ODT赛高！珂朵莉树！

珂朵莉树

ODT 来源 CF896C

适用范围

是一种支持以下操作的暴力算法:

• 区间修改
• 区间赋值
• 求区间内的第 $k$ 小
• 求区间 $k$ 次方和

ODT 本质上是建立在区间赋值的一种操作，有区间赋值的操作的话可以保证时间复杂度均摊为 $O(n\log \log n)$，但是如果没有区间赋值，而只有单点赋值，则将会导致 ODT 被卡到 $O(n^2)$， 同时如果是在出题人的精心设计的数据下，ODT也会被卡。但是数据如果是随机的（CCF的）就可以以近乎线性的优秀复杂度过掉。

code time(主要是不想讲原理了)

#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define rl register ll
#define fom(i, a) for(int i=a; i; -- i)
#define foa(i, a, b) for(int i=a; i < b; ++ i)
#define fos(i, a, b) for(int i=a; i <= b; ++ i)
#define fop(i, a, b) for(int i=a; i >= b; -- i)

namespace IO
{
    #define fdt(ch) for(; isdigit(ch); ch=gc())
    #define fsp(ch) for(; !isdigit(ch); ch=gc())
    int pp1=-1; const int pp2=(1<<21)-1; char buf[1<<21],*p1=buf,*p2=buf,buffer[1<<21];
    inline void flush() {fwrite(buffer,1,pp1+1,stdout); pp1=-1;}
    inline void pc(const char ch) {if(pp1==pp2) flush(); buffer[++pp1]=ch; }
    inline char gc(){ return p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++;}
    template <class T> inline void read(T &res){char ch=gc();bool f=0;res=0; fsp(ch) f|=ch=='-'; fdt(ch) res=(res<<1)+(res<<3)+(ch^48); res=f?~res+1:res;}
    template <class T> inline void ww(T x) { if(!x) pc('0'); else { static int stk[20]; int top=0;if(x<0)x=~x+1,pc('-');while(x) stk[top++]=x%10, x/=10; while(top--) pc(stk[top]^48);} }
}

#define ws IO::pc(' ')
#define wl IO::pc('\n')
#define ww(x) IO::ww(x)
#define read(x) IO::read(x)
#define flush() IO::flush()

const ll N = 1e5 + 2, P = 1e9 + 7; 

struct node
{
    ll l, r;
    mutable ll v;

    node (ll l, ll r = 0, ll v = 0) : l(l), r(r), v(v) {};
    
    bool operator <(const node &x) const { return l < x.l; }
};

ll n, m, seed, vmax, a[N];

set<node> s;

// #define swap(l, r) (l ^= r ^= l ^= r)

set<node>::iterator split(ll pos)
{
    auto it = s.lower_bound(node(pos));
    if(it != s.end() && it -> l == pos) return it;

    it --;
    // if(it -> r < pos) return s.end();
    ll l = it -> l, r = it -> r, v = it -> v;
    s.erase(it);
    s.insert((node){l, pos - 1, v});
    return s.insert(node(pos, r, v)).first;
}

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

inline void assign(ll l, ll r, ll x)
{
    auto itr = split(r + 1), itl = split(l);
    s.erase(itl, itr);
    s.insert(node(l, r, x));
}

struct Rank
{
    ll num, cnt;

    bool operator <(const Rank &x) const { return num < x.num; }

    Rank(ll num, ll cnt) : num(num), cnt(cnt) {}
};

inline ll rnk(ll l, ll r, ll x)
{
    auto itr = split(r + 1), itl = split(l);
    vector<Rank> v;
    for(auto it = itl; it != itr; ++ it) v.push_back(Rank(it -> v, (it -> r) - (it -> l) + 1));

    sort(v.begin(), v.end());

    // for(auto t : v)  if(t.cnt < x) x -= t.cnt; else return t.num;

    ll i;

    for(i = 0; i < v.size(); ++ i) if(v[i].cnt < x) x -= v[i].cnt; else break;

    return v[i].num;
}

inline ll qmi(ll a, ll k, ll p)
{
    ll res = 1; a %= p;

    while(k) 
    {
        if(k & 1) res = (a * res) % p;
        a = (a * a) % p, k >>= 1;
    }

    return res;
}

inline ll calcP(ll l, ll r, ll x, ll y)
{
    auto itr = split(r + 1), itl = split(l);
    ll ans = 0;
    for(auto it = itl; it != itr; ++ it ) ans = (ans + qmi(it -> v, x, y) * ((it -> r) - (it -> l) + 1) % y) % y;

    return ans;
}

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

int main()
{
    #ifndef ONLINE_JUDGE
    freopen("1.in", "r", stdin);
    #endif
    read(n), read(m), read(seed), read(vmax);

    // cin >> n >> m >> seed >> vmax;

    fos(i, 1, n) a[i] = (rnd() % vmax) + 1, s.insert((node){i, i, a[i]});

    s.insert((node)(n + 1, n + 1, 0));

    fos(i, 1, m) 
    {
        ll op, l, r, x, y;
        op = (rnd() % 4) + 1;
        l = (rnd() % n) + 1; r = (rnd() % n) + 1;
        if(l > r) swap(l, r);
        x = (op == 3) ? (rnd() % (r - l + 1)) + 1 : (rnd() % vmax) + 1;
        if(op == 4) y = (rnd() % vmax) + 1;

        // ww(op), wl;

        if(op == 1) add(l, r, x);
        else if(op == 2) assign(l, r, x);
        else if(op == 3) ww(rnk(l, r, x)), wl;
        else ww(calcP(l, r, x, y)), wl;
        // else if(op == 3) cout << rnk(l, r, x) << endl;
        // else cout << calcP(l, r, x, y) << endl;

        // cout << op << " " << l << " " << r << " " << x << " " << y << endl;
    }
    flush();
    return 0;
}

__EOF__

本文作者：carp
本文链接：https://www.cnblogs.com/carp-oier/p/17803305.html
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