Luo Gu P3273 / LOJ2020 / BZOJ4827 [AHOI2017 / HNOI2017] gift (FFT)

Because $ a_i + x, b_i + y $ and $ a_i + (xy), b_i $ are equivalent, so just discussed to $ a_i plus a $ $ [- m, m] $ to the number.

For the title equation:

$$\begin{aligned}\sum\limits^n_{i=1}(a_i+x-b_i)^2&=\sum\limits^n_{i=1}(a_i^2+b_i^2+x^2+2a_ix-2b_ix-2a_ib_i)\\&=\sum\limits^n_{i=1}a_i^2+\sum\limits^n_{i=1}b_i^2+2(\sum\limits^n_{i=1}a_i-\sum\limits^n_{i=1}b_i)x-2\sum\limits^n_{i=1}a_ib_i\end{aligned}$$

Can be found in the preceding item are given value, as long as the obtained $ \ sum \ limits ^ n_ {i = 1} maximum a_ib_i $ on it.

The $ A $, in turn, becomes $ \ sum \ limits ^ n_ {i = 1} a_ {n-i + 1} b_i $, which is typical of a convolution type, the $ A $ times longer, an FFT the convolution of the result in the coefficients is n + 1 ~ 2n entries n results, a maximum value extraction, violence enumerated at x, to obtain the minimum value.

#include<cstdio>
#include<cmath>
typedef long long ll;
const double pi=acos(-1.0);
const int N=300000;
char rB[1<<21],*rS,*rT;
inline char gc(){return rS==rT&&(rT=(rS=rB)+fread(rB,1,1<<21,stdin),rS==rT)?EOF:*rS++;}
inline int rd(){
    char c=gc();
    while(c<48||c>57)c=gc();
    int x=c&15;
    for(c=gc();c>=48&&c<=57;c=gc())x=(x<<3)+(x<<1)+(c&15);
    return x;
}
int r[N],sz=1,l=0;
ll val[50005];
struct cnum{
    double x,y;
    cnum(){}
    cnum(double x,double y):x(x),y(y){}
    inline cnum operator +(const cnum &b)const{return cnum(x+b.x,y+b.y);}
    inline cnum operator -(const cnum &b)const{return cnum(x-b.x,y-b.y);}
    inline cnum operator *(const cnum &b)const{return cnum(x*b.x-y*b.y,x*b.y+y*b.x);}
}a[N],b[N];
inline ll Min(ll a,ll b){return a<b?a:b;}
inline void FFT(cnum *a,short type){
    int i,j,M,R;
    cnum w,wn,t,tt;
    for(i=0;i<sz;++i)if(i<r[i]){t=a[i];a[i]=a[r[i]];a[r[i]]=t;}
    for(M=1;M<sz;M<<=1){
        wn=cnum(cos(pi/M),sin(pi/M)*type);
        for(i=0,R=M<<1;i<sz;i+=R){
            w=cnum(1,0);
            for(j=0;j<M;++j){
                t=a[i+j];tt=a[i+M+j]*w;
                a[i+j]=t+tt;a[i+M+j]=t-tt;
                w=w*wn;
            }
        }
    }
}
int main(){
    int n=rd(),m=rd(),i,x;
    ll sa2=0ll,sa=0ll,sb2=0ll,sb=0ll,res=0ll,ans=0x3f3f3f3f3f3f3f3f;
    for(i=1;i<=n;++i){
        sa+=(x=rd());sa2+=x*x;
        a[(n<<1)-i+1]=a[n-i+1]=cnum(x,0);
    }
    for(i=1;i<=n;++i){
        sb+=(x=rd());sb2+=x*x;
        b[i]=cnum(x,0);
    }
    for(;sz<=n*3;sz<<=1)++l;
    for(i=0;i<sz;++i)r[i]=(r[i>>1]>>1)|((i&1)<<l-1);
    FFT(a,1);FFT(b,1);
    for(i=0;i<sz;++i)a[i]=a[i]*b[i];
    FFT(a,-1);
    for(i=0;i<sz;++i)val[i]=(ll)(a[i].x/sz+0.5);
    for(i=n+1;i<=(n<<1);++i)if(val[i]>res)res=val[i];
    for(i=-m;i<=m;++i)ans=Min(ans,sa2+sb2+(ll)n*i*i+(sa-sb)*(i<<1)-(res<<1));
    printf("%lld",ans);
    return 0;
}

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