[CF1443E] Long Permutation

[CF1443E] Long Permutation - 康托展开

Description

你需要维护一个长度为(n)的排列(P)（初态下为 1,2,3...）和(2)种操作：

• (1 l r) 求出(sum_{i=l}^r P_i)
• (2 x) 将(P)替换为(P)的下(x)个排列

排列(P)初始为([1,2,3,cdots n])
总共有(q)次操作

数据范围：
(n,qleq2 imes 10^5,xleq 10^5)

Solution

显然会动的只有最后的不超过 15 个数，我们用康托展开和逆康托展开暴力处理即可

#include <bits/stdc++.h>
using namespace std;

#define int long long

int fac[20];

int Cantor(vector<int> &a, int n)
{
    int ans = 0;
    for (int i = 1; i <= n; i++)
    {
        int tmp = 0;
        for (int j = i + 1; j <= n; j++)
            if (a[j] < a[i])
                ++tmp;
        ans += tmp * fac[n - i];
    }
    return ans;
}

vector<int> ICantor(int x, int n)
{
    vector<int> ans(n + 2), a(n + 2), vec;
    for (int i = 1; i <= n; i++)
        vec.push_back(i);
    for (int i = 1; i <= n; i++)
    {
        a[i] = x / fac[n - i];
        x %= fac[n - i];
    }
    for (int i = 1; i <= n; i++)
    {
        ans[i] = vec[a[i]];
        vec.erase(vec.begin() + a[i]);
    }
    return ans;
}

struct Solver
{
    int n;
    vector<int> a;

    Solver(int n) : n(n)
    {
        a.resize(n + 2);
        for (int i = 1; i <= n; i++)
            a[i] = i;
    }

    void Add(int x)
    {
        int t = Cantor(a, n);
        t += x;
        a = ICantor(t, n);
    }

    int Sum(int l, int r)
    {
        int ans = 0;
        for (int i = l; i <= r; i++)
            ans += a[i];
        return ans;
    }
};

signed main()
{
    fac[0] = 1;
    for (int i = 1; i <= 16; i++)
        fac[i] = fac[i - 1] * i;

    int n, m;
    cin >> n >> m;
    Solver solver(min(n, 15ll));
    for (int i = 1; i <= m; i++)
    {
        int type;
        cin >> type;
        if (type == 1)
        {
            int ans = 0;
            int l, r;
            cin >> l >> r;
            if (n <= 15)
            {
                ans = solver.Sum(l, r);
            }
            else
            {
                int ll = n - 15 + 1, rr = n;
                ll = max(ll, l);
                rr = min(rr, r);
                if (ll > rr)
                    ans = (l + r) * (r - l + 1) / 2;
                if (ll <= rr)
                {
                    r = min(r, n - 15);
                    if (l <= r)
                        ans = (l + r) * (r - l + 1) / 2;
                }
                if (ll <= rr)
                    ans += (rr - ll + 1) * (n - 15);
                if (ll <= rr)
                    ans += solver.Sum(ll - (n - 15), rr - (n - 15));
            }
            cout << ans << endl;
        }
        else
        {
            int x;
            cin >> x;
            solver.Add(x);
        }
    }
}