所有的下标从 (0) 开始。
考虑枚举 (C) （第一个加上负的等于第二个加上其绝对值）和第二个手链的偏移量 (p)。答案就是
[sum_{i=0}^{n-1}(x_i+C-y_{(i+p) mod n})^2
]
复制一遍 (y) 数组就能去掉取模了，再展开就是
[sum_{i=0}^{n-1}((x_i+C)^2-2(x_i+C)y_{i+p}+y_{i+p}^2)
]
再展开就是
[sum_{i=0}^{n-1}(x_i+C)^2-2sum_{i=0}^{n-1}x_iy_{i+p}-2Csum_{i=0}^{n-1}y_i+sum_{i=0}^{n-1}y_i^2
]
发现除了第二项别的都可以预处理后在枚举 (C) 时 (O(1)) 得到，问题在于怎样快速求第二项。将 (x) 数组翻转就成了
[sum_{i=0}^{n-1}x_{n-i-1}y_{i+p}
]
显然 (p) 所对应的数就是 fft 后的第 (n+p-1) 项。fft 一次后枚举 (C,p) 即可。当然你也可以三分 (C)。
#include <iostream>
#include <cstdio>
#include <cmath>
using namespace std;
typedef long long ll;
int n, m, xx[50005], yy[50005], lim=1, tmpcnt, rev[300005];
ll ans=0x3f3f3f3f3f3f3f3f, sumxi, sumyi, sumxifang, sumyifang;
const double PI=acos(-1.0);
struct Complex{
double x, y;
Complex(double xx=0.0, double yy=0.0){
x = xx;
y = yy;
}
Complex operator+(const Complex &u)const{
return Complex(x+u.x, y+u.y);
}
Complex operator-(const Complex &u)const{
return Complex(x-u.x, y-u.y);
}
Complex operator*(const Complex &u)const{
return Complex(x*u.x-y*u.y, x*u.y+y*u.x);
}
}A[300005], B[300005];
void fft(Complex a[], int opt){
for(int i=0; i<lim; i++)
if(i<rev[i])
swap(a[i], a[rev[i]]);
for(int i=2; i<=lim; i<<=1){
Complex wn=Complex(cos(PI*2/i), opt*sin(PI*2/i));
int tmp=i>>1;
for(int j=0; j<lim; j+=i){
Complex w=Complex(1.0, 0.0);
for(int k=0; k<tmp; k++){
Complex tmp1=a[j+k], tmp2=w*a[j+k+tmp];
a[j+k] = tmp1 + tmp2;
a[j+k+tmp] = tmp1 - tmp2;
w = w * wn;
}
}
}
if(opt<0)
for(int i=0; i<lim; i++)
a[i].x /= lim;
}
int main(){
cin>>n>>m;
for(int i=0; i<n; i++){
scanf("%d", &xx[i]);
sumxi += xx[i];
sumxifang += xx[i] * xx[i];
A[n-i-1].x = xx[i];
}
for(int i=0; i<n; i++){
scanf("%d", &yy[i]);
sumyi += yy[i];
sumyifang += yy[i] * yy[i];
B[i+n].x = B[i].x = yy[i];
}
while(lim<=3*n) lim <<= 1, tmpcnt++;
for(int i=0; i<lim; i++)
rev[i] = (rev[i>>1]>>1) | ((i&1)<<(tmpcnt-1));
fft(A, 1);
fft(B, 1);
for(int i=0; i<lim; i++)
A[i] = A[i] * B[i];
fft(A, -1);
for(int C=-m; C<=m; C++){
for(int p=0; p<n; p++){
ll tmp=(ll)(A[n+p-1].x+0.5);
ll test=0;
for(int i=0; i<n; i++)
test += xx[i] * yy[(i+p)%n];
tmp = -2 * tmp;
tmp += sumxifang + (ll)2 * C * sumxi + (ll)C * C * n;
tmp -= (ll)2 * C * sumyi;
tmp += sumyifang;
ans = min(ans, tmp);
}
}
cout<<ans<<endl;
return 0;
}