【[AH2017/HNOI2017]礼物】

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(-100sim100)
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)-2sum_{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);
}