题目描述
#include<iostream> #include<cstdio> #include<cmath> #define N 200002 using namespace std; typedef long long ll; const double pai=acos(-1.0); ll ans,sum,sum2,l,L,c[N]; int rev[N],n,m; inline ll rd(){ ll x=0;char c=getchar();bool f=0; while(!isdigit(c)){if(c==‘-‘)f=1;c=getchar();} while(isdigit(c)){x=(x<<1)+(x<<3)+(c^48);c=getchar();} return f?-x:x; } struct fs{ double x,y; fs(){x=y=0;} fs(double xx,double yy){x=xx;y=yy;} fs operator +(const fs &b)const{return fs{x+b.x,y+b.y};} fs operator -(const fs &b)const{return fs{x-b.x,y-b.y};} fs operator *(const fs &b)const{return fs{x*b.x-y*b.y,x*b.y+y*b.x};} }a[N],b[N]; inline void FFT(fs *a,int tag){ for(int i=0;i<l;++i)if(i>rev[i])swap(a[i],a[rev[i]]); for(int i=1;i<l;i<<=1){ fs wn(cos(pai/i),tag*sin(pai/i)); for(int j=0;j<l;j+=(i<<1)){ fs w(1,0); for(int k=0;k<i;++k,w=w*wn){ fs x=a[j+k],y=w*a[i+j+k]; a[j+k]=x+y;a[i+j+k]=x-y; } } } } int main(){ n=rd();m=rd(); for(int i=1;i<=n;++i)a[n-i+1].x=rd(); for(int i=1;i<=n;++i){ b[i].x=rd(); sum+=a[i].x*a[i].x+b[i].x*b[i].x; sum2+=a[i].x-b[i].x; } l=1;L=0; while(l<(n<<1))l<<=1,L++; for(int i=1;i<l;++i)rev[i]=(rev[i>>1]>>1)|((i&1)<<(L-1)); FFT(a,1);FFT(b,1); for(int i=0;i<l;++i)a[i]=a[i]*b[i]; FFT(a,-1); for(int i=0;i<l;++i)a[i].x=(ll)(a[i].x/l+0.1); for(int i=1;i<=n;++i)a[i+n].x+=a[i].x; for(int i=n+1;i<=n<<1;++i){ ans=max(ans,(ll)a[i].x); } ll x=1e18; for(int i=sum2/n-6;i<=sum2/n+6;++i)x=min(x,n*i*i+2*i*sum2); printf("%lld",sum-2*ans+x); return 0; }
原文:https://www.cnblogs.com/ZH-comld/p/10257433.html