题意

传送门 Codeforces 896C

题解

对于区间操作类型随机且包含区间赋值操作，同时数据随机的数据结构题，可以考虑应用珂朵莉树进行求解。使用 std::set \text{std::set} std::set 实现，初始化时将各元素依次插入，时间复杂度为 O ( N log ⁡ N ) O(N\log N) O(NlogN)；其余操作总时间复杂度 O ( N log ⁡ log ⁡ N ) O(N\log\log N) O(NloglogN)。

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
struct node
{
    int l, r;
    mutable ll v;
    bool operator<(const node &o) const { return l < o.l; }
};
int N, M, seed, vmax;
set<node> tree;

set<node>::iterator split(int p)
{
    auto it = tree.lower_bound(node{p, 0, 0});
    if (it != tree.end() && it->l == p)
        return it;
    --it;
    int l = it->l, r = it->r;
    ll v = it->v;
    tree.erase(it);
    tree.insert(node{l, p, v});
    return tree.insert(node{p, r, v}).first;
}

void assign(int l, int r, int x)
{
    auto ed = split(r), bg = split(l);
    tree.erase(bg, ed);
    tree.insert(node{l, r, x});
}

void add(int l, int r, int x)
{
    auto ed = split(r), bg = split(l);
    for (auto it = bg; it != ed; ++it)
        it->v += x;
}

ll kth(int l, int r, int k)
{
    auto ed = split(r), bg = split(l);
    vector<pair<ll, int>> tmp;
    for (auto it = bg; it != ed; ++it)
        tmp.push_back({it->v, it->r - it->l});
    sort(tmp.begin(), tmp.end());
    for (auto p : tmp)
        if ((k -= p.second) <= 0)
            return p.first;
}

ll pw(ll x, int n, int mod)
{
    ll res = 1;
    x %= mod;
    while (n)
    {
        if (n & 1)
            res = res * x % mod;
        x = x * x % mod, n >>= 1;
    }
    return res;
}

ll sum_pw(int l, int r, int x, int m)
{
    ll sum = 0;
    auto ed = split(r), bg = split(l);
    for (auto it = bg; it != ed; ++it)
        sum += pw(it->v, x, m) * (it->r - it->l);
    return sum % m;
}

int rnd()
{
    int ret = seed;
    seed = ((ll)seed * 7 + 13) % 1000000007;
    return ret;
}

int main()
{
    scanf("%d%d%d%d", &N, &M, &seed, &vmax);
    for (int i = 0; i < N; ++i)
        tree.insert(node{i, i + 1, rnd() % vmax + 1});
    while (M--)
    {
        int op = rnd() % 4 + 1, l = rnd() % N + 1, r = rnd() % N + 1, x, y;
        if (l > r)
            swap(l, r);
        --l;
        if (op == 3)
            x = rnd() % (r - l) + 1;
        else
            x = rnd() % vmax + 1;
        if (op == 4)
            y = rnd() % vmax + 1;
        if (op == 1)
            add(l, r, x);
        else if (op == 2)
            assign(l, r, x);
        else if (op == 3)
            printf("%lld\n", kth(l, r, x));
        else
            printf("%lld\n", sum_pw(l, r, x, y));
    }
    return 0;
}