题意:带修改的区间第k大
题解:树状数组套主席树,开n个主席树,用树状数组维护,先离散化,然后动态开点按权值插入到树状数组访问到的节点,然后修改也是同样的修改,查询需要在主席树上二分,同时维护树状数组所访问的那些节点,在主席树上跑即可
//#pragma GCC optimize(2)
//#pragma GCC optimize(3)
//#pragma GCC optimize(4)
//#pragma GCC optimize("unroll-loops")
//#pragma comment(linker, "/stack:200000000")
//#pragma GCC optimize("Ofast,no-stack-protector")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#include<bits/stdc++.h>
#define fi first
#define se second
#define db double
#define mp make_pair
#define pb push_back
#define pi acos(-1.0)
#define ll long long
#define vi vector<int>
#define mod 998244353
#define ld long double
//#define C 0.5772156649
//#define ls l,m,rt<<1
//#define rs m+1,r,rt<<1|1
#define pll pair<ll,ll>
#define pil pair<int,ll>
#define pli pair<ll,int>
#define pii pair<int,int>
#define ull unsigned long long
//#define base 1000000000000000000
#define fin freopen("a.txt","r",stdin)
#define fout freopen("a.txt","w",stdout)
#define fio ios::sync_with_stdio(false);cin.tie(0)
inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
inline void sub(ll &a,ll b){a-=b;if(a<0)a+=mod;}
inline void add(ll &a,ll b){a+=b;if(a>=mod)a-=mod;}
template<typename T>inline T const& MAX(T const &a,T const &b){return a>b?a:b;}
template<typename T>inline T const& MIN(T const &a,T const &b){return a<b?a:b;}
inline ll qp(ll a,ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;a=a*a%mod,b>>=1;}return ans;}
inline ll qp(ll a,ll b,ll c){ll ans=1;while(b){if(b&1)ans=ans*a%c;a=a*a%c,b>>=1;}return ans;}
using namespace std;
const ull ba=233;
const db eps=1e-8;
const ll INF=0x3f3f3f3f3f3f3f3f;
const int N=200000+10,maxn=200000+10,inf=0x3f3f3f3f;
int n,m,cnt;
int qq[N],top=0;
struct bit_seg{
int root[N],ls[N*50],rs[N*50],sum[N*50],res,te[2][20],num[2];
bit_seg(){res=0;}
void update(int &o,int pos,int v,int l,int r)
{
if(!o)
{
if(top)o=qq[top--];
else o=++res;
}
sum[o]+=v;
if(l==r)return ;
int m=(l+r)>>1;
if(pos<=m)update(ls[o],pos,v,l,m);
else update(rs[o],pos,v,m+1,r);
if(!sum[o])qq[++top]=o,o=0;
}
int query(int k,int l,int r)
{
if(l==r)return l;
int m=(l+r)>>1,ans=0;
for(int i=1;i<=num[1];i++)ans+=sum[ls[te[1][i]]];
for(int i=1;i<=num[0];i++)ans-=sum[ls[te[0][i]]];
// printf("%d %d %d\n",l,r,ans);
if(ans>=k)
{
for(int i=1;i<=num[1];i++)te[1][i]=ls[te[1][i]];
for(int i=1;i<=num[0];i++)te[0][i]=ls[te[0][i]];
return query(k,l,m);
}
else
{
for(int i=1;i<=num[1];i++)te[1][i]=rs[te[1][i]];
for(int i=1;i<=num[0];i++)te[0][i]=rs[te[0][i]];
return query(k-ans,m+1,r);
}
}
int bitquery(int L,int R,int k)
{
num[0]=num[1]=0;
for(int i=R;i;i-=i&(-i))te[1][++num[1]]=root[i];
for(int i=L-1;i;i-=i&(-i))te[0][++num[0]]=root[i];
return query(k,1,cnt);
}
void bitupdate(int i,int pos,int v)
{
for(;i<=cnt;i+=i&(-i))update(root[i],pos,v,1,cnt);
}
}s;
struct node{int op,x,y,z;}q[N];
int Hash[N],a[N];
char o[5];
int main()
{
scanf("%d%d",&n,&m);
for(int i=1;i<=n;i++)
{
scanf("%d",&a[i]);
Hash[cnt++]=a[i];
}
for(int i=1;i<=m;i++)
{
scanf("%s%d%d",o,&q[i].x,&q[i].y);
q[i].op=(o[0]=='C');
if(q[i].op==0)scanf("%d",&q[i].z);
if(q[i].op==1)Hash[cnt++]=q[i].y;
}
sort(Hash,Hash+cnt);cnt=unique(Hash,Hash+cnt)-Hash;
for(int i=1;i<=n;i++)
{
a[i]=lower_bound(Hash,Hash+cnt,a[i])-Hash+1;
s.bitupdate(i,a[i],1);
}
for(int i=1;i<=m;i++)
{
if(q[i].op==0)printf("%d\n",Hash[s.bitquery(q[i].x,q[i].y,q[i].z)-1]);
else
{
q[i].y=lower_bound(Hash,Hash+cnt,q[i].y)-Hash+1;
s.bitupdate(q[i].x,a[q[i].x],-1);a[q[i].x]=q[i].y;s.bitupdate(q[i].x,a[q[i].x],1);
}
}
return 0;
}
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