禮物

• 這估計是最水，最無腦的一道題了
• 首先發現總和最接近時答案最小
• 發現答案就是\((\sum_{i=1}^{n}a[i]^2+b[i]^2)-2*max(\sum_{i=1}^{n}a[i]*b[i+j])(0<=j<=n-1)\)
• 前面隨便算，主要是後面那個式子，其實就是兩個數列錯位相乘加起來最大值

• 把\(b\)反過來就變成\(\sum_{i=1}^{n}a[i]*b[n-i-j])(0<=j<=n-1)\)，直接就多項式卷積，FFT一算就行了。

  // luogu-judger-enable-o2
  #include<bits/stdc++.h>
  using namespace
   
   std;
  typedef int sign;
  typedef long long ll;
  #define For(i,a,b) for(register sign i=(sign)a;i<=(sign)b;++i)
  #define Fordown(i,a,b) for(register sign i=(sign)a;i>=(sign)b;--i)
  const int N=5e4+5;
  bool cmax(ll &a,ll b){return (a<b)?a=b,1:0;}
  bool cmin(ll &a,ll b){return (a>b)?a=b,1:0
   
  ;}
  template<typename T>inline T read()
  {
  T f=1,ans=0;
  char ch=getchar();
  while(!isdigit(ch)&&ch!=‘-‘)ch=getchar();
  if(ch==‘-‘)f=-1,ch=getchar();
  while(isdigit(ch))ans=(ans<<3)+(ans<<1)+(ch-‘0‘),ch=getchar();
  return ans*f;
  }
  template<typename T>inline void write(T x,char
   
   y)
  {
  if(x==0)
  {
      putchar(‘0‘);putchar(y);
      return;
  }
  if(x<0)
  {
      putchar(‘-‘);
      x=-x;
  }
  static char wr[20];
  int top=0;
  for(;x;x/=10)wr[++top]=x%10+‘0‘;
  while(top)putchar(wr[top--]);
  putchar(y);
  }
  void file()
  {
  #ifndef ONLINE_JUDGE
      freopen("3723.in","r",stdin);
      freopen("3723.out","w",stdout);
  #endif
  }
  struct comp
  {
  double r,v;
  comp operator + (const comp &x) const {return (comp){r+x.r,v+x.v};}
  comp operator - (const comp &x) const {return (comp){r-x.r,v-x.v};}
  comp operator * (const comp &x) const {return (comp){r*x.r-v*x.v,r*x.v+v*x.r};} 
  };
  comp a[N<<2],b[N<<2];
  int m,n,t;
  int rev[N<<2];
  int sum1,sum2;
  void input()
  {
  n=read<int>();t=read<int>();m=n-1;
  For(i,0,m)a[i].r=read<int>(),sum1+=a[i].r;
  For(i,0,m)b[m-i].r=read<int>(),sum2+=b[m-i].r;
  }
  void init()
  {
  if(sum1>sum2)swap(a,b),swap(sum1,sum2);
  int temp=(sum2-sum1)/n;
  if(sum1+(temp+1)*n-sum2<sum2-sum1-temp*n)temp++;
  For(i,0,m)a[i].r+=temp;
  m<<=1;
  int cnt=0;
  for(n=1;n<=m;n<<=1)cnt++;
  for(int i=0;i<n;i++)rev[i]=(rev[i>>1]>>1)|((i&1)*(1<<(cnt-1)));
  }
  const double pi=acos(-1.0);
  ll ans;
  void FFT(comp *p,int type)
  {
  int i,j,k,num;
  for(i=0;i<n;i++)if(i<rev[i])swap(p[i],p[rev[i]]);
  for(i=2;i<=n;i<<=1)
  {
      comp wi={cos(pi*2/i),sin(pi*2/i)*type};
      num=i>>1;
      for(j=0;j<n;j+=i)
      {
          comp x={1,0};
          for(k=0;k<num;k++,x=x*wi)
          {
              comp u=p[j+k],v=p[j+k+num]*x;
              p[j+k]=u+v;
              p[j+k+num]=u-v;
          }
      }
  }
  }
  int w[N<<2];
  void work()
  {
  //For(i,0,m>>1)write((int)a[i].r,i==m/2?‘\n‘:‘ ‘);
  //For(i,0,m>>1)write((int)b[i].r,i==m/2?‘\n‘:‘ ‘);
  ll res=0;
  For(i,0,m>>1)ans+=a[i].r*a[i].r+b[i].r*b[i].r;
  FFT(a,1);FFT(b,1);
  for(int i=0;i<n;i++)a[i]=a[i]*b[i];
  FFT(a,-1);
  for(int i=0;i<n;i++)w[i]=(int)(a[i].r/n+0.5);
  m>>=1;
  For(i,0,m-1)cmax(res,w[i]+w[i+m+1]);
  write(ans-res*2,‘\n‘);
  }
  int main()
  {
  file();
  input();
  init();
  work();
  return 0;
  }

HNOI2017禮物