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.

線段樹的每個節點維護其所表示的區間的 有序序列。
程式碼

#include<cstring>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<vector>
using namespace std;
#define LL long long

const int N = 100005 ;

vector<int>T[N<<2];

void Build(int o,int le,int ri){
    if(le==ri){
        int
 
 val;scanf("%d",&val);
        T[o].push_back(val);
        return ;
    }
    int mid=(le+ri)>>1;
    Build(o<<1,le,mid);
    Build(o<<1|1,mid+1,ri);
    T[o].resize(ri-le+1);//  注意
    merge(T[o<<1].begin(),T[o<<1].end(),T[o<<1|1].begin(),T[o<<1|1].end(),T[o].begin());
}

int
 
 Query(int o,int L,int R,int le,int ri,int x){
    if(L==le&&R==ri) return upper_bound(T[o].begin(),T[o].end(),x)-T[o].begin();
    int mid=(L+R)>>1;
    if(ri<=mid) return Query(o<<1,L,mid,le,ri,x);
    else if(le>mid) return Query(o<<1|1,mid+1,R,le,ri,x);
    else return Query(o<<1,L,mid,le,mid,x)+Query(o<<1|1,mid+1,R,mid+1,ri,x);
}

int main(){
    int n,m; scanf("%d%d",&n,&m);
    Build(1,1,n);
    while(m--){
        int x,y,z;scanf("%d%d%d",&x,&y,&z);
        int le,ri;
        le=0; ri=n-1; int ans;
        while(le<=ri){
            int mid=(le+ri)>>1;
            if(Query(1,1,n,x,y,T[1][mid]) >= z) {
                ans=mid; ri= mid-1;
            }else le=mid+1;
        }
        printf("%d\n",T[1][ans]);
    }
return 0;
}

/*


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

*/