bzoj 4827: [Hnoi2017]礼物

 

枚举对一个环整体加上的值c=-m~m，则答案

$\sum_{i=0}^{n-1}{(a_i-b_{(i+d)\%n}+c)^2}$

$=\sum_{i=0}^{n-1}{(a_i^2+b_i^2+c^2-2cb_i+2ca_i-2a_ib_{(i+d)\%n})}$

其中只有$\sum_{i=0}^{n-1}{a_ib_{(i+d)\%n}}$一项不能直接计算，但可以看出这是循环卷积的形式，用fft求出对每个d的答案，再取最大值即可

#include<cstdio>
#include<cmath>
#include<algorithm>
#define G *++ptr
char buf[2000000],*ptr=buf-1;
int _(){
    int x=0,c=G;
    while(c<48)c=G;
    while(c>47)x=x*10+c-48,c=G;
    return x;
}
const int M=133333;
const double _2pi=acos(-1)*2;
struct cplx{
    double a,b;
    cplx(double _a=0,double _b=0):a(_a),b(_b){}
    cplx operator+(cplx x){return cplx(a+x.a,b+x.b);}
    cplx operator-(cplx x){return cplx(a-x.a,b-x.b);}
    cplx operator*(cplx x){return cplx(a*x.a-b*x.b,a*x.b+b*x.a);}
}A[M],B[M],e[2][M];
int N,P;
int n,m,r[M],a[50007],b[50007],a2=0,a1=0,b2=0,b1=0,ans=0x7fffffff,ab=0;
void dft(cplx*a,int t){
    for(int i=0;i<N;++i)if(i<r[i])std::swap(a[i],a[r[i]]);
    for(int i=1;i<N;i<<=1){
        for(int j=0,z=N/(i<<1);j<N;j+=i<<1){
            cplx*b=a+j,*c=b+i;
            for(int k=0;k<i;++k){
                cplx x=b[k],y=c[k]*e[t][k*z];
                b[k]=x+y;
                c[k]=x-y;
            }
        }
    }
    if(t)for(int i=0;i<N;++i)a[i].a/=N;
}
int main(){
    fread(buf,1,sizeof(buf),stdin)[buf]=0;
    n=_();m=_();
    for(int i=0;i<n;++i)a[i]=_(),a2+=a[i]*a[i],a1+=a[i];
    for(int i=0;i<n;++i)b[i]=_(),b2+=b[i]*b[i],b1+=b[i];
    for(N=2,P=0;N<n*2+3;N<<=1,++P);
    for(int i=1;i<N;++i)r[i]=r[i>>1]>>1|(i&1)<<P;
    for(int i=0;i<N;++i){
        e[0][i]=cplx(cos(i*_2pi/N),sin(i*_2pi/N));
        e[1][i]=cplx(cos(i*_2pi/N),-sin(i*_2pi/N));
    }
    for(int i=0;i<n;++i)A[i]=a[n-1-i],B[i]=b[i];
    dft(A,0);dft(B,0);
    for(int i=0;i<N;++i)A[i]=A[i]*B[i];
    dft(A,1);
    for(int i=n;i<n*2;++i){
        int v=int(A[i].a+A[i-n].a+0.49);
        if(v>ab)ab=v;
    }
    for(int c=-m;c<=m;++c){
        int v=a2+b2+c*c*n-2*b1*c+2*a1*c-2*ab;
        if(v<ans)ans=v;
    }
    printf("%d\n",ans);
    return 0;
}