一道神题233。。。其实要是把思路想通了也不是很难。
正文部分：
设经过了各种操作之后的数列为\(A\)与\(B\)，总增加量为\(x\)。
那么第\(i\)项所造成的影响就是
\((A_i+x-B_i)^2=A_i^2+x^2+B_i^2+2A_ix-2A_iB_i-2xB_i^2\)
然而\(A\)手环加上相等于\(B\)手环减去，所以\(m\)的范围变为\(-100\sim100\)
那么我们可以轻松的得到这两个数列的总影响：
\(\sum_{i=1}^{i=n}(A_i+x-B_i)^2\)
尝试把这个式子展开，成为：
\(\sum_{i=1}^{i=n}A_i^2+\sum_{i=1}^{i=n}B_i^2+n*x^2+2x(\sum_{i=1}^{i=n}A_i-\sum_{i=1}^{i=n}B_i)-2\sum_{i=1}^{i=n}A_iB_i\)
这时发现除了最后一项其他项全是固定的，所以我们只需要让最后一项最大即可，可以用\(fft\)来求解：
My code
#include <bits/stdc++.h>
#define il inline
#define gc getchar
#define FI "gift"
const int MAXN = 4e6 + 10;
const int INF = 0x7f7f7f7f;
const double pi = acos(-1);
using namespace std;
il int read() {
int res = 0;char c;bool sign = 0;
for(c = gc();!isdigit(c);c = gc()) sign |= c == '-';
for(;isdigit(c);c = gc()) res = (res << 1) + (res << 3) + (c ^ 48);
return sign ? -res : res;
}
il void FileIO() {
freopen(FI".in","r",stdin);
freopen(FI".out","w",stdout);
return;
}
struct Cpx {
double r,i;
Cpx(){}
Cpx(double x,double y) {
r = x,i = y;
}
Cpx operator + (const Cpx& x) const {
return Cpx(r + x.r,i + x.i);
}
Cpx operator - (const Cpx& x) const {
return Cpx(r - x.r,i - x.i);
}
Cpx operator * (const Cpx& x) const {
return Cpx(r * x.r - i * x.i,r * x.i + i * x.r);
}
void operator *= (Cpx& x) {
*this = *this * x;
return;
}
}a[MAXN],b[MAXN];
int n,m,i,j,k,R[MAXN],_a[MAXN],_b[MAXN];
int sqr_a,sqr_b,sum_a,sum_b,li,x,ans = INF;
il void fft(Cpx a[MAXN],int f) {
int i,j,k;
for(i = 0;i < li;i++) {
if(i < R[i]) swap(a[i],a[R[i]]);
}
for(i = 1;i < li;i <<= 1) {
Cpx wn(cos(pi / i),f * sin(pi / i));
for(j = 0;j < li;j += (i << 1)) {
Cpx w(1,0);
for(k = 0;k < i;k++,w *= wn) {
Cpx x = a[j + k],y = w * a[j + k + i];
a[j + k] = x + y;a[j + k + i] = x - y;
}
}
}
return;
}
int main() {
// FileIO();
n = read();m = read();
for(i = 1;i <= n;i++) {
_a[i] = read();
sqr_a += _a[i] * _a[i];
sum_a += _a[i];
}
for(i = 1;i <= n;i++) {
_b[i] = read();
sqr_b += _b[i] * _b[i];
sum_b += _b[i];
}
for(i = 1;i <= n;i++) {
a[i].r = a[i + n].r = _a[i];
b[i].r = _b[n - i + 1];
}
int t = 0;li = 1;while(li <= n*3) li <<= 1,t++;
for(i = 0;i <= li;i++) {
R[i] =(R[i >> 1] >> 1 | ((i & 1) << (t - 1)));
}
fft(a,1);fft(b,1);
for(i = 0;i <= li;i++) a[i] *= b[i];
fft(a,-1);
for(i = 0;i <= li;i++) {
a[i].r = (int)(a[i].r / li + 0.5);
}
for(i = 1;i <= n;i++) {
for(x = -m;x <= m;x++) {
int tmp = sqr_a + sqr_b + n * x * x;
tmp += 2 * x *(sum_a - sum_b);
ans = min(ans,tmp - 2 * (int)a[i + n].r);
}
}
printf("%lld",1LL * ans);
}