mydcwfy's Blog

AHOI2022 排列

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Symbols count in article: 2.5k Reading time ≈ 5 mins.

有趣的数论题目，似乎说不上用了什么高级知识。

题意：一个排列的权值定义为最少置换次数使得出现循环。给出排列，求将任意两个不在一个置换环的数交换后排列的权值和。$n\leq 5\times 10 ^ 5$。

首先，容易发现一个排列的取值就是虽有置换环长度的 $\text{lcm}$。

$\text{lcm}$ 可能会炸，所以需要考虑枚举所有的质数，求质数幂次的最大值，最后合并答案。

考虑暴力枚举两个不同的置换环，我们需要将这两个环对于 $\text{lcm}$ 的贡献去掉，然后将两个置换环长度和加入整个 $\text{lcm}$ 的贡献。去掉两个数可能会产生删除最大值两次，于是我们需要维护质数幂的前三大值以便维护。回滚的时候记录下所有改变的值即可。

暴力枚举置换环最坏是 $O(n ^ 2)$ 的，还没有算计算 $\text{lcm}$ 的复杂度，但是容易发现只有 $O(\sqrt n)$ 个不同的置换环长度，而贡献又只与长度相关。这样枚举复杂度降为 $O(n)$，后面直接暴力枚举涉及到的质数即可通过。不太会算复杂度，据说是 $O(n\log n)$？

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void work()
{
    read(n);
    all.clear();
    for (int i = 1; i <= n; ++ i) read(a[i]);
    for (int i = 1; i <= n; ++ i) vis[i] = false;
    for (int i = 1; i <= n; ++ i) cnt[i] = 0;
    for (int i = 1; i <= n; ++ i) mx1[i] = mx2[i] = mx3[i] = 0;
    for (int i = 1; i <= n; ++ i) {
        if (vis[i]) continue;
        int j = i, cir = 0;
        while (!vis[j]) cir ++, vis[j] = true, j = a[j];
        if (!cnt[cir] ++) all.push_back(cir);
        // printf("Cir %d\n", cir);
    }
    for (int c : all) {
        int cur = c, t = 0, p;
        while (cur ^ 1) {
            p = fac[cur].front(), t = 0;
            while (cur % p == 0) t ++, cur /= p;
            for (int cs = 1; cs <= cnt[c]; ++ cs)
                if (t > mx1[p]) mx3[p] = mx2[p], mx2[p] = mx1[p], mx1[p] = t;
                else if (t > mx2[p]) mx3[p] = mx2[p], mx2[p] = t;
                else chkmax(mx3[p], t);
        }
    }
    int lcm = 1, res = 0;
    for (int i = 1; i < N; ++ i)
        if (mx1[i]) lcm = (LL) lcm * pw[i][mx1[i]] % Mod;
    // printf("Lcm = %d\n", lcm);
    for (int c1 : all)
        for (int c2 : all) {
            if (c1 == c2 && cnt[c1] == 1) continue;
            int cur = lcm;
            auto del = [&](int x) {
                int p, t;
                while (x ^ 1) {
                    p = fac[x].front(), t = 0;
                    if (!havbac[p])
                        havbac[p] = true, bac1[p] = mx1[p], bac2[p] = mx2[p], bac3[p] = mx3[p];
                    while (x % p == 0) t ++, x /= p;
                    if (t == mx1[p])
                        cur = (LL) cur * inv[p][mx1[p] - mx2[p]] % Mod, mx1[p] = mx2[p], mx2[p] = mx3[p];
                    else if (t == mx2[p]) mx2[p] = mx3[p];
                }
            };
            auto ins = [&](int x) {
                int p, t;
                while (x ^ 1) {
                    p = fac[x].front(), t = 0;
                    if (!havbac[p])
                        havbac[p] = true, bac1[p] = mx1[p], bac2[p] = mx2[p], bac3[p] = mx3[p];
                    while (x % p == 0) t ++, x /= p;
                    if (t > mx1[p])
                        cur = (LL) cur * pw[p][t - mx1[p]] % Mod, mx3[p] = mx2[p], mx2[p] = mx1[p], mx1[p] = t;
                    else if (t > mx2[p]) mx3[p] = mx2[p], mx2[p] = t;
                    else chkmax(mx3[p], t);
                }
            };
            auto bac = [&](int x) {
                int p, t;
                while (x ^ 1) {
                    p = fac[x].front(), t = 0;
                    while (x % p == 0) t ++, x /= p;
                    if (havbac[p])
                        havbac[p] = false, mx1[p] = bac1[p], mx2[p] = bac2[p], mx3[p] = bac3[p];
                }
            };
            del(c1), del(c2), ins(c1 + c2);
            bac(c1), bac(c2), bac(c1 + c2);
            // printf("%d %d %d", c1, c2, cur);
            if (c1 == c2) res = (res + (LL) cur * cnt[c1] % Mod * (cnt[c1] - 1) % Mod * c1 % Mod * c1) % Mod;
            else res = (res + (LL) cur * cnt[c1] % Mod * cnt[c2] % Mod * c1 % Mod * c2) % Mod;
            // printf(" : %d\n", res);
        }
    printf("%d\n", res);
}