嘟嘟嘟
感觉fft的题重点在于推式子……
因为(n leqslant 5e4, m leqslant 100)，所以可以枚举旋转的位置和增加的亮度，然后想办法在(O(1))时间内得到答案。
令枚举到第(i)个位置时(A, B)两个手环的序列为(A_i, B_i)，此时(B_i)整体怎加了(x)亮度，则：
[ans = sum_{i = 0} ^ {n - 1} (a_i - b_i - x) ^ 2
]
然后拆开((a_i - b_i - x) ^ 2)：
({a_i} ^ 2 + {b_i} ^ 2 + x ^ 2 - 2a_i b_i - 2a_i x + 2b_i x).
于是
[ans = sum_{i = 0} ^ {n - 1} {a_i} ^ 2 + sum_{i = 0} ^ {n - 1} {b_i} ^ 2 + nx ^ 2 - 2x(sum_{i = 0} ^ {n - 1} b_i - sum_{i = 0} ^ {n - 1} a_i) - 2 sum_{i = 0} ^ {n - 1} a_i b_i
]
然后就发现只有最后一项是不知道的，而且只要把(A)翻转一下，就可以卷积了！
但是如果在每一次枚举都求一遍卷积，(O(mn ^ 2 log{n}))时间肯定撑不住，而且求的时候也有很多重复的部分。
因此，我们可以断环为链，得到一个翻转后的长度为(2n)的(A)的序列。然后和(B)卷积。
枚举到第(i)个位置的时候，根据卷积下标相加，(c(n + i))就是答案。
#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<vector>
#include<stack>
#include<queue>
using namespace std;
#define enter puts("")
#define space putchar(' ')
#define Mem(a, x) memset(a, x, sizeof(a))
#define rg register
typedef long long ll;
typedef double db;
const int INF = 0x3f3f3f3f;
const db eps = 1e-8;
const int maxn = 8e6 + 5;
const db PI = acos(-1);
inline ll read()
{
ll ans = 0;
char ch = getchar(), last = ' ';
while(!isdigit(ch)) last = ch, ch = getchar();
while(isdigit(ch)) ans = (ans << 1) + (ans << 3) + ch - '0', ch = getchar();
if(last == '-') ans = -ans;
return ans;
}
inline void write(ll x)
{
if(x < 0) x = -x, putchar('-');
if(x >= 10) write(x / 10);
putchar(x % 10 + '0');
}
int n, m, len = 1;
ll a2 = 0, b2 = 0, suma = 0, sumb = 0;
ll Max = 0, ans = INF;
struct Comp
{
db x, y;
inline Comp operator + (const Comp& oth)const
{
return (Comp){x + oth.x, y + oth.y};
}
inline Comp operator - (const Comp& oth)const
{
return (Comp){x - oth.x, y - oth.y};
}
inline Comp operator * (const Comp& oth)const
{
return (Comp){x * oth.x - y * oth.y, x * oth.y + oth.x * y};
}
inline friend void swap(Comp& a, Comp& b)
{
swap(a.x, b.x); swap(a.y, b.y);
}
}a[maxn], b[maxn], omg[maxn], inv[maxn];
void init()
{
omg[0] = inv[0] = (Comp){1, 0};
omg[1] = inv[len - 1] = (Comp){cos(2 * PI / len), sin(2 * PI / len)};
for(int i = 2; i < len; ++i) omg[i] = inv[len - i] = omg[i - 1] * omg[1];
}
void fft(Comp* a, Comp* omg)
{
int lim = 0;
while((1 << lim) < len) lim++;
for(int i = 0; i < len; ++i)
{
int t = 0;
for(int j = 0; j < lim; ++j) if((i >> j) & 1) t |= (1 << (lim - j - 1));
if(i < t) swap(a[i], a[t]);
}
for(int l = 2; l <= len; l <<= 1)
{
int q = l >> 1;
for(Comp* p = a; p != a + len; p += l)
for(int i = 0; i < q; ++i)
{
Comp t = omg[len / l * i] * p[i + q];
p[i + q] = p[i] - t, p[i] = p[i] + t;
}
}
}
int main()
{
n = read(); m = read();
for(int i = 0; i < n; ++i)
{
db x = read();
a[i] = a[n + i] = (Comp){x, 0};
a2 += x * x; suma += x;
}
for(int i = 0; i < n; ++i)
{
db x = read();
b[n - i - 1] = (Comp){x, 0};
b2 += x * x; sumb += x;
}
while(len < (n << 1) + n) len <<= 1;
init();
fft(a, omg); fft(b, omg);
for(int i = 0; i < len; ++i) a[i] = a[i] * b[i];
fft(a, inv);
for(int i = 0; i < len; ++i) a[i].x = (ll)(a[i].x / len + 0.5);
for(int i = 0; i < n; ++i)
for(int j = -m; j <= m; ++j)
ans = min(ans, a2 + b2 + (ll)n * j * j - (((sumb - suma) * j) << 1) - ((ll)a[i + n].x << 1));
write(ans), enter;
return 0;
}