礼物「AHOI / HNOI2017」

题意

有两个手环，手环上均有\(n\)个珠子，每个珠子有一个值。你可以给第二个手环每个珠子加上同一个值。求\(\sum(x[i]-y[i])^2\)的最小值。

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思路

这道题还是比较young and simple的。

设加上的值为\(c\)，那么所求式子等价于\(\sum(c+x[i]-y[i])^2=\sum(c^2+(x[i]-y[i])^2+2*c*(x[i]-y[i]))\)

二次函数顶点公式可以得到\(c\)的最优取值为\(\frac{\sum(y[i]-x[i])}{n}\)，注意这里会出现精度问题。

FFT求一下\(\sum x[i]y[i]\)即可。

代码

#include <bits/stdc++.h> 

using namespace std;

namespace StandardIO {
	
	template<typename T> inline void read (T &x) {
		x=0;T f=1;char c=getchar();
		for (; c<'0'||c>'9'; c=getchar()) if (c=='-') f=-1;
		for (; c>='0'&&c<='9'; c=getchar()) x=x*10+c-'0';
		x*=f; 
	}
	template<typename T> inline void write (T x) {
		if (x<0) putchar('-'),x=-x;
		if (x>=10) write(x/10);
		putchar(x%10+'0');
	}
	
}

using namespace StandardIO;

namespace Solve {
	
	const int N=600000;
	const double pi=acos(-1);
	typedef complex<double> cx;
	
	int n,m,q,L;
	double ans=0x7fffffff;
	int x[N],y[N],R[N];
	cx a[N],b[N];
	int Sx,Sy,S;
	
	void fft (cx *tmp,int type) {
		for (register int i=0; i<n; ++i) if (i<R[i]) swap(tmp[i],tmp[R[i]]);
		for (register int i=1; i<n; i<<=1) {
			cx wn(cos(pi/i),type*sin(pi/i));
			for (register int j=0; j<n; j+=(i<<1)) {
				cx w(1,0);
				for (register int k=0; k<i; ++k,w*=wn) {
					cx s=tmp[j+k],t=tmp[j+k+i]*w;
					tmp[j+k]=s+t,tmp[j+k+i]=s-t;
				}
			}
		} 
	}
	
	inline void MAIN () {
		read(n),read(q);
		for (register int i=1; i<=n; ++i) read(x[i]);
		for (register int i=1; i<=n; ++i) read(y[i]);
		for (register int i=1; i<=n; ++i) Sx+=x[i]*x[i],Sy+=y[i]*y[i],S+=(x[i]-y[i]);
		for (register int i=1; i<=n; ++i) a[i]=x[i];
		for (register int i=n+1; i<=2*n; ++i) a[i]=x[i-n];
		for (register int i=1; i<=n; ++i) b[i]=y[n-i+1];
		m=2*n;
		for (n=1; n<=m; n<<=1) ++L;
		for (register int i=0; i<=n; ++i) R[i]=(R[i>>1]>>1)|((i&1)<<(L-1));
		fft(a,1),fft(b,1);
		for (register int i=0; i<=n; ++i) a[i]*=b[i];
		fft(a,-1);
		int c=static_cast<int>(-static_cast<double>(S)/static_cast<double>(m/2)+0.5);
		for (register int ddd=c-5; ddd<=c+5; ++ddd) {
		 for (register int i=m/2+1; i<=m; ++i) {
			ans=min(ans,(double)Sx+(double)Sy+2.0*S*ddd+(double)m/2.0*ddd*ddd-2.0*static_cast<int>(a[i].real()/n+0.5));
		}   
		}
		
		write(static_cast<int>(ans+0.5));
	}
	
}

int main () {
	Solve::MAIN();
}