poj--2104--K-th Number（伴随数组）

K-th Number

Time Limit: 20000MS Memory Limit: 65536KTotal Submissions: 45681 Accepted: 15185Case 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 31 5 2 6 3 7 42 5 34 4 11 7 3

Sample Output

563

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

给了n个数，求对应区间中对应中对应的数字。

开结构体数组，记录id，根据数字大小先对整体进行排序，然后根据id找出对应的区间数字，即使都排序了，但是因为有id在，我们还是可以确定一个数是不是我们所求区间的数字

#include<cstdio>#include<cstring>#include<queue>#include<algorithm>using namespace std;struct node{int x,id;}p[101000];int n,m;bool cmp(node s1,node s2){return s1.x<s2.x;}int main(){while(scanf("%d%d",&n,&m)!=EOF){for(int i=0;i<n;i++){scanf("%d",&p[i].x);p[i].id=i;}sort(p,p+n,cmp);while(m--){int x,y,z;scanf("%d%d%d",&x,&y,&z);x--,y--;for(int i=0;i<n;i++){if(p[i].id>=x&&p[i].id<=y){z--;if(!z){printf("%d\n",p[i].x);break;}}}}}return 0;}