珂朵莉树(ODT老司机树)

题目链接https://vjudge.net/problem/CodeForces-896C

关于珂朵莉树的讲解可以参考https://blog.csdn.net/niiick/article/details/83062256

按照题意,完成加,赋值,求第K小,区间x次方和四种操作

直接上代码

#include<bits/stdc++.h>
#define ll long long
#define IT set<node>::iterator
#define I_like_Chtholly_forever 1000000007
using namespace std;
struct node
{
    ll l,r;
    mutable ll val;
    node(ll L,ll R=-1,ll V=0): l(L), r(R), val(V) {}
    bool operator < (const node& tt)const
    {
        return l<tt.l;   //以区间左端点排序
    }
};
set<node> st;

IT split(ll pos)
{
    IT it=st.lower_bound(node(pos));//二分找到第一个左端点不小于pos的区间
    if(it!=st.end()&&it->l==pos) return it;//pos本身就是某个区间的左端点，不用分裂
    --it;//否则上一个区间才是包含pos的区间
    ll l=it->l,r=it->r,val=it->val;
    st.erase(it);//删除原结点
    st.insert(node(l,pos-1,val));
    return st.insert(node(pos,r,val)).first;//这里.first返回的是迭代器
}
void assign_val(ll l,ll r,ll val)
{
    IT itr=split(r+1),itl=split(l);
    st.erase(itl,itr);
    st.insert(node(l,r,val));
}
#define pir pair<ll,ll>//前一个记录值，后一个记录出现次数
ll kth(ll l,ll r,ll k)
{
    vector<pir> vec;
    IT itr=split(r+1),itl=split(l);

    for(; itl!=itr; ++itl)
        vec.push_back( pir(itl->val,itl->r-itl->l+1) );
    sort(vec.begin(),vec.end());//全部存下来排序就好

    for(vector<pir>::iterator it=vec.begin(); it!=vec.end(); ++it)
    {
        k-=it->second;
        if(k<=0) return it->first;
    }
    return -1;
}
void add(ll l,ll r,ll val)
{
    IT itr=split(r+1),itl=split(l);
    for(; itl!=itr; ++itl)
    {
        itl->val+=val;
    }
}
ll quick_pow(ll a, ll b, ll mod)
{
    ll res = 1;
    ll ans = a % mod;
    while (b)
    {
        if (b&1)
            res = res * ans % mod;
        ans = ans * ans % mod;
        b>>=1;
    }
    return res;
}
ll sum(ll l,ll r,ll en,ll mod)
{
    IT itr=split(r+1),itl=split(l);
    ll res=0;
    for(; itl!=itr; ++itl)
    {
        res = (res + (itl->r-itl->l +1) * quick_pow(itl->val,en,mod))%mod;
    }
    return res;
}
ll n,m,vmax,seed;
ll rnd()
{
    ll ret = seed;
    seed = (seed * 7 + 13) %  I_like_Chtholly_forever;
    return ret;
}
ll a[100005];
int main()
{
    cin>>n>>m>>seed>>vmax;
    for (int i = 1; i<= n; i++)
    {
        a[i] = (rnd() % vmax) + 1;
        st.insert(node(i,i,a[i]));
    }
    ll op,l,r,x,y;
    for (int i=1; i<=m; i++ )
    {
        op = (rnd() % 4) + 1;
        l = (rnd() % n) + 1;
        r = (rnd() % n) + 1;
        if (l > r)
            swap(l, r);
        if (op == 3)
            x = (rnd() % (r - l + 1)) + 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_val(l, r, x);
        else if (op == 3)
            cout<<kth(l,r,x)<<endl;
        else
            cout<<sum(l,r,x,y)<<endl;
    }
    return 0;
}