2104 -- K-th Number

Description

You are working for Macrohard company in data structures department. After failing your previous task about key insertion you were asked to write a new data structure that would be able to return quickly k-th order statistics in the array segment. 
That is, given an array a[1...n] of different integer numbers, your program must answer a series of questions Q(i, j, k) in the form: "What would be the k-th number in a[i...j] segment, if this segment was sorted?" 
For example, consider the array a = (1, 5, 2, 6, 3, 7, 4). Let the question be Q(2, 5, 3). The segment a[2...5] is (5, 2, 6, 3). If we sort this segment, we get (2, 3, 5, 6), the third number is 5, and therefore the answer to the question is 5.

Input

The first line of the input file contains n --- the size of the array, and m --- the number of questions to answer (1 <= n <= 100 000, 1 <= m <= 5 000). 
The second line contains n different integer numbers not exceeding 10⁹ by their absolute values --- the array for which the answers should be given. 
The following m lines contain question descriptions, each description consists of three numbers: i, j, and k (1 <= i <= j <= n, 1 <= k <= j - i + 1) and represents the question Q(i, j, k).

Output

For each question output the answer to it --- the k-th number in sorted a[i...j] segment.

Sample Input

7 3
1 5 2 6 3 7 4
2 5 3
4 4 1
1 7 3

Sample Output

5
6
3

---

趁热赶紧来道主席树裸题，感觉有些明白了
主席树，可持久化线段树，核心思想就是维护线段树的历史状态
在这道题中，我们以数字大小为位置，位置上存的是出没出现过（1/0），统计加和
我们依次将数字加入线段树，这样就会形成n个历史版本
再加上我们维护的数据具有可减性，所以对于(l,r)，每个节点位置上只要用r这颗线段树减去l线段树对应的值，就是相当于只维护了(l,r)这个区间内的元素的线段树，自然就可以在上面愉快的跳跃了~
这次的主席树大部分是自己写的了2333，看看吧~

 #include<cstdio>
 #include<algorithm>
 #define N 100005
 using namespace std;
 int n,m,siz,tot;
 int a[N],b[N],rt[N];
 struct node{
     int l,r,sum;
 }t[*N];
 int insert(int k,int x,int l,int r){
     t[++tot]=t[k];k=tot;
     if(l==r){
         t[k].l=t[k].r=;
         t[k].sum++;
         return tot;
     }
     int mid=(l+r)>>;
     if(x<=mid) t[k].l=insert(t[k].l,x,l,mid);
     else t[k].r=insert(t[k].r,x,mid+,r);
     t[k].sum=t[t[k].l].sum+t[t[k].r].sum;
     return k;
 }
 int query(int lk,int rk,int x,int l,int r){
     if(l==r)return l;
     int mid=(l+r)>>;
     if(t[t[rk].l].sum-t[t[lk].l].sum>=x)return query(t[lk].l,t[rk].l,x,l,mid);
     else return query(t[lk].r,t[rk].r,x-(t[t[rk].l].sum-t[t[lk].l].sum),mid+,r);
 }
 int main(){
     scanf("%d%d",&n,&m);
     for(int i=;i<=n;i++)scanf("%d",&a[i]),b[i]=a[i];
     sort(b+,b++n);
     siz=unique(b+,b++n)-b-;
     for(int i=;i<=n;i++)a[i]=lower_bound(b+,b++n,a[i])-b;
     tot=;
     for(int i=;i<=n;i++)
         rt[i]=insert(rt[i-],a[i],,n);
     for(int i=,l,r,k;i<=m;i++){
         scanf("%d%d%d",&l,&r,&k);
         printf("%d\n",b[query(rt[l-],rt[r],k,,n)]);
     }
     return ;
 }