P3723 [AH2017/HNOI2017]禮物

式子化簡

$$\sum _{i=1}^n(x_i?y_j{)}^2$$

我們對第一個手環$+c$，相當于$(x_i+c?y_i{)}^2$，對第二個手環$+c$相當于$(x_i?y_i?c{)}^2$

也就是$(x_i?y_j+c{)}^2$,$c\in [?m,m]$

接下來我們就是要求
$$\begin{gathered} \sum _{i=1}^n(x_i?y_i+c{)}^2 \\ \sum _{i=1}^n(x_i^2+y_i^2+2c(x_i?y_i)+c^2?2x_i{y}_i) \end{gathered}$$
除去最后一項，前面的都是定值，無非就是枚舉$c$，check一下$\sum _{i=1}^{n}2c(x_i?y_i)+c^2$的最小值嘛，

但是不難發現這是一個開口向上得二次函數，所以最小值可以直接通過對稱軸$O(1)$求得，

所以我們要讓$\sum _{i=1}^n{x}_i{y}_i$最大，也就是$\sum _{i=1}^n{x}_i{y}_{i+t}$最大

我們翻轉$x$，得到$\sum _{i=1}^n{x}_{n?i+1}{y}_{i+t}$最大，這就是一個多項式卷積的形式了

代碼

/*Author : lifehappy */ #include <bits/stdc++.h>using namespace std;typedef long long ll;const double pi = acos(-1.0);const int N = 1e6 + 10;struct Complex {double r, i;Complex(double _r = 0, double _i = 0) : r(_r), i(_i) {}}a[N], b[N];Complex operator + (const Complex & a, const Complex & b) {return Complex(a.r + b.r, a.i + b.i); }Complex operator - (const Complex & a, const Complex & b) {return Complex(a.r - b.r, a.i - b.i); }Complex operator * (const Complex & a, const Complex & b) {return Complex(a.r * b.r - a.i * b.i, a.r * b.i + a.i * b.r); }int r[N];void fft(Complex * f, int lim, int rev) {for(int i = 0; i < lim; i++) {if(r[i] < i) {swap(f[i], f[r[i]]);}}for(int i = 1; i < lim; i <<= 1) {Complex wn = Complex(cos(pi / i), rev * sin(pi / i));for(int p = i << 1, j = 0; j < lim; j += p) {Complex w = Complex(1, 0);for(int k = 0; k < i; k++, w = w * wn) {Complex x = f[j + k], y = w * f[i + j + k];f[j + k] = x + y, f[i + j + k] = x - y;}}}if(rev == -1) {for(int i = 0; i < lim; i++) {f[i].r /= lim;}} }void get_r(int lim) {for(int i = 0; i < lim; ++i) {r[i] = (i & 1) * (lim >> 1) + (r[i >> 1] >> 1);} }int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);ll ans = 0, res = 0;int n, m, lim = 1;scanf("%d %d", &n, &m);for(int i = 1; i <= n; i++) {scanf("%lf", &a[n - i + 1].r); }for(int i = 1; i <= n; i++) {scanf("%lf", &b[i].r);b[i + n]= b[i];}for(int i = 1; i <= n; i++) {ans += a[i].r * a[i].r + b[i].r * b[i].r;res += a[i].r - b[i].r;}ll c1 = floor(res * 1.0 / n), c2 = ceil(res * 1.0 / n);ans += min(n * c1 * c1 - 2 * c1 * res, n * c2 * c2 - 2 * c2 * res);n <<= 2;while(lim < n) lim <<= 1;n >>= 2;get_r(lim);fft(a, lim, 1);fft(b, lim, 1);for(int i = 0; i < lim; i++) {a[i] = a[i] * b[i];}fft(a, lim, -1);res = 0;for(int i = 1; i <= n; i++) {res = max(res, ll(a[i + n].r + 0.5));}printf("%lld\n", ans - 2 * res);return 0; }

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