2013
11-10

# K-th Number

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.

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).

For each question output the answer to it — the k-th number in sorted a[i...j] segment.

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

5
6
3

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

//* @author: 82638882@163.com
import java.io.*;
import java.util.Arrays;
class Main
{
public static void main(String[] args) throws IOException
{
int a=Integer.parseInt(ss[0]);
int k=Integer.parseInt(ss[1]);
my[] p=new my[a];
for(int i=0;i< a;i++)
p[i]=new my(i,Integer.parseInt(ss[i]));
Arrays.sort(p);
while((k--)!=0)
{
int l=Integer.parseInt(ss[0])-1;
int r=Integer.parseInt(ss[1]);
int cnt=Integer.parseInt(ss[2]);
for(int i=0;i< a;i++)
{
if(p[i].num>=l&&p[i].num< r)
{
cnt--;
if(cnt==0)
{
System.out.println(p[i].t);
break;
}
}
}
}
}

}

class my implements Comparable< my>
{
int num,t;
public my(int a,int b)
{
num=a;
t=b;
}
public int compareTo(my arg0) {
return t-arg0.t;
}
}

1. 第一题是不是可以这样想，生了n孩子的家庭等价于n个家庭各生了一个1个孩子，这样最后男女的比例还是1:1

2. 学算法中的数据结构学到一定程度会乐此不疲的，比如其中的2－3树，类似的红黑树，我甚至可以自己写个逻辑文件系统结构来。

3. 有两个重复的话结果是正确的，但解法不够严谨，后面重复的覆盖掉前面的，由于题目数据限制也比较严，所以能提交通过。已更新算法