P3723 【[AH2017/HNOI2017]礼物】

被某大佬指出这是多项式板子！？

我们假设我们原始数列是\(a_i, c_i\)， 旋转后的数列是\(a_i, b_i\)，我们的增加量为x

\[\sum_{i = 1}^n(a_i - b_i + x)^2
\]

拆开平方得：

\[\sum_{i = 1}^na_i^2+b_i^2+x^2+2*x*a_i-2*x*b_i-2*a_i*b_1
\]

把这些东西分下类：

\[x^2*n+(\sum_{i=1}^na_i^2+b_i^2)+2*x*(\sum_{i = 1}^n a_i+b_i)+2*(\sum_{i = 1}^na_i*b_i)
\]

发现\(x\)只有\([-100, 100]\)，我们考虑枚举x，然后就只有最后一堆是未知的

我们考虑怎么求最后一堆：两个值乘在一起的和，是不是和多项式有关系呢？

但这并不是一个卷积的形式，但我们考虑把b反向，原式就变成了：\(\sum_{i=1}^na_i*b_{n-i+1}\)

于是我们就可与愉快的用多项式来做这道题了。不会多项式？戳戳看？

把a数组倍长，把b数组反向，于是这道题的式子就成了：\(\sum_{i=1}^na_{i+k}c_{n-i+1}\)，然后多项式的第\(n+1-2*n\)就分别代表\(k\)取\(0-n-1\)的值了

\(Code:\)

#include<bits/stdc++.h>
using namespace std;
#define il inline
#define re register
#define int long long
const double pi = acos(-1);
#define inf 12345678900000000
il int read() {
    re int x = 0, f = 1; re char c = getchar();
    while(c < '0' || c > '9') { if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - 48, c = getchar();
    return x * f;
}
#define rep(i, s, t) for(re int i = s; i <= t; ++ i)
#define maxn 300005
struct node {
	double x, y;
}a[maxn], b[maxn];
il node operator + (node a, node b) { return (node){a.x + b.x, a.y + b.y}; }
il node operator - (node a, node b) { return (node){a.x - b.x, a.y - b.y}; }
il node operator * (node a, node b) { return (node){a.x * b.x - a.y * b.y, a.x * b.y + a.y * b.x}; }
int n, m, lim, r[maxn], ans = inf, sum1, sum2, Ans = -inf;
il void FFT(node *a, int f, int len) {
	rep(i, 0, len - 1) if(r[i] > i) swap(a[r[i]], a[i]);
	for(re int mid = 1; mid < len; mid <<= 1) {
		node base = (node){cos(pi / mid), f * sin(pi / mid)};
		for(re int p = mid * 2, j = 0; j < len; j += p) {
			node w = (node){1, 0};
			for(re int k = 0; k < mid; ++ k, w = w * base) {
				node x = a[j + k], y = a[j + k + mid] * w;
				a[j + k] = x + y, a[j + k + mid] = x - y;
			}
		}
	}
}
signed main() {
	n = read(), m = read();
	rep(i, 1, n) a[i].x = a[i + n].x = read();
	rep(i, 1, n) b[n - i + 1].x = read();
	rep(i, 1, n) sum1 += a[i].x * a[i].x + b[i].x * b[i].x, sum2 += a[i].x - b[i].x;
	while((1 << lim) <= 3 * n) ++ lim;
	rep(i, 0, (1 << lim)) r[i] = (r[i >> 1] >> 1) | ((i & 1) << (lim - 1));
	FFT(a, 1, (1 << lim)), FFT(b, 1, (1 << lim));
	rep(i, 0, (1 << lim)) a[i] = a[i] * b[i];
	FFT(a, -1, (1 << lim));
	rep(x, -m, m) ans = min(ans, x * x * n + sum1 + 2 * x * sum2);
	rep(i, n + 1, 2 * n) Ans = max(Ans, (int)(a[i].x / (1 << lim) + 0.5));
	printf("%lld", ans - 2 * Ans);
	return 0;
}