两个数列的加和移动其实可以看成是一个数列的加减和移动。(想一想，为什么)

我们设第一个数列加的值为k，

则
\[ \displaystyle \sum _{i=1}^{n}(x_i+k-y_i)^2\\=\displaystyle \sum_{i=1}^n(x_i^2+k^2+y_i^2+2kx_i-2ky_i-2x_iy_i)\\=\displaystyle \sum_{i=1}^nx_i^2+\displaystyle \sum_{i=1}^ny_i^2 +nk^2+2kSum_x-2kSum_y-2\displaystyle \sum_{i=1}^nx_iy_i\\=\displaystyle \sum_{i=1}^nx_i^2+\displaystyle \sum_{i=1}^ny_i^2 +nk^2+2(Sum_x-Sum_y)k-2\displaystyle \sum_{i=1}^nx_iy_i \]
冷静分析.jpg

首先 \(\displaystyle \sum_{i=1}^nx_i^2+\displaystyle \sum_{i=1}^ny_i^2\)还是不会变的，所以我们先加上。

然后\(nk^2+2(Sum_x-Sum_y)k\)和怎么移动无关，通过二次函数的知识我们知道\(k\displaystyle =\frac{Sum_x-Sum_y}{n}\)时最优，注意这时候k不一定是整数，需要考虑一下。

最后就是\(2\displaystyle \sum_{i=1}^nx_iy_i\)，当然我们希望这个值越大越好，并且这次和k无关。到我们发扬人类智慧的时候了，我们将x数组反转，然后再复制一倍，我们就惊奇的发现卷积后的数组从\(n+1\)到\(2n\)的每一个值就是移动相应个数后的\(2\displaystyle \sum_{i=1}^nx_iy_i\)，取个max就好啦。

三个加起来就是我们要的答案啦。

时间复杂度O(n log n).

#include<iostream>
#include<cstdio>
#define LL long long
using namespace std;
int n, m;
LL ans, suma, sumb, zh_AK = -1e18, k = 1e18;
const int N = 400010, mod = 998244353, G = 3, Ginv = (mod + 1) / 3;
int r[N];
LL a[N], b[N];
inline int read() 
{
    int res = 0; char ch = getchar(); bool XX = false;
    for (; !isdigit(ch); ch = getchar())(ch == '-') && (XX = true);
    for (; isdigit(ch); ch = getchar())res = (res << 3) + (res << 1) + (ch ^ 48);
    return XX ? -res : res;
}
LL ksm(LL a, LL b, LL mod) 
{
    LL res = 1;
    for (; b; b >>= 1, a = a * a % mod)
        if (b & 1)res = res * a % mod;
    return res;
}
void NTT(LL *A, int lim, int opt) 
{
    for (int i = 0; i < lim; ++i)
        r[i] = (r[i >> 1] >> 1) | ((i & 1) ? (lim >> 1) : 0);
    for (int i = 0; i < lim; ++i)
        if (i < r[i])swap(A[i], A[r[i]]);
    int len;
    LL wn, w, x, y;
    for (int mid = 1; mid < lim; mid <<= 1) 
    {
        len = mid << 1;
        wn = ksm(opt == 1 ? G : Ginv, (mod - 1) / len, mod);
        for (int j = 0; j < lim; j += len) 
        {
            w = 1;
            for (int k = j; k < j + mid; ++k, w = w * wn % mod) 
            {
                x = A[k]; y = A[k + mid] * w % mod;
                A[k] = (x + y) % mod;
                A[k + mid] = (x - y + mod) % mod;
            }
        }
    }
    if (opt == 1)return;
    int ni = ksm(lim, mod - 2, mod);
    for (int i = 0; i < lim; ++i)A[i] = A[i] * ni % mod;
}
void MUL(LL *A, int n, LL *B, int m) 
{
    int lim = 1;
    while (lim <= (n + m))lim <<= 1;
    NTT(A, lim, 1); NTT(B, lim, 1);
    for (int i = 0; i < lim; ++i)A[i] = A[i] * B[i] % mod;
    NTT(A, lim, -1);
}
signed main() 
{
    cin >> n >> m;
    for (int i = 1; i <= n; ++i)a[i] = read(), suma += a[i];
    for (int i = 1; i <= n; ++i)b[i] = read(), sumb += b[i];
    for (int i = 1; i <= n; ++i)ans += a[i] * a[i] + b[i] * b[i];
    for (int i = -201; i <= 201; ++i)k = min(k, n * i * i + 2 * (suma - sumb) * i);
    ans += k;
    for (int i = 1; i <= n / 2; ++i)swap(a[i], a[n - i + 1]);
    for (int i = n + 1; i <= n + n; ++i)a[i] = a[i - n];
    MUL(a, n + n, b, n);
    for (int i = n + 1; i <= n + n; ++i)zh_AK = max(zh_AK, a[i]);
    cout << ans - 2 * zh_AK;
    return 0;
}