了解了。。。。

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <math.h>
#include <string>
#include <iostream>
#include <algorithm>
using namespace std;
#include <queue>
#include <stack>
#include <vector>
#include <deque>
#include <set>
#include <map>
typedef long long LL;
const int MAXN = 100999;//点数的最大值
const int MAXM = 1000010;//边数的最大值
const LL INF = 1152921504;
/*
* 划分树(查询区间第k小)
*/
int tree[20][MAXN];//表示每层每个位置的值
int sorted[MAXN];//已经排序好的数
int toleft[20][MAXN];//toleft[p][i]表示第p层从1到i有数分入左边
void build(int l,int r,int dep)
{if(l == r)return;int mid = (l+r)>>1;int same = mid - l + 1;//表示等于中间值而且被分入左边的个数for(int i = l; i <= r; i++) //注意是l,不是oneif(tree[dep][i] < sorted[mid])same--;int lpos = l;int rpos = mid+1;for(int i = l; i <= r; i++){if(tree[dep][i] < sorted[mid])tree[dep+1][lpos++] = tree[dep][i];else if(tree[dep][i] == sorted[mid] && same > 0){tree[dep+1][lpos++] = tree[dep][i];same--;}elsetree[dep+1][rpos++] = tree[dep][i];toleft[dep][i] = toleft[dep][l-1] + lpos - l;}build(l,mid,dep+1);build(mid+1,r,dep+1);
}
//查询区间第k小的数,[L,R]是大区间,[l,r]是要查询的小区间
int query(int L,int R,int l,int r,int dep,int k)
{if(l == r)return tree[dep][l];int mid = (L+R)>>1;int cnt = toleft[dep][r] - toleft[dep][l-1];//区间内有多少个被分到左区间if(cnt >= k)//左儿子里的树>＝ｋ,在左边找{int newl = L + toleft[dep][l-1] - toleft[dep][L-1];//改变区间，使得新区间包含l ,r 里面的数int newr = newl + cnt - 1;return query(L,mid,newl,newr,dep+1,k);}else{int newr = r + toleft[dep][R] - toleft[dep][r];//右边有(toleft[dep][R] - toleft[dep][r])个在左区间，可以向右移动这么多int newl = newr - (r-l-cnt);return query(mid+1,R,newl,newr,dep+1,k-cnt);}
}
int main()
{int n,m;// freopen("in.txt","r",stdin);while(scanf("%d%d",&n,&m)!=EOF){memset(tree,0,sizeof(tree));for(int i = 1; i <= n; i++){scanf("%d",&tree[0][i]);sorted[i] = tree[0][i];}sort(sorted+1,sorted+n+1);build(1,n,0);while(m--){int l,r,k;scanf("%d%d%d",&l,&r,&k);printf("%d\n",query(1,n,l,r,0,k));}}return 0;}