P3723 [AH2017/HNOI2017]礼物（FFT）

P3723 [AH2017/HNOI2017]礼物

式子化简

∑i=1n(xi−yj)2\sum_{i = 1} ^{n} (x_i- y_j) ^2\\ i=1∑n(xi−yj)2

我们对第一个手环+c+c+c，相当于(xi+c−yi)2(x_i + c - y_i) ^ 2(xi+c−yi)2，对第二个手环+c+c+c相当于(xi−yi−c)2(x_i - y_i - c) ^2(xi−yi−c)2

也就是(xi−yj+c)2(x_i - y_j + c) ^2(xi−yj+c)2,c∈[−m,m]c \in[-m, m]c∈[−m,m]

接下来我们就是要求
∑i=1n(xi−yi+c)2∑i=1n(xi2+yi2+2c(xi−yi)+c2−2xiyi)\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_iy_i)\\ i=1∑n(xi−yi+c)2i=1∑n(xi2+yi2+2c(xi−yi)+c2−2xiyi)
除去最后一项，前面的都是定值，无非就是枚举ccc，check一下∑i=1n2c(xi−yi)+c2\sum\limits_{i = 1} ^{n} 2c(x_i - y_i) + c ^ 2i=1∑n2c(xi−yi)+c2的最小值嘛，

但是不难发现这是一个开口向上得二次函数，所以最小值可以直接通过对称轴O(1)O(1)O(1)求得，

所以我们要让∑i=1nxiyi\sum\limits_{i = 1} ^{n}x_i y_ii=1∑nxiyi最大，也就是∑i=1nxiyi+t\sum\limits_{i = 1} ^{n}x_iy_{i + t}i=1∑nxiyi+t最大

我们翻转xxx，得到∑i=1nxn−i+1yi+t\sum\limits_{i = 1} ^{n} x_{n - i + 1} y_{i + t}i=1∑nxn−i+1yi+t最大，这就是一个多项式卷积的形式了

代码

/*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;
}