題目：https://www.lydsy.com/JudgeOnline/problem.php?id=4827

式子就是 \sum_{i=0}^{n-1}(a[ i ] - b[ i+k ] + c)^2 。把 b 翻成兩倍後卷積即可。關於 c 的部分是一個二次函式，注意 c 只能是整數！

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#define db double
using namespace
 
 std;
const int N=5e4+5,M=N*6;
const db pi=acos(-1);
int n,m,sm,len,ca,cb,r[M];
struct cpl{db x,y;}a[M],b[M],I;
cpl operator+ (cpl a,cpl b){return (cpl){a.x+b.x,a.y+b.y};}
cpl operator- (cpl a,cpl b){return (cpl){a.x-b.x,a.y-b.y};}
cpl operator* (cpl a,cpl b){return (cpl){a.x*b.x-a.y*b.y,a.x*b.y+a.y*b.x};}
 
int rdn()
{
  int ret=0;bool fx=1;char ch=getchar();
  while(ch>'9'||ch<'0'){if(ch=='-')fx=0;ch=getchar();}
  while(ch>='0'&&ch<='9') ret=(ret<<3)+(ret<<1)+ch-'0',ch=getchar();
  return fx?ret:-ret;
}
void fft(cpl *a,bool fx)
{
  for(int i=0;i<len;i++)
    if(i<r[i])swap(a[i],a[r[i]]);
  
 
for(int R=2;R<=len;R<<=1)
    {
      int m=R>>1;
      cpl Wn=(cpl){ cos(pi/m),fx?-sin(pi/m):sin(pi/m) };
      for(int i=0;i<len;i+=R)
    {
      cpl w=I;
      for(int j=0;j<m;j++,w=w*Wn)
        {
          cpl tmp=w*a[i+m+j];
          a[i+m+j]=a[i+j]-tmp;
          a[i+j]=a[i+j]+tmp;
        }
    }
    }
}
int main()
{
  n=rdn();m=rdn(); I.x=1;
  for(int i=n-1;i>=0;i--)
    {
      a[i].x=rdn();sm+=a[i].x*a[i].x;ca+=a[i].x;
    }
  for(int i=0;i<n;i++)
    {
      b[i].x=b[i+n].x=rdn();sm+=b[i].x*b[i].x;cb+=b[i].x;
    }
  int c=floor((db)(cb-ca)/n),tmp=n*c*c+2*(ca-cb)*c;
  c++; tmp=min(tmp,n*c*c+2*(ca-cb)*c);
  sm+=tmp;
  len=1;
  for(;len<=n*3;len<<=1);
  for(int i=0;i<len;i++)r[i]=(r[i>>1]>>1)+((i&1)?len>>1:0);
  fft(a,0); fft(b,0);
  for(int i=0;i<len;i++)a[i]=a[i]*b[i];
  fft(a,1); tmp=0;
  for(int i=(n<<1)-1;i>=n-1;i--)
    tmp=max(tmp,int(a[i].x/len+0.5));
  sm-=tmp<<1;
  printf("%d\n",sm);
  return 0;
}