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CF 896C Willem, Chtholly and Seniorious

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题目描述

https://codeforces.com/problemset/problem/896/C

简要题意：给定一个长度为 $n$ 的序列 $a_i$，现在有 $m$ 个操作，操作有四种，第一种操作给定区间 $[l,r]$ 和权值 $v$，将 $[l,r]$ 所有数加上 $v$；第二种操作给定区间 $[l,r]$ 和权值 $v$，将 $[l,r]$ 内所有数变成 $v$；第三种操作给定区间 $[l,r]$ 和 $k$，求 $[l,r]$ 内第 $k$ 小的数；第四种操作给定区间 $[l,r]$ 和 $x,y$，求 $(\sum_{i=1}^na_i^k)\bmod y$

$n,m\le 10^5$，数据随机

Solution

因为数据随机，所以我们直接使用珂朵莉树维护权值相同的段即可，据说随机数据情况下，只有 $O(\log \log n)$ 段

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#include <iostream>
#include <set>
#include <algorithm>
#include <vector>
#define maxn 100010
#define ll long long
using namespace std;

ll pow_mod(ll x, ll n, int p) {
    ll s = 1;
    for (; n; n >>= 1, x = x * x % p)
        if (n & 1) s = s * x % p;
    return s; 
}

struct Chtolly {
    int l, r;
    mutable ll v;

    Chtolly(int l = 0, int r = 0, ll v = 0) : l(l), r(r), v(v) {}
    friend bool operator < (const Chtolly &u, const Chtolly &v) { return u.l < v.l; } 
}; set<Chtolly> S;

set<Chtolly>::iterator split(int k) {
    auto it = S.lower_bound(Chtolly(k));
    if (it != S.end() && it -> l == k) return it;
    it = prev(it); Chtolly t = *it;
    if (it -> r < k) return S.end();
    return S.erase(it), S.emplace(t.l, k - 1, t.v), S.emplace(k, t.r, t.v).first;
}

void assign(int l, int r, int v) {
    auto itr = split(r + 1), itl = split(l);
    S.erase(itl, itr);
    S.emplace(l, r, v); 
}

void add(int l, int r, int v) {
    auto itr = split(r + 1), itl = split(l);
    for (auto it = itl; it != itr; ++it) it -> v += v;
}

ll query_rk(int l, int r, int k) {
    auto itr = split(r + 1), itl = split(l);
    vector<pair<ll, int>> vec;
    for (auto it = itl; it != itr; ++it)
        vec.emplace_back(it -> v, it -> r - it -> l + 1);
    sort(vec.begin(), vec.end()); 
    for (auto [v, cnt] : vec) { 
        if (k <= cnt) return v;
        k -= cnt;
    } return 0;
}

ll query_sum(int l, int r, int k, int p) {
    auto itr = split(r + 1), itl = split(l);
    ll ans = 0; 
    for (auto it = itl; it != itr; ++it) 
        ans = (ans + (it -> r - it -> l + 1) * pow_mod(it -> v % p , k, p)) % p;
    return ans;
}

int seed, vmax; const int p = 1000000007;
inline int Rand() {
    int res = seed;
    seed = (seed * 7ll + 13) % p;
    return res;
}

int n, m, a[maxn];

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr); cout.tie(nullptr);

    cin >> n >> m >> seed >> vmax;
    for (int i = 1; i <= n; ++i) {
        a[i] = Rand() % vmax + 1;
        S.emplace(i, i, a[i]); 
    }
    for (int i = 1; i <= m; ++i) {
        int opt = Rand() % 4 + 1, l = Rand() % n + 1, r = Rand() % n + 1, x, y;
        if (l > r) swap(l, r);
        if (opt == 3) x = Rand() % (r - l + 1) + 1;
        else x = Rand() % vmax + 1;
        if (opt == 4) y = Rand() % vmax + 1;

        if (opt == 1) add(l, r, x);
        else if (opt == 2) assign(l, r, x);
        else if (opt == 3) cout << query_rk(l, r, x) << "\n";
        else cout << query_sum(l, r, x, y) << "\n";
    }
    return 0;
}