整体二分（模板二）动态区间第K大

这才是更一般的二分写法--HDU5412

 #define IOS ios_base::sync_with_stdio(0); cin.tie(0);
 #include <cstdio>//sprintf islower isupper
 #include <cstdlib>//malloc  exit strcat itoa system("cls")
 #include <iostream>//pair
 #include <fstream>//freopen("C:\\Users\\13606\\Desktop\\Input.txt","r",stdin);
 #include <bitset>
 //#include <map>
 //#include<unordered_map>
 #include <vector>
 #include <stack>
 #include <set>
 #include <string.h>//strstr substr strcat
 #include <string>
 #include <time.h>// srand(((unsigned)time(NULL))); Seed n=rand()%10 - 0~9;
 #include <cmath>
 #include <deque>
 #include <queue>//priority_queue<int, vector<int>, greater<int> > q;//less
 #include <vector>//emplace_back
 //#include <math.h>
 #include <cassert>
 #include <iomanip>
 //#include <windows.h>//reverse(a,a+len);// ~ ! ~ ! floor
 #include <algorithm>//sort + unique : sz=unique(b+1,b+n+1)-(b+1);+nth_element(first, nth, last, compare)
 using namespace std;//next_permutation(a+1,a+1+n);//prev_permutation
 //******************
 clock_t __START,__END;
 double __TOTALTIME;
 void _MS(){__START=clock();}
 void _ME(){__END=clock();__TOTALTIME=(double)(__END-__START)/CLOCKS_PER_SEC;cout<<"Time: "<<__TOTALTIME<<" s"<<endl;}
 //***********************
 #define rint register int
 #define fo(a,b,c) for(rint a=b;a<=c;++a)
 #define fr(a,b,c) for(rint a=b;a>=c;--a)
 #define mem(a,b) memset(a,b,sizeof(a))
 #define pr printf
 #define sc scanf
 #define ls rt<<1
 #define rs rt<<1|1
 typedef pair<int,int> PII;
 typedef vector<int> VI;
 typedef long long ll;
 const double E=2.718281828;
 const double PI=acos(-1.0);//
 const ll INF=(1LL<<);
 const int inf=(<<);
 const double ESP=1e-;
 const int mod=(int)1e9+;
 const int N=(int)4e6+;

 int ans[N],a[N];
 int c[N],maxn;
 int lowbit(int n){return n&(-n);}
 void add(int i,int t)
 {
     while(i<=maxn)
     {
         c[i]+=t;
         i+=lowbit(i);
     }
 }
 int Sum(int i)
 {
     int sum=;
     while(i)
     {
         sum+=c[i];
         i-=lowbit(i);
     }
     return sum;
 }
 struct node
 {
     int l,r,k,id,currk,op;
 }q[N],temp[N];
 void Init()
 {
     for(int i=;i<=N-;++i)
         c[i]=;
 }

 void solve(int l,int r,int ql,int qr)
 {
     if(ql>qr)return;
     int L=ql,R=qr,mid=l+r>>;

     if(l==r)
     {
         for(int i=ql;i<=qr;++i)
             if(q[i].op==)
                 ans[q[i].id]=l;//答案
         return;
     }
     for(int i=ql;i<=qr;++i)
     {
         if(q[i].op!=)temp[q[i].r<=mid?L++:R--]=q[i];//分位置

         if(q[i].op==&&q[i].r<=mid)add(q[i].l,);//q.l才是要加的位置,q.r是值
         else if(q[i].op==&&q[i].r<=mid)add(q[i].l,-);
         else if(q[i].op==)
         {
             int sum=Sum(q[i].r)-Sum(q[i].l-);//二分是保证按时间顺序来的,直接计算就行
             if(q[i].currk+sum>=q[i].k)temp[L++]=q[i];//分位置
             else q[i].currk+=sum,temp[R--]=q[i];
         }
     }
     for(int i=ql;i<=qr;++i)//清空-树状数组
     {
         if(q[i].op==&&q[i].r<=mid)add(q[i].l,-);
         else if(q[i].op==&&q[i].r<=mid)add(q[i].l,);
     }
     //还原
     for(int i=ql;i<=L-;++i)
         q[i]=temp[i];
     for(int i=qr,j=L;j<=qr;--i,++j)
         q[j]=temp[i];

     //递归下去
     if(L>ql)solve(l,mid,ql,L-);
     if(R<qr)solve(mid+,r,R+,qr);
 }

 int main()
 {
     int n,m;
     while(~sc("%d",&n))
     {
     //    Init();
         maxn=n;
         for(int i=;i<=n;++i)
             sc("%d",&q[i].r),q[i].l=i,a[i]=q[i].r,q[i].op=;
         sc("%d",&m);
         int ask=;
         for(int i=;i<=m;++i)
         {
             int J;
             sc("%d",&J);
             if(J==)
             {
                 int pos,x;
                 sc("%d%d",&pos,&x);
                 ++n;
                 q[n].l=pos,q[n].r=a[pos],q[n].op=;//del
                 ++n,a[pos]=x;
                 q[n].l=pos,q[n].r=a[pos],q[n].op=;//add
             }
             else
                 ++n,sc("%d%d%d",&q[n].l,&q[n].r,&q[n].k),q[n].currk=,q[n].id=++ask,q[n].op=;//query
         }
         solve(,inf,,n);
         for(int i=;i<=ask;++i)
             pr("%d\n",ans[i]);
     }
     return ;
 }

 /**************************************************************************************/