P3723 [AH2017/HNOI2017]礼物

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给两个序列$x,y$。求可以移动$y$,即向右移动，$i\rightarrow i+1,n\rightarrow 1$。

求最小值$\sum(x_i+c-y_i)^2$

无脑展开

一个定值，一个二次函数，一个卷积。

$\sum x_{i+k} y_i=h_{n+k+1}(x)=\sum x_{i+k} fy_{n-i+1}$，$fy$为$y$翻转。

由于项数比较小，$FFT$即可

代码 
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#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define pii pair<int, int>
#define mk make_pair
const int N = 1e6 + 10;
const int mod = 1e9 + 7;

const double pi = acos(-1.0);

int read()
{
    int x = 0, f = 1;
    char c = getchar();
    while (c < '0' || c > '9')
    {
        if (c == '-')
            f = -1;
        c = getchar();
    }
    while (c >= '0' && c <= '9')
        x = (x << 1) + (x << 3) + c - '0', c = getchar();
    return x * f;
}

struct Complex
{
    double x, y;
    Complex(double _x = 0.0, double _y = 0.0)
    {
        x = _x;
        y = _y;
    }

    Complex operator-(const Complex &b) const
    {
        return Complex(x - b.x, y - b.y);
    }

    Complex operator+(const Complex &b) const
    {
        return Complex(x + b.x, y + b.y);
    }

    Complex operator*(const Complex &b) const
    {
        return Complex(x * b.x - y * b.y, x * b.y + y * b.x);
    }
};
int rev[N];
void FFT(Complex *A, int n, int inv)
{
    for (int i = 0; i < n; i++)
        if (i < rev[i])
            swap(A[i], A[rev[i]]);
    for (int l = 1; l < n; l <<= 1)
    {
        Complex temp(cos(pi / l), inv * sin(pi / l));
        for (int i = 0; i < n; i += (l << 1))
        {
            Complex omega(1, 0);
            for (int j = 0; j < l; j++, omega = omega * temp)
            {
                Complex x = A[i + j], y = omega * A[i + j + l];
                A[i + j] = x + y;
                A[i + j + l] = x - y;
            }
        }
    }

    if (inv == -1)
        for (int i = 0; i < n; i++)
            A[i].x /= n;
}

void FFTX(Complex *a, int n, Complex *b, int m, Complex *ans)
{
    int ML = 1, bit = 0;
    while (ML < n + m)
        ML <<= 1, bit++;
    for (int i = 0; i < ML; i++)
        rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (bit - 1));
    FFT(a, ML, 1);
    FFT(b, ML, 1);
    for (int i = 0; i < ML; i++)
        ans[i] = a[i] * b[i];
    FFT(ans, ML, -1);
}
Complex a[N], b[N], c[N];
ll ans;
ll suma, sumb;
int main()
{
    int n = read(), m = read();
    for (int i = 1; i <= n; i++)
        a[i].x = read(), suma += a[i].x, ans += a[i].x * a[i].x;
    for (int i = 1; i <= n; i++)
        b[i].x = read(), sumb += b[i].x, ans += b[i].x * b[i].x;
    for (int i = 1; i <= n; i++)
        b[i + n] = b[i];
    // if (-(suma - sumb) < 0)
    //     ans += 0;
    // else
    // {
    //     ll cc = -(suma - sumb) / n;

    //     ans += min(n * cc * cc + 2 * cc * (suma - sumb), n * (cc + 1) * (cc + 1) + 2 * (cc + 1) * (suma - sumb));
    // }
    int t = sumb - suma;
    int c1 = floor(t * 1.0 / n), c2 = ceil(t * 1.0 / n);
    ans += min(n * c1 * c1 - 2 * c1 * t, n * c2 * c2 - 2 * c2 * t);
    reverse(a + 1, a + 1 + n);
    // for (int i = 1; i <= 2 * n; i++)
    //     cout << a[i].x << " ";
    // cout << endl;
    // for (int i = 1; i <= 2 * n; i++)
    //     cout << b[i].x << " ";
    // cout << endl;

    FFTX(a, 2 * n, b, 2 * n, c);

    int mx = 0;
    for (int i = 0; i <= n; i++)
        mx = max(int(c[n + i + 1].x + 0.5), mx);
    ans -= 2 * mx;
    printf("%lld\n", ans);
}