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暴力康托展开/逆康托展开即可。时间复杂度

#include <algorithm>
#include <array>
#include <bitset>
#include <iostream>
#include <numeric>
#include <tuple>
#include <vector>

using namespace std;

using ll = long long;

int n;
int k;

void P() {
    ll a;
    cin >> a;
    a--;
    vector<int> rk(n);
    for (int i = 1; i <= n; i++) {
        rk[n - i] = a % i;
        a /= i;
    }
    vector<int> remain(n);
    iota(remain.begin(), remain.end(), 1);
    for (int i = 0; i < n; i++) {
        cout << remain[rk[i]] << ' ';
        remain.erase(remain.begin() + rk[i]);
    }
    cout << '\n';
}

void Q() {
    vector<int> remain(n);
    iota(remain.begin(), remain.end(), 1);
    ll res = 0;
    for (int i = 0; i < n; i++) {
        int a;
        cin >> a;
        a = lower_bound(remain.begin(), remain.end(), a) - remain.begin();
        res = res * (n - i) + a;
        remain.erase(remain.begin() + a);
    }
    cout << res + 1 << '\n';
}

void Solve() {
    char op;
    cin >> n >> k;
    while (k--) {
        cin >> op;
        if (op == 'P') {
            P();
            continue;
        }
        Q();
    }
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    Solve();
    return 0;
}