CF896C Willem, Chtholly and Seniorious

Posted 2022-07-01 ppxppx

洛咕

题面:请你写一种奇怪的数据结构，支持：

- \(1\) \(l\) \(r\) \(x\) ：将\([l,r]\) 区间所有数加上\(x\)

- \(2\) \(l\) \(r\) \(x\) ：将\([l,r]\) 区间所有数改成\(x\)

- \(3\) \(l\) \(r\) \(x\) ：输出将\([l,r]\) 区间从小到大排序后的第\(x\) 个数是的多少(即区间第\(x\) 小，数字大小相同算多次，保证 \(1\leq\) \(x\) \(\leq\) \(r-l+1\) )

- \(4\) \(l\) \(r\) \(x\) \(y\) ：输出\([l,r]\) 区间每个数字的\(x\) 次方的和模\(y\) 的值(即(\(\sum^r_i=la_i^x\) ) \(\mod y\) )

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<queue>
#include<map>
#include<set>
#define ll long long
#define IT set<node>::iterator
using namespace std;
inline ll read()
    ll x=0,o=1;char ch=getchar();
    while(ch!='-'&&(ch<'0'||ch>'9'))ch=getchar();
    if(ch=='-')o=-1,ch=getchar();
    while(ch>='0'&&ch<='9')x=x*10+ch-'0',ch=getchar();
    return x*o;

ll n,m,seed,vmax,a[100005];
struct node
    int l,r;mutable ll val;
    node(int L,int R=-1,ll V=0)l=L,r=R,val=V;
    bool operator <(const node &x)const
        return l<x.l;
    
;
set<node>s;
inline ll rnd()
    ll res=seed;
    seed=(seed*7+13)%1000000007;
    return res;

inline IT split(int pos)
    IT it=s.lower_bound(node(pos));
    if(it!=s.end()&&it->l==pos)return it;
    --it;
    int l=it->l,r=it->r;ll val=it->val;
    s.erase(it);
    s.insert(node(l,pos-1,val));
    return s.insert(node(pos,r,val)).first;

inline void add(int l,int r,ll val)
    IT itr=split(r+1),itl=split(l);
    for(;itl!=itr;++itl)itl->val+=val;
    return;

inline void assign(int l,int r,ll val)
    IT itr=split(r+1),itl=split(l);
    s.erase(itl,itr);
    s.insert(node(l,r,val));

inline ll kth(int l,int r,int k)
    vector<pair<ll,int> >q;
    IT itr=split(r+1),itl=split(l);
    for(;itl!=itr;++itl)
        q.push_back(pair<ll,int>(itl->val,itl->r-itl->l+1));
    sort(q.begin(),q.end());
    for(vector<pair<ll,int> >::iterator it=q.begin();it!=q.end();++it)
        k-=it->second;
        if(k<=0)return it->first;
    

inline ll ksm(ll a,ll b,ll c)
    ll cnt=1;a%=c;
    while(b)
        if(b&1)cnt=(1ll*cnt*a)%c;
        a=(1ll*a*a)%c;
        b>>=1;
    
    return cnt;

inline ll ask_sum(int l,int r,ll x,ll y)
    ll ans=0;
    IT itr=split(r+1),itl=split(l);
    for(;itl!=itr;++itl)
        ans+=(ksm(itl->val,x,y)*((itl->r-itl->l+1)%y))%y,ans%=y;
    return ans;

int main()
    n=read();m=read();seed=read();vmax=read();
    for(int i=1;i<=n;++i)
        a[i]=(rnd()%vmax)+1;
        s.insert(node(i,i,a[i]));
    
    for(int i=1;i<=m;++i)
        int opt=(rnd()%4)+1;
        int l=(rnd()%n)+1,r=(rnd()%n)+1;
        ll x,y;
        if(l>r)swap(l,r);
        if(opt==3)x=(rnd()%(r-l+1))+1;
        else x=(rnd()%vmax)+1;
        if(opt==4)y=(rnd()%vmax)+1;
        if(opt==1)add(l,r,x);
        else if(opt==2)assign(l,r,x);
        else if(opt==3)printf("%lld\n",kth(l,r,x));
        else if(opt==4)printf("%lld\n",ask_sum(l,r,x,y));
    
    return 0;