>
#include <ctime>
#include <cmath>
#include <queue>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define F(i,j,k) for (int i=j;i<=k;++i)
#define D(i,j,k) for (int i=j;i>=k;--i)
#define ll long long
#define double long double
#define llinf 10000000000000000LL
#define maxn 500005
#define eps 1e-6

struct Complex{
    double x,y;
    Complex (){}
    Complex (double _x,double _y){x=_x;y=_y;}
    Complex operator + (Complex a) {return Complex(x+a.x,y+a.y);}
    Complex operator - (Complex a) {return Complex(x-a.x,y-a.y);}
    Complex operator * (Complex a) {return Complex(x*a.x-y*a.y,x*a.y+y*a.x);}
}A[maxn],B[maxn];

const double pi=acos(-1.0);
int rev[maxn];
ll ans=llinf,res[maxn],sumA2=0,sumB2=0,sumA=0,sumB=0;

void FFT(Complex *x,int n,int flag)
{
    F(i,0,n-1) if (rev[i]>i) swap(x[rev[i]],x[i]);
    for (int m=2;m<=n;m<<=1)
    {
        Complex wn=Complex(cos(2*pi/m),flag*sin(2*pi/m));
        for (int i=0;i<n;i+=m)
        {
            Complex w=Complex(1.0,0);
            for (int j=0;j<(m>>1);++j)
            {
                Complex u=x[i+j],v=x[i+j+(m>>1)]*w;
                x[i+j]=u+v;x[i+j+(m>>1)]=u-v;
                w=w*wn;
            }
        }
    }
}

int n,m,L=0;

int main()
{
    scanf("%d%d",&n,&m);
    F(i,0,n-1)
    {
        int x;scanf("%d",&x);
        A[i].x=x;
        sumA+=x;
        sumA2+=(ll)x*x;
    }
    D(i,n-1,0)
    {
        int x;scanf("%d",&x);
        B[i].x=x;
        sumB+=x;
        sumB2+=(ll)x*x;
        B[i+n].x=B[i].x;
    }
    for(m=1;m<=4*n;m<<=1);while(!(m>>L&1))L++;
    F(i,0,m-1)rev[i]=(rev[i>>1]>>1)|((i&1)<<(L-1));
    FFT(A,m,1);FFT(B,m,1);F(i,0,m-1)A[i]=A[i]*B[i];FFT(A,m,-1);
    F(i,0,m-1) res[i]=(A[i].x+0.4)/m;
    F(i,-100,100)
    {
        ll tmp=2*i*(sumA-sumB)+n*i*i;
        F(j,n-1,2*n-1) ans=min(ans,sumA2+sumB2+tmp);
    }
    ll tmp=-llinf;
    F(i,n-1,2*n-1) tmp=max(tmp,res[i]);
    ans-=2*tmp;
    printf("%lld\n",ans);
}