ODT入门_odt是什么

struct node
{ 
   
    int l,r; //区间的左端点、右端点
    mutable LL v; //数值
    node(int L, int R=-1, LL V=0):l(L), r(R), v(V) { 
   }  //初始化
    bool operator<(const node& o) const     //重载运算符
    { 
   
        return l < o.l;//从小到大
    }
};

#define IT set<node>::iterator //position在【l,r】的范围里，把它分成【l,pos-1】和【pos,r】两个区间
IT split(int pos)
{ 

IT it = s.lower_bound(node(pos));   //找到首个不小于pos的set
if (it != s.end() && it->l == pos)   //无需，直接返回
return it;
--it;  //否则一定在前一个区间中
int L = it->l, R = it->r;  //【l,r】就是要分裂的区间
LL V = it->v;  //取出值
s.erase(it);   //删除原集合
s.insert(node(L, pos-1, V));  //构建前半段的新结合
return s.insert(node(pos, R, V)).first;  //构建后半段的新集合并且返回地址
}

void add(int l, int r, LL val)
{ 

IT itr = split(r+1),itl = split(l); //左闭右开的区间
for (; itl != itr; ++itl) 
itl->v += val;
}//对所有三元组进行维护

void assign_val(int l, int r, LL val)
{ 

IT itr = split(r+1),itl = split(l);  //求出要被摊平区间的收尾地址
s.erase(itl, itr);  //删除原集合
s.insert(node(l, r, val));  //添加新集合
}

LL rank(int l, int r, int k)
{ 

vector<pair<LL, int> > vp;
IT itr = split(r+1),itl = split(l);
vp.clear();
for (; itl != itr; ++itl)
vp.push_back(pair<LL,int>(itl->v, itl->r - itl->l + 1));
sort(vp.begin(), vp.end());
for (vector<pair<LL,int> >::iterator it=vp.begin();it!=vp.end();++it)
{ 

k -= it->second;
if (k <= 0) 
return it->first;
}
}

LL pown(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(int l, int r, int ex, int mod)
{ 

IT itr = split(r+1),itl = split(l);
LL res = 0;
for (; itl != itr; ++itl)
res = (res + (LL)(itl->r - itl->l + 1) * pown(itl->v, LL(ex), LL(mod))) % mod;//快速幂
return res;
}

#include<iostream>
#include<cstdio>
#include<cstring>
#include<map>
#include<set>
#include<algorithm>
#include<cmath>
#include<string.h>
#include<queue>
#include<vector>
using namespace std;
typedef long long LL;
#define IT set<node>::iterator 
const int MAXN=1e5+7;
const int MOD7=1e9+7;
struct node{ 

int l,r;
mutable LL v;
node(int L,int R=-1,LL V=0):l(L),r(R),v(V){ 
}
bool operator<(const node &o)const{ 

return l<o.l;
}
};
LL qpow(LL a,LL b,LL mod){ 
//快速幂 
LL ans=1;
LL x=a%mod;
while(b){ 

if(b&1)ans=ans*x%mod;
x=x*x%mod;
b>>=1;
}
return ans;
}
set<node>s;
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 V=it->v;
s.erase(it);
s.insert(node(L,pos-1,V));
return s.insert(node(pos,R,V)).first;
}
void add(int l,int r,LL val){ 
//给l到r所有数加上val 
IT itr=split(r+1),itl=split(l);
for(;itl!=itr;++itl)
itl->v+=val;
}
void assign_val(int l,int r,LL val){ 
//把l到r所有数改为val 
IT itr=split(r+1),itl=split(l);
s.erase(itl,itr);
s.insert(node(l,r,val));
}
LL ranks(int l,int r,int k){ 
//询问l到r区间第k小 
vector<pair<LL,int> >vp;
IT itr=split(r+1),itl=split(l);
vp.clear();
for(;itl!=itr;++itl)
vp.push_back(pair<LL,int>(itl->v,itl->r - itl->l +1));
sort(vp.begin(),vp.end());
for(vector<pair<LL,int> >::iterator it=vp.begin();it!=vp.end();++it){ 

k-=it->second;
if(k<=0)return it->first;
}
}
LL sum(int l,int r,int ex,int mod){ 
//询问l到r每个数字的x次方模y的和 
IT itr=split(r+1),itl=split(l);
LL res=0;
for(;itl!=itr;++itl)
res=(res+(LL)(itl->r - itl->l+1)*qpow(itl->v,LL(ex),LL(mod))) %mod;
return res;
}
int n,m;
LL seed,vmax;
LL rd(){ 

LL ret=seed;
seed=(seed*7+13)%MOD7;
return ret;
}
LL a[MAXN];
int main(){ 

cin>>n>>m>>seed>>vmax;
for(int i=1;i<=n;i++){ 

a[i]=(rd()%vmax)+1;
s.insert(node(i,i,a[i]));
}
s.insert(node(n+1,n+1,0));
int lines=0;
for(int i=1;i<=m;i++){ 

//----------------------------- 
int op=int(rd()%4)+1;
int l=int(rd()%n)+1;
int r=int(rd()%n)+1;
if(l>r)swap(l,r);
int x,y;
if(op==3)x=int(rd()%(r-l+1))+1;
else x=int(rd()%vmax)+1;
if(op==4)y=int(rd()%vmax)+1;
//------------------------------以上为题目要求生成数据 
if(op==1)add(l,r,LL(x));
else if(op==2)assign_val(l,r,LL(x));
else if(op==3)cout<<ranks(l,r,x)<<endl;
else cout<<sum(l,r,x,y)<<endl;
}
return 0;
}