http://codeforces.com/problemset/problem/896/C
题意：
对于一个随机序列，执行以下操作：
区间赋值
区间加
区间求第k小
区间求k次幂的和
对于随机序列，可以使用Old Driver Tree
就是将序列中，连续的相同值域合并为一段
然后暴力操作
#include<set> #include<vector> #include<cstdio> #include<iostream> #include<algorithm> using namespace std; typedef long long LL; #define N 100001 int n,m,seed,vmax,ret; int a[N]; struct node { int l,r; mutable LL val; bool operator < (node p) const { return l<p.l; } node(int l=0,int r=0,LL val=0):l(l),r(r),val(val) { } }; set<node>s; typedef set<node> :: iterator seti; vector<pair<LL,int> >par; void read(int &x) { x=0; char c=getchar(); while(!isdigit(c)) c=getchar(); while(isdigit(c)) { x=x*10+c-'0'; c=getchar(); } } int rnd() { ret=seed; seed=((LL)seed*7+13)%1000000007; return ret; } void split(int pos) { seti it=s.lower_bound(node(pos,-1,-1)); if(it==s.end() || it->l>pos) { --it; int l=it->l,r=it->r; LL val=it->val; s.erase(it); s.insert(node(l,pos-1,val)); s.insert(node(pos,r,val)); } } LL quickpow(LL a,LL x,LL mod) { LL res=1; for(;x;x>>=1,a=a*a%mod) if(x&1) res=res*a%mod; return res; } int main() { read(n); read(m); read(seed); read(vmax); for(int i=1;i<=n;++i) a[i]=rnd()%vmax+1; int r; for(int i=1;i<=n;) { r=i+1; while(a[r]==a[i]) r++; s.insert(node(i,r-1,(LL)a[i])); i=r; } int op,l,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; split(l); if(r<n) split(r+1); seti itl=s.lower_bound(node(l,-1,-1)); seti itr=s.upper_bound(node(r,-1,-1)); if(op==1) { for(seti it=itl;it!=itr;++it) it->val+=x; } else if(op==2) { s.erase(itl,itr); s.insert(node(l,r,x)); } else if(op==3) { par.clear(); for(seti it=itl;it!=itr;++it) par.push_back(make_pair(it->val,it->r-it->l+1)); sort(par.begin(),par.end()); for(int i=0;i<par.size();++i) { x-=par[i].second; if(x<=0) { cout<<par[i].first<<' '; break; } } } else { LL ans=0; for(seti it=itl;it!=itr;++it) { LL val=quickpow(it->val%y,x,y); val=val*(it->r-it->l+1)%y; ans=(ans+val)%y; } cout<<ans<<' '; } } return 0; }
— Willem...
— What's the matter?
— It seems that there's something wrong with Seniorious...
— I'll have a look...
Seniorious is made by linking special talismans in particular order.
After over 500 years, the carillon is now in bad condition, so Willem decides to examine it thoroughly.
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 thatl ≤ 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.
.
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.
For each operation of types 3 or 4, output a line containing the answer.
10 10 7 9
2
1
0
3
10 10 9 9
1
1
3
3
In the first example, the initial array is {8, 9, 7, 2, 3, 1, 5, 6, 4, 8}.
The operations are:
- 2 6 7 9
- 1 3 10 8
- 4 4 6 2 4
- 1 4 5 8
- 2 1 7 1
- 4 7 9 4 4
- 1 2 7 9
- 4 5 8 1 1
- 2 5 7 5
- 4 3 10 8 5