【文文殿下】[AH2017/HNOI2017]礼物

题解

二项式展开，然后暴力FFT就好了。会发现有一个卷积与c无关，我们找一个最小的项就行了。

Tips:记得要倍长其中一个数组，防止FFT出锅

代码如下：

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 5e4+10;
const double pi = acos(-1.0);
struct Complex{
	double r,i;
	Complex(double r,double i):r(r),i(i){}
	Complex(){}
} A[maxn<<4],B[maxn<<4];
Complex operator + (Complex a,Complex b) {
	return Complex(a.r+b.r,a.i+b.i);
}
Complex operator - (Complex a,Complex b) {
	return Complex(a.r-b.r,a.i-b.i);
}
Complex operator * (Complex a,Complex b) {
	return Complex(a.r*b.r-a.i*b.i,a.r*b.i+a.i*b.r);
}
void operator *= (Complex &a,Complex b) {
	a=a*b;
}
void fft(Complex *a,int n,int inv) {
	for(int i = 1,j=n>>1;i<n-1;++i) {
		if(i<j) swap(a[i],a[j]);
		int k = n>>1;
		while(j>=k) j-=k,k>>=1;
		j+=k;
	}
	for(int j = 2;j<=n;j<<=1) {
		Complex wn(cos(2*pi/j*inv),sin(2*pi/j*inv));
		for(int i = 0;i<n;i+=j) {
			Complex w(1,0);
			for(int k = i;k<i+(j>>1);++k) {
				Complex u(a[k]),t(a[k+(j>>1)]*w);
				a[k]=u+t;
				a[k+(j>>1)]=u-t;
				w*=wn;
			}
		}
	}
	if(inv == -1) 
		for(int i = 0;i<n;++i) a[i].r/=n;
}
int n,m;
int a[maxn],b[maxn];
ll aa,bb,sa,sb;
int main() {
	ios::sync_with_stdio(false);
	cin.tie(0);cout.tie(0);
	cin>>n>>m;
	for(int i = 0;i<n;++i) cin>>a[i];
	for(int i = 0;i<n;++i) cin>>b[i];
	for(int i = 0;i<n;++i) {
		aa+=a[i]*a[i];
		bb+=b[i]*b[i];
		sa+=a[i];
		sb+=b[i];
	}
	for(int i = 0;i<n;++i) A[n-i].r=a[i],B[i].r=B[i+n].r=b[i];
	int lmt = 1;
	while(lmt<=2*n) lmt<<=1;
	fft(A,lmt,1);fft(B,lmt,1);
	for(int i = 0;i<lmt;++i) A[i]*=B[i];
	fft(A,lmt,-1);
	ll mn = 0;
	for(int i = 0;i<2*n;++i) {
		mn = max(mn , (ll)(A[i].r+0.5));
	}
	ll ans = 10000000000000000LL;
	for(int c = -m;c<=m;++c) {
		ll cc = 1LL*n*c*c;
		ans = min(ans , aa+bb+cc+2LL*sa*c-2LL*sb*c-2LL*mn);
	}
	cout<<ans<<endl;
	return 0;
}