解法一

数据量比较小，直接按照题意模拟即可，可以开设 1000 个动态数组(链表)，然后模拟合并操作。时间复杂度为 \Omicron(n^2) 。具体实现细节参见代码(C++)：

#define n 1000
#define m 580
ll fa[n + 2], sum[n + 2], num[n + 2], ansi;
vector<ll> g[n + 2];

signed main()
{
    for (ll i = 1; i <= n; ++i)
    {
        fa[i] = i;
        g[i].emplace_back(i);
    }

    for (ll i = 1; i <= m; ++i)
    {
        ll j = (i * i + i * m) % n + 1, fj = fa[j];
        for (ll k : g[fj])
        {
            fa[k] = fa[i];
            g[fa[i]].emplace_back(k);
        }
        g[fj].clear();
    }

    ll w = 0;
    for (ll i = 1; i <= n; ++i)
    {
        for (ll j : g[i])
        {
            sum[i] += j;
        }
        num[i] = g[i].size();
        w += g[i].size();
    }

    for (ll i = 1; i <= n; ++i)
    {
        if (num[i] > num[ansi] || (num[i] == num[ansi] && sum[i] > sum[ansi]))
        {
            ansi = i;
        }
    }
    printf("flag{%lld}", num[ansi] * sum[ansi]);
    return 0;
}

解法二

学过算法的选手应该很快能看出，这就是一道并查集的模板题。直接用并查集知识解之即可。时间复杂度为 \Omicron(n) 。具体实现细节参见代码(C++)：

#define n 1000
#define m 580

ll fa[n + 2], num[n + 2], sum[n + 2], ansi;
ll finds(ll k)
{
    while (k != fa[k])
    {
        k = fa[k];
    }
    return k;
}

signed main()
{
    for (ll i = 1; i <= n; ++i)
    {
        fa[i] = i;
    }
    for (ll i = 1; i <= m; ++i)
    {
        ll j = (i * i + i * m) % n + 1;
        fa[finds(j)] = finds(i);
    }
    for (ll i = 1; i <= n; ++i)
    {
        ll k = finds(i);
        num[k]++;
        sum[k] += i;
    }
    for (ll i = 1; i <= n; ++i)
    {
        if (num[i] > num[ansi] || (num[i] == num[ansi] && sum[i] > sum[ansi]))
        {
            ansi = i;
        }
    }
    printf("flag{%lld}", num[ansi] * sum[ansi]);
    return 0;
}