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

求區間中第k大的數。

#include <iostream>
#include <algorithm>
#define MAXN 100010
using namespace std;
int T, n, m, tot, a[MAXN], b[MAXN], cnt;
//T為測試樣例數，n為元陣列個數，m為查詢個數，tot為建樹的新節點
//a為原數列，b用來離散化的數列，其unique後元素的個數為cnt
struct N {
	int ls, rs, w;
	//ls左結點下標，rs右結點下標，w區間元素個數
}tree[30*MAXN];
int root[MAXN];//記錄每棵樹的根節點下標
int Build_tree(int l, int r) {//建樹
	int newnode = tot++;
	tree[newnode].w = 0;
	if(l != r) {
		int mid = (l + r) >> 1;
		tree[newnode].ls = Build_tree(l, mid);
		tree[newnode].rs = Build_tree(mid + 1, r);		
	}
	return newnode;
}
int update(int rt, int pos, int  val) {
	int newnode = tot++;
	int temp = newnode;
	tree[newnode].w = tree[rt].w + val;
	int l = 1, r = cnt;
	while(l < r) {
		int mid = (l + r) >> 1;
		if(pos <= mid) {
			tree[newnode].ls = tot++;
			tree[newnode].rs = tree[rt].rs;
			newnode = tree[newnode].ls;
			rt = tree[rt].ls;
			r = mid;
		} else {
			tree[newnode].ls = tree[rt].ls;
			tree[newnode].rs = tot++;
			newnode = tree[newnode].rs;
			rt = tree[rt].rs;
			l = mid + 1;
		}
		tree[newnode].w = tree[rt].w + val;
	}
	return temp;
}
int query(int rt1, int rt2, int k) {
	int l = 1, r = cnt;
	while(l < r) {
		int mid = (l + r) >> 1;
		int tmp = tree[tree[rt2].ls].w - tree[tree[rt1].ls].w;
		if(tmp >= k) {
			rt1 = tree[rt1].ls;
			rt2 = tree[rt2].ls;
			r= mid;
		} else {
			k -= tmp;
			rt1 = tree[rt1].rs;
			rt2 = tree[rt2].rs;
			l = mid + 1;
		}
	}
	return l;
}
int main() {
		scanf("%d %d", &n, &m);
		for(int i = 1; i <= n; i++) {
 			scanf("%d", &a[i]);
 			b[i - 1] = a[i];
		}
		sort(b, b + n);
		cnt = unique(b, b + n) - b;
		tot = 0;
		root[0] = Build_tree(1, cnt);
		for(int i = 1; i <= n; i++) {
			int tmp = (int)(lower_bound(b, b + cnt, a[i]) - b) + 1;
			root[i] = update(root[i - 1], tmp, 1);
		}
		int l, r, k;
		for(int i = 0; i < m; i++) {
			scanf("%d %d %d", &l, &r, &k);
			int tmp = query(root[l - 1], root[r], k);
			printf("%d\n", b[tmp - 1]);
		}
	return 0;
}