4827: [Hnoi2017]礼物

 

4827: [Hnoi2017]礼物

链接

分析：

　　求最小的$\sum_{i=1}^{n}(x_i-y_i)^2$

　　设旋转了j位，每一位加上了c。

　　$\sum\limits_{i=1}^{n}(x_{i+j}+c-y_i)^2$

　　$=\sum\limits_{i=1}^{n}x_{i+j}^2+y_i^2+c^2+2x_{i+j}c-2y_ic-2x_{i+j}y_i$

　　$=\sum x_i^2+\sum y_i^2+nc^2+2c \sum (x_i-y_i)-2\sum x_{i+j}y_i$

　　发现只有最后一项是要求的。

　　$\sum x_{i+j}y_i$ 然后把y翻转一下，就是$\sum x_{i+j}y_{n - i +1}$，再把x加倍即可。

　　考虑为什么这样就解决了，如果不旋转，那么$a[1],a[2],a[3]$与$b[1],b[2],b[3]$相乘，那么得到的$c[4]=a[1] \times b[3]+a[2]\times b[2]+a[3]\times b[1]$，所以如果翻转b，c[4]就是旋转0位的答案。在a后面加倍一次，得到那么f[5]就是旋转1位的答案，同理f[6]是旋转2位的答案。

　　而且c是可以直接求出的，可以不用枚举，戳这。

代码：

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>
#include<cmath>
#include<cctype>
#include<set>
#include<queue>
#include<vector>
#include<map>
using namespace std;
typedef long long LL;

inline int read() {
    int x=0,f=1;char ch=getchar();for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-1;
    for(;isdigit(ch);ch=getchar())x=x*10+ch-'0';return x*f;
}

const int N = 200005, INF = 1e9;
const double Pi = acos(-1.0), eps = 1e-10;
int rev[N], f[N], x[N], y[N];
struct Com{ 
    double x, y; 
    Com(double _x = 0.0,double _y = 0) { x = _x, y = _y; }
}a[N], b[N];
Com operator + (const Com &A,const Com &B) { return Com(A.x + B.x, A.y + B.y); }
Com operator - (const Com &A,const Com &B) { return Com(A.x - B.x, A.y - B.y); }
Com operator * (const Com &A,const Com &B) { return Com(A.x * B.x - A.y * B.y, A.x * B.y + A.y * B.x); }

void FFT(Com *a,int n,int ty) {
    for (int i = 0; i < n; ++i) if (i < rev[i]) swap(a[i], a[rev[i]]);
    Com w1, w, t, u;
    for (int m = 2; m <= n; m <<= 1) {
        w1 = Com(cos(2 * Pi / m), ty * sin(2 * Pi / m));
        for (int i = 0; i < n; i += m) {
            w = Com(1, 0);
            for (int k = 0; k < (m >> 1); ++k) {
                t = w * a[i + k + (m >> 1)];
                u = a[i + k];
                a[i + k] = u + t;
                a[i + k + (m >> 1)] = u - t;
                w = w * w1;
            }
        }
    }
}
void mul(Com *a,Com *b,int len) {
    FFT(a, len, 1);
    FFT(b, len, 1);
    for (int i = 0; i <= len; ++i) a[i] = a[i] * b[i];
    FFT(a, len, -1);
    for (int i = 0; i <= len; ++i) f[i] = (int)(a[i].x / (double)len + 0.5);
}
int main() {
    int n = read(), m = read(), len = 1, lg = 0, Sx = 0, Sy = 0, S = 0;
    while (len <= n + n) len <<= 1, lg ++;
    for (int i = 1; i <= len; ++i) 
        rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (lg - 1));
    for (int i = 0; i < n; ++i) 
        x[i] = read(), Sx += (x[i] * x[i]), S += 2 * x[i];    
    for (int i = 0; i < n; ++i) 
        y[i] = read(), Sy += (y[i] * y[i]), S -= 2 * y[i];
    for (int i = 0; i < n + n; ++i) a[i] = Com(x[i % n], 0);
    for (int i = 0; i < n; ++i) b[i] = Com(y[n - i - 1], 0);
    mul(a, b, len);
    int ans = INF;
    for (int c = -m; c <= m; ++c) 
        for (int i = n; i < n + n; ++i) 
            ans = min(ans, Sx + Sy + c * S + n * c * c - 2 * f[i]);
    cout << ans;
    return 0;
}