HNOI2017礼物

礼物

• 这估计是最水，最无脑的一道题了
• 首先发现总和最接近时答案最小
• 发现答案就是((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))
• 前面随便算，主要是后面那个式子，其实就是两个数列错位相乘加起来最大值
• 把(b)反过来就变成(sum_{i=1}^{n}a[i]*b[n-i-j])(0<=j<=n-1))，直接就多项式卷积，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?'
':' ');
    //For(i,0,m>>1)write((int)b[i].r,i==m/2?'
':' ');
    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,'
');
}
int main()
{
    file();
    input();
    init();
    work();
    return 0;
}