bzoj4827: [Hnoi2017]礼物

这个m的范围很有欺骗性啊。。。以为是要枚举C的

不管C的话其实就是要一个min｛sigema (Y(i)-X(i+k))^2 ｝

把Y翻转过来，这样旋转k步的圈就是第n+k位

把它拆开的话其实只要算一个X*Y，就是fft的事儿了

再看看C怎么处理，我们带着C拆开，那么会多一个这个东西：sigema (x+C-y)*C == (sumx-sumy+C)*C

C取(sumx-sumy)/n就可以了，保险起见多算几次。

#include<cstdio>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<cstdlib>
#include<algorithm>
#include<cmath>
using namespace std;
typedef long long LL;
LL sqr(LL x){return x*x;}

const double pi=acos(-1.0);
struct complex
{
    double r,i;
    complex(){}
    complex(double R,double I){r=R,i=I;}
    friend complex operator +(complex x,complex y){return complex(x.r+y.r,x.i+y.i);}
    friend complex operator -(complex x,complex y){return complex(x.r-y.r,x.i-y.i);}
    friend complex operator *(complex x,complex y){return complex(x.r*y.r-x.i*y.i,x.r*y.i+x.i*y.r);}
}A[410000],B[410000],C[410000]; int Re[410000];
void fft(complex *a,int n,int op)
{
    for(int i=0;i<n;i++)
        if(i<Re[i])swap(a[i],a[Re[i]]);
        
    for(int i=1;i<n;i<<=1)
    {
        complex wn(cos(pi/i),sin(op*pi/i));
        for(int j=0;j<n;j+=(i<<1))
        {
            complex w(1,0);
            for(int k=0;k<i;k++,w=w*wn)
            {
                complex t1=a[j+k],t2=a[j+k+i]*w;
                a[j+k]=t1+t2;a[j+k+i]=t1-t2;
            }
        }
    }
}

int len,n,aa[51000],bb[51000];
LL solve(int CC)
{
    memset(A,0,sizeof(A));
    memset(B,0,sizeof(B));
    LL sum=0;
    for(int i=0;i<len;i++)
    {
        A[i].r=A[len+i].r=aa[i]+CC, B[i].r=bb[len-1-i];
        sum+=sqr(aa[i]+CC)+sqr(bb[len-1-i]);
    }
    
    fft(A,n,1),fft(B,n,1);
    for(int i=0;i<n;i++)C[i]=A[i]*B[i];
    fft(C,n,-1);
    
    LL ret=(1LL<<62);
    for(int i=len-1;i<=2*(len-1);i++)
        ret=min(ret,sum-2*LL(C[i].r/n+0.5));
    return ret;
}
int main()
{
    freopen("a.in","r",stdin);
    freopen("a.out","w",stdout);
    int M; LL sum=0;
    scanf("%d%d",&len,&M);
    for(int i=0;i<len;i++)scanf("%d",&aa[i]),sum+=aa[i];
    for(int i=0;i<len;i++)scanf("%d",&bb[i]),sum-=bb[i];
    sum/=len;
    
    int m=len*3,L=0;
    for(n=1;n<=m;n*=2)L++;
    for(int i=1;i<=n;i++)Re[i]=(Re[i>>1]>>1)|((i&1)<<(L-1));
    
    LL mmin=(1LL<<62);
    for(int C=sum-1;C<=sum+1;C++)
        mmin=min(mmin,solve(C));
    printf("%lld
",mmin);
    
    return 0;
}