Luogu 3723 [AH2017/HNOI2017]礼物

BZOJ 4827

$$\sum_{i = 1}^{n}(x_i - y_i + c)^2 = \sum_{i = 1}^{n}(x_i^2 + y_i^2 + c^2 - 2 * x_iy_i + 2c * x_i - 2c * y_i) = \sum_{i = 1}^{n}x_i^2 + \sum_{i = 1}^{n}y_i^2 + nc^2 + (2\sum_{i = 1}^{n}(x_i -y_i))c - 2 * \sum_{i = 1}^{n}x_iy_i$$

发现第一项和第二项是一个定值，而第三项和第四项构成了一个开口向上的二次函数，当$c$最靠近对称轴的时候最小，唯一要处理的是最后一项$\sum_{i = 1}^{n}x_iy_i$。

把$y$序列翻转一下，变成$\sum_{i = 1}^{n}x_iy_{n - i + 1}$，这是一个卷积的形式，可以使用$NTT$来加速。

而题目中要求可以旋转一个序列，那么把$x$序列倍长然后当作多项式和翻转后的$y$序列乘起来。

这一项就是在乘起来之后的第$n + 1$项到$2 * n$项中取个最大值。

时间复杂度$O(nlogn)$。

Code：

#include <cstdio>
#include <cstring>
using namespace std;
typedef long long ll;

const int N = 3e5 + 5;
const ll P = 998244353LL;
const ll inf = 1LL << 60;

int n, m, lim = 1, pos[N];
ll a[N], b[N];

template <typename T>
inline void read(T &X) {
    X = 0; char ch = 0; T op = 1;
    for (; ch > '9' || ch < '0'; ch = getchar())
        if (ch == '-') op = -1;
    for (; ch >= '0' && ch <= '9'; ch = getchar())
        X = (X << 3) + (X << 1) + ch - 48;
    X *= op;
}

template <typename T>
inline void swap(T &x, T &y) {
    T t = x; x = y; y = t;
}

template <typename T>
inline void chkMin(T &x, T y) {
    if(y < x) x = y;
}

inline ll fpow(ll x, ll y) {
    ll res = 1LL;
    for (; y > 0; y >>= 1) {
        if (y & 1) res = res * x % P;
        x = x * x % P;
    }
    return res;
}

inline void prework() {
    int l = 0;
    for (; lim <= 3 * n; ++l, lim <<= 1);
    for (int i = 0; i < lim; i++) 
        pos[i] = (pos[i >> 1] >> 1) | ((i & 1) << (l - 1));
}

inline void ntt(ll *c, int opt) {
    for (int i = 0; i < lim; i++)
        if (i < pos[i]) swap(c[i], c[pos[i]]);
    for (int i = 1; i < lim; i <<= 1) {
        ll wn = fpow(3, (P - 1) / (i << 1));
        if(opt == -1) wn = fpow(wn, P - 2);
        for (int len = i << 1, j = 0; j < lim; j += len) {
            ll w = 1LL;
            for (int k = 0; k < i; k++, w = w * wn % P) {
                ll x = c[j + k], y = w * c[j + k + i] % P;
                c[j + k] = (x + y) % P, c[j + k + i] = (x - y + P) % P; 
            }
        }
    }
    
    if (opt == -1) {
        ll inv = fpow(lim, P - 2);
        for (int i = 0; i < lim; i++)
            c[i] = c[i] * inv % P;
    }
}

int main() {
    read(n), read(m);
    ll suma = 0LL, sqra = 0LL, sumb = 0LL, sqrb = 0LL; 
    for (int i = 0; i < n; i++) {
        read(a[i]);
        suma += a[i], sqra += a[i] * a[i];
    }
    for (int i = 0; i < n; i++) {
        read(b[i]);
        sumb += b[i], sqrb += b[i] * b[i];
    }
    
    for (int i = 0; i < n; i++) a[i + n] = a[i];
    for (int i = 0; i < (n / 2); i++) swap(b[i], b[n - i - 1]);
    
    prework();
    ntt(a, 1), ntt(b, 1);
    for (int i = 0; i < lim; i++) a[i] = a[i] * b[i] % P;
    ntt(a, -1);
    
/*    for (int i = 0; i < lim; i++)
        printf("%lld%c", a[i], i == (lim - 1) ? '\n' : ' ');   */
    
    ll ans = inf;
    for (int i = 0; i < n; i++) {
        for (int j = -m; j <= m; j++) {
            ll res = sqra + sqrb + 1LL * n * j * j;
            res += 2LL * j * (suma - sumb) - 2LL * a[i + n - 1];
            chkMin(ans, res);
        }
    }
    
    printf("%lld\n", ans);
    return 0;
}