HNOI2017 礼物

一道简单的\(FFT\)题
题目链接

题意简述

把一个数组全部加上一个整数\(k\),然后要使\(\sum_{i=1}^n(a_i-b_i)^2\)最小.
注意\(a,b\)这两个数组均可旋转.

---

解析

我们把式子展开.
\(\sum_{i=1}^n(a_i-b_i+x)^2\)
\(=\sum_{i=1}^n(a_i^2+b_i^2+x^2+2a_ix-2b_ix-2a_ib_i)\)
\(=\sum_{i=1}^n(a_i^2+b_i^2)+nx^2+2x\sum_{i=1}^n(a_i-b_i)-\sum_{i=1}^na_ib_i\)
考虑到\(x\)的取值范围很小,为\((-100,100)\),因此可以直接枚举.
所以除了最后一项之外,其他的都知道了.
考虑把\(a\)倍长,再把\(b\)反转.那么,乘积\(c\)的第\(n+1,n+2...2n\)就是答案.枚举一个\(i\)取个\(min\)即可.
计算一下时间复杂度.
\(FFT\)是\(O(n*log\ n)\)的
枚举是\(O(n*m)\)的.
因此总时间复杂度是\(O(n*log\ n+n*m)\)的,可以通过此题.

代码如下

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<vector>
#define N (400010)
#define inf (1e16)
#define rg register int
#define Label puts("NAIVE")
#define spa print(' ')
#define ent print('\n')
#define rand() (((rand())<<(15))^(rand()))
typedef long double ld;
typedef long long LL;
typedef unsigned long long ull;
using namespace std;
inline char read(){
	static const int IN_LEN=1000000;
	static char buf[IN_LEN],*s,*t;
	return (s==t?t=(s=buf)+fread(buf,1,IN_LEN,stdin),(s==t?-1:*s++):*s++);
}
template<class T>
inline void read(T &x){
	static bool iosig;
	static char c;
	for(iosig=false,c=read();!isdigit(c);c=read()){
		if(c=='-')iosig=true;
		if(c==-1)return;
	}
	for(x=0;isdigit(c);c=read())x=((x+(x<<2))<<1)+(c^'0');
	if(iosig)x=-x;
}
inline char readchar(){
	static char c;
	for(c=read();!isalpha(c);c=read())
	if(c==-1)return 0;
	return c;
}
const int OUT_LEN = 10000000;
char obuf[OUT_LEN],*ooh=obuf;
inline void print(char c) {
	if(ooh==obuf+OUT_LEN)fwrite(obuf,1,OUT_LEN,stdout),ooh=obuf;
	*ooh++=c;
}
template<class T>
inline void print(T x){
	static int buf[30],cnt;
	if(x==0)print('0');
	else{
		if(x<0)print('-'),x=-x;
		for(cnt=0;x;x/=10)buf[++cnt]=x%10+48;
		while(cnt)print((char)buf[cnt--]);
	}
}
inline void flush(){fwrite(obuf,1,ooh-obuf,stdout);}
const ld PI=3.14159265359;
struct com{
	ld a,b;
	com(){a=b=0;}
	com operator +(com x){
		com res;
		res.a=a+x.a,res.b=b+x.b;
		return res;
	}
	com operator -(com x){
		com res;
		res.a=a-x.a,res.b=b-x.b;
		return res;
	}
	com operator *(com x){
		com res;
		res.a=a*x.a-b*x.b,res.b=a*x.b+b*x.a;
		return res;
	}
}a[N],b[N];
int n,m,Lim,len,rev[N];
LL a2,sa,b2,sb;
void FFT(com *a,int tp){
	for(int i=0;i<Lim;i++)
	if(i<rev[i])swap(a[i],a[rev[i]]);
	for(int i=1;i<Lim;i<<=1){
		com w; w.a=cos(PI/i),w.b=tp*sin(PI/i);
		for(int R=i<<1,j=0;j<Lim;j+=R){
			com p; p.a=1;
			for(int k=j;k<j+i;k++,p=p*w){
				com x=a[k],y=p*a[k+i];
				a[k]=x+y,a[k+i]=x-y;
			}
		}
	}
	if(tp==-1)
	for(int i=0;i<Lim;i++)
	a[i].a/=(ld)Lim;
}
int main(){
	read(n),read(m);
	for(int i=1;i<=n;i++){
		int x; read(x);
		a2+=(LL)x*x,sa+=x;
		a[i].a=a[i+n].a=x;
	}
	for(int i=1;i<=n;i++){
		int x; read(x);
		b2+=(LL)x*x,sb+=x;
		b[n-i+1].a=x;
	}
	for(Lim=1;Lim<=n*3;Lim<<=1)len++;
	for(int i=0;i<Lim;i++)
	rev[i]=(rev[i>>1]>>1)|((i&1)<<(len-1));
	FFT(a,1),FFT(b,1);
	for(int i=0;i<Lim;i++)a[i]=(a[i]*b[i]);
	FFT(a,-1);
	LL ans=inf;
	for(int i=n+1;i<=2*n;i++)
	for(int j=-m;j<=m;j++)
	ans=min(ans,a2+b2+2ll*(LL)j*(sa-sb)+(LL)n*(LL)j*(LL)j-(LL)round(a[i].a)*2ll);
	printf("%lld\n",ans);
}