P3723 [AH2017/HNOI2017] NTT-优快云博客

本文链接：https://blog.youkuaiyun.com/neweryyy/article/details/124092641

本文解析了AH2017/HNOI2017竞赛中的礼物分配问题，通过数学建模解决c和k的最小化问题，涉及二次函数优化和循环卷积操作，展示了如何在O(nlogn)的时间复杂度下找到最优解。

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题意

传送门 P3723 [AH2017/HNOI2017]礼物

题解

增加的值 $c\geq 0$ ，需要讨论 $x_i+c$ 与 $y_i+c$ 的情况。仅讨论第一种。
$\min\limits_{c,k}\sum\limits_{i=0}^{n-1} (x_i + c - y_{i+k \mod n})^2$
将式子拆开得到
$\min\limits_{c,k}\Big\{\sum\limits_{i=0}^{n-1} (x_i^2+ y_i^2) + \sum\limits_{i=0}^{n-1}[c^2 + 2c(x_i - y_i)] - 2\sum\limits_{i=0}^{n-1}x_iy_{i+k \mod n}\Big\}$ 后两项是独立的，倒数第二项可以通过求一元二次函数极值得到，最后一项做循环卷积后枚举 $k$ 即可。总时间复杂度 $O(n\log n)$ 。

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
constexpr ll MOD = 998244353, PRT = 3;
ll qpow(ll x, ll n)
{
    ll res = 1;
    while (n > 0)
    {
        if (n & 1)
            res = res * x % MOD;
        x = x * x % MOD, n >>= 1;
    }
    return res;
}
vector<int> rev;
struct Poly : vector<ll>
{
    Poly() {}
    Poly(int n) : vector<ll>(n) {}
    Poly(const initializer_list<ll> &list) : vector<ll>(list) {}
    void fft(int n, bool inverse)
    {
        if ((int)rev.size() != n)
        {
            rev.resize(n);
            for (int i = 0; i < n; ++i)
                rev[i] = rev[i >> 1] >> 1 | (i & 1 ? n >> 1 : 0);
        }
        resize(n);
        for (int i = 0; i < n; ++i)
            if (i < rev[i])
                std::swap(at(i), at(rev[i]));

        for (int m = 1; m < n; m <<= 1)
        {
            int m2 = m << 1;
            ll _w = qpow(inverse ? qpow(PRT, MOD - 2) : PRT, (MOD - 1) / m2);
            for (int i = 0; i < n; i += m2)
                for (int w = 1, j = 0; j < m; ++j, w = w * _w % MOD)
                {
                    ll &x = at(i + j), &y = at(i + j + m), t = w * y % MOD;
                    y = x - t;
                    if (y < 0)
                        y += MOD;
                    x += t;
                    if (x >= MOD)
                        x -= MOD;
                }
        }
    }
    void dft(int n) { fft(n, 0); };
    void idft(int n)
    {
        fft(n, 1);
        for (int i = 0, inv = qpow(n, MOD - 2); i < n; ++i)
            at(i) = at(i) * inv % MOD;
    }
    Poly operator*(const Poly &p) const
    {
        auto a = *this, b = p;
        int k = 1, n = a.size() + b.size() - 1;
        while (k < n)
            k <<= 1;
        a.dft(k), b.dft(k);
        for (int i = 0; i < k; ++i)
            a[i] = a[i] * b[i] % MOD;
        a.idft(k);
        a.resize(n);
        return a;
    }
};
constexpr int MAXN = 5E4 + 5;
int N, M;
int A[MAXN], B[MAXN];

int main()
{
    ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    cin >> N >> M;
    for (int i = 0; i < N; ++i)
        cin >> A[i];
    for (int i = 0; i < N; ++i)
        cin >> B[i];
    ll sa2 = 0, sb2 = 0;
    ll sa = 0, sb = 0;
    for (int i = 0; i < N; ++i)
        sa += A[i], sa2 += A[i] * A[i];
    for (int i = 0; i < N; ++i)
        sb += B[i], sb2 += B[i] * B[i];
    Poly f(N), g(N);
    for (int i = 0; i < N; ++i)
        f[i] = A[i], g[i] = B[i];
    reverse(f.begin(), f.end());
    f = f * g;
    reverse(f.begin(), f.end());
    ll conv = 0;
    f.resize(N * 2);
    for (int i = 0; i < N; ++i)
        conv = max(conv, f[i] + f[N + i]);
    ll res = sa2 + sb2 - 2 * conv;
    ll c = max((ll)(-(sa - sb) / N), 0ll);
    ll sc = 0;
    for (ll _c = c - 1; _c <= c + 1; ++_c)
        sc = min(sc, N * _c * _c + 2 * _c * (sa - sb));
    c = max((ll)((sa - sb) / N), 0ll);
    for (ll _c = c - 1; _c <= c + 1; ++_c)
        sc = min(sc, N * _c * _c + 2 * _c * (sa - sb));
    res += sc;
    cout << res << '\n';
    return 0;
}