【ODT】cf896C - Willem, Chtholly and Seniorious

仿佛没用过std::set

Seniorious has n pieces of talisman. Willem puts them in a line, the i-th of which is an integer ai.

In order to maintain it, Willem needs to perform m operations.

There are four types of operations:

• 1 l r x: For each i such that l ≤ i ≤ r, assign ai + x to ai.
• 2 l r x: For each i such that l ≤ i ≤ r, assign x to ai.
• 3 l r x: Print the x-th smallest number in the index range [l, r], i.e. the element at the x-th position if all the elements ai such that l ≤ i ≤ r are taken and sorted into an array of non-decreasing integers. It's guaranteed that 1 ≤ x ≤ r - l + 1.
• 4 l r x y: Print the sum of the x-th power of ai such that l ≤ i ≤ r, modulo y, i.e. .

Input

The only line contains four integers n, m, seed, vmax (1 ≤ n, m ≤ 105, 0 ≤ seed < 109 + 7, 1 ≤ vmax ≤ 109).

The initial values and operations are generated using following pseudo code:

def rnd():

    ret = seed
    seed = (seed * 7 + 13) mod 1000000007
    return ret

for i = 1 to n:

    a[i] = (rnd() mod vmax) + 1

for i = 1 to m:

    op = (rnd() mod 4) + 1
    l = (rnd() mod n) + 1
    r = (rnd() mod n) + 1

    if (l > r): 
         swap(l, r)

    if (op == 3):
        x = (rnd() mod (r - l + 1)) + 1
    else:
        x = (rnd() mod vmax) + 1

    if (op == 4):
        y = (rnd() mod vmax) + 1

Here op is the type of the operation mentioned in the legend.

Output

For each operation of types 3 or 4, output a line containing the answer.

---

题目分析

ODT的入门例题。

ODT实际上是将区间缩成点，用平衡树来维护区间的过程。这个东西的“复杂度”只能够依赖于数据随机。

具体可以参考：【毒瘤Warning】Chtholly Tree珂朵莉树详解

 #include<bits/stdc++.h>
 typedef long long ll;
 const int maxn = ;

 struct node
 {
     int l,r;
     mutable ll val;
     node(int a=, int b=, ll c=):l(a),r(b),val(c) {}
     bool operator < (node a) const
     {
         return l < a.l;
     }
 };
 typedef std::set<node>::iterator itr;
 int n,m,p,seed,vmax,a[maxn];
 std::set<node> s;

 int rand()
 {
     int ret = seed;
     seed = (seed*7ll+)%;
     return ret;
 }
 ll qmi(ll a, ll b)
 {
     ll ret = ;
     for (a%=p; b; b>>=,a=1ll*a*a%p)
         if (b&) ret = 1ll*ret*a%p;
     return ret;
 }
 itr split(int pos)
 {
     itr loc = s.lower_bound(node(pos));
     if (loc!=s.end()&&(*loc).l==pos) return loc;
     --loc;
     int l = (*loc).l, r = (*loc).r;
     ll val = (*loc).val;
     s.erase(loc);
     s.insert(node(l, pos-, val));
     return s.insert(node(pos, r, val)).first;
 }
 void merge(int l, int r, int val)
 {
     itr rpos = split(r+), lpos = split(l);
     s.erase(lpos, rpos);
     s.insert(node(l, r, val));
 }
 void modify(int l, int r, int val)
 {
     itr rpos = split(r+), lpos = split(l);
     for (; lpos!=rpos; ++lpos) (*lpos).val += val;
 }
 ll getRank(int l, int r, int k)
 {
     itr rpos = split(r+), lpos = split(l);
     std::vector<std::pair<ll, int> > mp;
     for (; lpos!=rpos; ++lpos)
         mp.push_back(std::make_pair((*lpos).val, (*lpos).r-(*lpos).l+));
     std::sort(mp.begin(), mp.end());
     for (int i=,mx=mp.size(); i<mx; i++)
     {
         k -= mp[i].second;
         if (k <= ) return mp[i].first;
     }
     return -;
 }
 int calc(int l, int r, int x)
 {
     int ret = ;
     itr rpos = split(r+), lpos = split(l);
     for (; lpos!=rpos; ++lpos)
         ret = (ret+1ll*qmi((*lpos).val, x)*((*lpos).r-(*lpos).l+)%p)%p;
     return ret;
 }
 int main()
 {
     scanf("%d%d%d%d",&n,&m,&seed,&vmax);
     for (int i=; i<=n; i++)
     {
         a[i] = rand()%vmax+;
         s.insert(node(i, i, a[i]));
     }
     s.insert(node(n+, n+, ));
     for (int i=,x; i<=m; i++)
     {
         int opt = rand()%+, l = rand()%n+, r = rand()%n+;
         if (l > r) std::swap(l, r);
         if (opt==) x = rand()%(r-l+)+;
         else x = (rand()%vmax)+;
         if (opt==) modify(l, r, x);
         else if (opt==) merge(l, r, x);
         else if (opt==) printf("%lld\n",getRank(l, r, x));
         else{
             p = rand()%vmax+;
             printf("%d\n",calc(l, r, x));
         }
     }
     return ;
 }

END