一个长度为N的整数序列，编号0 - N - 1。进行Q次查询，查询编号i至j的所有数中，第K大的数是多少。
例如: 1 7 6 3 1。i = 1, j = 3，k = 2，对应的数为7 6 3，第2大的数为6。
Input
第1行：1个数N，表示序列的长度。(2 <= N <= 50000)
第2 - N + 1行：每行1个数，对应序列中的元素。(0 <= S[i] <= 10^9)
第N + 2行：1个数Q，表示查询的数量。(2 <= Q <= 50000)
第N + 3 - N + Q + 2行：每行3个数，对应查询的起始编号i和结束编号j，以及k。(0 <= i <= j <= N - 1,1 <= k <= j - i + 1)
Output
共Q行，对应每一个查询区间中第K大的数。
Input示例
5
1
7
6
3
1
3
0 1 1
1 3 2
3 4 2
Output示例
7
6
1 区间第k大，主席树写一发，使用java写的，顺便复习
吐槽---java真是慢，一定要离散化，c++随便写都过
而且没有sort，没有map，只有hashmap，真逊啊。。。。
package p1175; import java.io.*;
import java.util.*; import javax.management.Query; public class Main
{
public static class Node
{
int sum;
Node lc, rc;
public Node(Node t)
{
if(t == null)
{
lc = rc = null;
sum = 0;
}
else
{
lc = t.lc;
rc = t.rc;
sum = t.sum;
}
}
} static final int N = 50010; public static void main(String[] args) throws IOException
{
// TODO Auto-generated method stub
BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
BufferedWriter writer = new BufferedWriter(new OutputStreamWriter(System.out));
int n = getInt(reader);
int[] arr = new int[n], store = new int[n];
for(int i = 0; i < n; i++) store[i] = arr[i] = getInt(reader);
mergeSort(store, 0, n - 1);
int len = 1;
for(int i = 1; i < n; i++)
if(store[i] != store[i - 1]) store[len++] = store[i];
HashMap<Integer, Integer> myMap = new HashMap<Integer, Integer>();
for(int i = 0; i < len; i++)
myMap.put(store[i], i);
for(int i = 0; i < n; i++)
arr[i] = myMap.get(arr[i]); Node[] root = new Node[n + 1];
Node last = null;
for(int i = 0; i < n; i++)
{
root[i] = new Node(last);
buildTree(root[i], arr[i], 0, len);
last = root[i];
} for(int m = getInt(reader); m > 0; m--)
{
int l = getInt(reader), r = getInt(reader), k = getInt(reader);
int ans = queryKth(l > 0 ? root[l - 1] : null, root[r], k, 0, len);
writer.write(store[ans] + "\r\n");
writer.flush();
}
} public static int querySum(Node t)
{
if(t == null) return 0;
return t.sum;
} public static int queryKth(Node l, Node r, int kth, int left, int right)
{
if(left == right) return left;
int mid = (left + right) >> 1, ret = -1;
int lCnt = l == null ? 0 : querySum(l.rc), rCnt = r == null ? 0 : querySum(r.rc);
if(rCnt - lCnt >= kth) ret = queryKth(l == null ? null : l.rc, r == null ? null : r.rc, kth, mid + 1, right);
else ret = queryKth(l == null ? null : l.lc, r == null ? null : r.lc, kth - (rCnt - lCnt), left, mid);
return ret;
} public static void buildTree(Node x, int val, int left, int right)
{
if(left < right)
{
int mid = (left + right) >> 1;
if(val <= mid)
{
x.lc = new Node(x.lc);
buildTree(x.lc, val, left, mid);
}
else
{
x.rc = new Node(x.rc);
buildTree(x.rc, val, mid + 1, right);
}
}
x.sum++;
} static int[] tmp = new int[N];
public static void mergeSort(int[] arr, int l, int r)
{
if(l >= r) return;
int mid = (l + r) >> 1;
mergeSort(arr, l, mid);
mergeSort(arr, mid + 1, r);
int itL = l, itR = mid + 1, now = l;
while(itL <= mid && itR <= r)
{
if(arr[itL] < arr[itR]) tmp[now++] = arr[itL++];
else tmp[now++] = arr[itR++];
}
while(itL <= mid) tmp[now++] = arr[itL++];
while(itR <= r) tmp[now++] = arr[itR++];
for(int i = l; i <= r; i++) arr[i] = tmp[i];
} public static int getInt(BufferedReader reader) throws IOException
{
char ch = ' ';
for( ; !(ch >= '0' && ch <= '9'); ch = (char) reader.read()) ;
int ret = 0;
for( ; ch >= '0' && ch <= '9'; ch = (char) reader.read())
ret = ret * 10 + ch - '0';
return ret;
} }