K-th Number(主席树)

K-th Number

Time Limit: 20000MS		Memory Limit: 65536K
Total Submissions: 59327		Accepted: 20660
Case Time Limit: 2000MS

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

Hint

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

Source

Northeastern Europe 2004, Northern Subregion

//题意: n 个数，m 次询问，求 l,r,中第 k 小的数是什么

题解:有很多方法可以写，线段树配上归并，我用来当做主席树模板来做，学习！

 # include <cstdio>
 # include <cstring>
 # include <cstdlib>
 # include <iostream>
 # include <vector>
 # include <queue>
 # include <stack>
 # include <map>
 # include <bitset>
 # include <sstream>
 # include <set>
 # include <cmath>
 # include <algorithm>
 # pragma  comment(linker,"/STACK:102400000,102400000")
 using namespace std;
 # define LL          long long
 # define pb          push_back
 # define pr          pair
 # define mkp         make_pair
 # define lowbit(x)   ((x)&(-x))
 # define PI          acos(-1.0)
 # define INF         0x3f3f3f3f3f3f3f3f
 # define eps         1e-
 # define MOD         

 inline int scan() {
     int x=,f=; char ch=getchar();
     while(ch<''||ch>''){if(ch=='-') f=-; ch=getchar();}
     while(ch>=''&&ch<=''){x=x*+ch-''; ch=getchar();}
     return x*f;
 }
 inline void Out(int a) {
     if(a<) {putchar('-'); a=-a;}
     if(a>=) Out(a/);
     putchar(a%+'');
 }
 # define MX
 /**************************/
 struct Node
 {
     int l,r;
     int sum;
 }tree[MX*];

 int n,m;
 int newn,tcnt;
 int dat[MX];
 int root[MX];
 int cpy[MX];
 int getid(int x){return lower_bound(cpy+,cpy++n,x)-cpy;}

 int update(int l,int r,int y,int pos)
 {
     tcnt++;
     tree[tcnt].sum = tree[y].sum+;
     int sav = tcnt;

     if (l==r) return sav;
     int mid = (l+r)/;
     if (pos<=mid)
     {
         tree[sav].l = update(l,mid,tree[y].l,pos);
         tree[sav].r = tree[y].r;
     }
     else
     {
         tree[sav].l = tree[y].l;
         tree[sav].r = update(mid+,r,tree[y].r,pos);
     }
     return sav;
 }

 int inqy(int l,int r,int x,int y,int k)
 {
     if (l==r) return l;
     int all =tree[ tree[y].l ].sum -  tree[ tree[x].l ].sum;
     int mid = (l+r)/;
     if (k<=all) return inqy(l,mid,tree[x].l,tree[y].l,k);
     else return inqy(mid+,r,tree[x].r,tree[y].r,k-all);

 }

 int main ()
 {
     while (scanf("%d%d",&n,&m)!=EOF)
     {
         for (int i=;i<=n;i++)
         {
             dat[i] = scan();
             cpy[i] = dat[i];
         }
         sort(cpy+,cpy++n);
         newn = unique(cpy+,cpy++n)-cpy;
         tcnt=;
         for (int i=;i<=n;i++) root[i]=update(,n,root[i-],getid(dat[i]));
         while (m--)
         {
             int a = scan();
             int b = scan();
             int k = scan();
             printf("%d\n",cpy[inqy(,n,root[a-],root[b],k)]);
         }
     }
     return ;
 }