文章目錄

• 題意
• 解法

題意

$區間加,區間推平,區間第k大,區間乘方和取模.$

解法

區間推平操作+資料隨機.用珂朵莉樹維護,均攤的時間複雜度是$logn$. 謝謝大家.

#include<bits/stdc++.h> //Ithea Myse Valgulious
namespace chtholly{
typedef long long ll;
#define re0 register int
#define
 
 rel register ll
#define rec register char
#define gc getchar
//#define gc() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<23,stdin),p1==p2)?-1:*p1++)
#define pc putchar
#define p32 pc(' ')
#define pl puts("")
/*By Citrus*/
char buf[1<<23],*p1=buf,*p2=buf;
inline int read(){
  int x=0,f=1;char
 
 c=gc();
  for (;!isdigit(c);c=gc()) f^=c=='-';
  for (;isdigit(c);c=gc()) x=(x<<3)+(x<<1)+(c^'0');
  return f?x:-x;
  }
template <typename mitsuha>
inline bool read(mitsuha &x){
  x=0;int f=1;char c=gc();
  for (;!isdigit(c)&&~c;c=gc()) f^=c=='-';
  if (!~c) return 0;
  for
 
 (;isdigit(c);c=gc()) x=(x<<3)+(x<<1)+(c^'0');
  return x=f?x:-x,1;
  }
template <typename mitsuha>
inline int write(mitsuha x){
  if (!x) return 0&pc(48);
  if (x<0) pc('-'),x=-x;
  int bit[20],i,p=0;
  for (;x;x/=10) bit[++p]=x%10;
  for (i=p;i;--i) pc(bit[i]+48);
  return 0;
  }
inline char fuhao(){
  char c=gc();
  for (;isspace(c);c=gc());
  return c;
  }
}using namespace chtholly;
using namespace std;
int n,q,vmax,seed;

int rnd() {
int ret=seed;
seed=(seed*7ll+13)%1000000007;
return ret;
}

ll kasumi(ll a,ll b,ll mod){
ll s=1;
for (a%=mod;b;b>>=1,a=a*a%mod) if (b&1) s=s*a%mod;
return s;
}/*快速冪寫萎了......a一定要先模mod!*/

struct chtholly_tree{
struct node{
  int l,r; mutable ll v;
  node(int nl,int nr=-1,ll val=0):l(nl),r(nr),v(val) {}
  bool operator <(const node &b) const{
    return l<b.l;
    }
  };
typedef set<node>::iterator it;
set<node> s;
it split(int pos) {
  it p=s.lower_bound(pos);
  if (p!=s.end()&&p->l==pos) return p; --p;
  int l=p->l,r=p->r; ll v=p->v;
  s.erase(p);
  s.insert(node(l,pos-1,v));
  return s.insert(node(pos,r,v)).first;
  }
void assign(int l,int r,int v) {
  it itr=split(r+1),itl=split(l);
  s.erase(itl,itr);
  s.insert(node(l,r,v));
  }
void add(int l,int r,int v){
  it p,itr=split(r+1),itl=split(l);
  for (p=itl;p!=itr;++p) p->v+=v;
  }
ll kth(int l,int r,int k,int i=0) {
  it p,itr=split(r+1),itl=split(l);
  vector<node> llx;
  for (p=itl;p!=itr;++p) llx.push_back(*p);
  sort(llx.begin(),llx.end(),[](node a,node b){return a.v<b.v;});
  /*所以我過載cmp為什麼不給過啊!一定要我用lambda表示式!*/
  for (i=0;i<llx.size()&&k>0;++i)
    k-=llx[i].r-llx[i].l+1;
  return llx[i-1].v;
  }
ll query(int l,int r,int x,int y){
  it p,itr=split(r+1),itl=split(l);
  ll zxy=0;
  for (p=itl;p!=itr;++p) 
    zxy=(zxy+1ll*(p->r-p->l+1)*kasumi(p->v,x,y))%y;
  return zxy;
  }
void init(int n) {
  for (int i=1;i<=n;++i)
    s.insert(node(i,i,rnd()%vmax+1));
  }
}my_;

int main() {
read(n),read(q);
read(seed),read(vmax);
my_.init(n);
int cnt=0;
for (;q--;) {
  int op=rnd()%4+1,l,r,x,y;
  l=rnd()%n+1,r=rnd()%n+1;
  if (l>r) swap(l,r);
  switch(op) {
    case 3: 
      x=rnd()%(r-l+1)+1;
      write(my_.kth(l,r,x)),pl;
      break;
    case 4:
      x=rnd()%vmax+1;
      y=rnd()%vmax+1;
      write(my_.query(l,r,x,y)),pl;
      break;
    case 1:
      x=rnd()%vmax+1;
      my_.add(l,r,x);
      break;
    case 2:
      x=rnd()%vmax+1;
      my_.assign(l,r,x);
      break;
    }
  }
}