K-th Number （主席树第K大模板）

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 10 ⁹ 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

Hint

This problem has huge input,so please use c-style input(scanf,printf),or you may got time limit exceed.

 

#include <cstdio>
#include <vector>
#include <algorithm>
using namespace std;

const int N=1e5+5;
int a[N];///输入的数组
int root[N];///root[i]表示数组前i个元素组成的线段树的根结点
struct node{
    int l,r;///线段树左右子节点
    int sum;///统计子树上的元素个数
}T[N*40];///开20倍或者40倍
int num=0;///节点编号
vector<int> vec;

int getid(int k){
    return lower_bound(vec.begin(),vec.end(),k)-vec.begin()+1;
}

void update(int l,int r,int &now,int bef,int k){///更新操作
    T[++num]=T[bef];///复制上一个线段树
    now=num;///更新当前线段树的根结点
    T[num].sum++;
    if(l==r) return;
    int mid=(l+r)>>1;
    if(k<=mid){
        update(l,mid,T[now].l,T[bef].l,k);
    }
    else{
        update(mid+1,r,T[now].r,T[bef].r,k);
    }
}

int query(int l,int r,int x,int y,int k){///查询区间[x,y]第k大的数
    if(l==r) return l;
    int mid=(l+r)>>1;
    int cnt=T[T[y].l].sum-T[T[x].l].sum;///第y颗树比第x颗树在左子树上多的结点数
    if(cnt>=k){///在左子树上
        return query(l,mid,T[x].l,T[y].l,k);
    }
    else{
        return query(mid+1,r,T[x].r,T[y].r,k-cnt);
    }
}

int main(){
    int n,m;
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++){
        scanf("%d",&a[i]);
        vec.push_back(a[i]);
    }
    sort(vec.begin(),vec.end());
    ///unique返回去重后容器中不重复序列的最后一个元素的下一个元素
    ///erase删除[左,右)的元素
    vec.erase(unique(vec.begin(),vec.end()),vec.end());
    for(int i=1;i<=n;i++){
        update(1,n,root[i],root[i-1],getid(a[i]));
    }
    while(m--){
        int x,y,k;
        scanf("%d%d%d",&x,&y,&k);
        printf("%d
",vec[query(1,n,root[x-1],root[y],k)-1]);
    }
}