UVA - 10057 - A mid-summer night's dream. (中位を探せ!)

UVA - 10057
A mid-summer night's dream.
Time Limit: 30000MS
Memory Limit: Unknown
64bit IO Format: %lld & %llu
Submit Status
Description
Problem C
A mid-summer night’s dream
Input: standard input
Output: standard output
 
This is year 2200AD. Science has progressed a lot in two hundred years. Two hundred years is mentioned here because this problem is being sent back to 2000AD with the help of time machine. Now it is possible to establish direct connection between man and computer CPU. People can watch other people’s dream on 3D displayer (That is the monitor today) as if they were watching a movie. One problem in this century is that people have become so dependent on computers that their analytical ability is approaching zero. Computers can now read problems and solve them automatically. But they can solve only difficult problems. There are no easy problems now. Our chief scientist is in great trouble as he has forgotten the number of his combination lock. For security reasons computers today cannot solve combination lock related problems. In a mid-summer night the scientist has a dream where he sees a lot of unsigned integer numbers flying around. He records them with the help of his computer, Then he has a clue that if the numbers are (X1, X2,   …  , Xn) he will have to find an integer number A (This A is the combination lock code) such that
             
             (|X1-A| + |X2-A| + … … + |Xn-A|) is minimum.
 
Input
Input will contain several blocks. Each block will start with a number n (0 
Output
For each set of input there will be one line of output. That line will contain the minimum possible value for A. Next it will contain how many numbers are there in the input that satisfy the property of A (The summation of absolute deviation from A is minimum). And finally you have to print how many possible different integer values are there for A (these values need not be present in the input). These numbers will be separated by single space.
 
Sample Input:
2 10 10 4 1 2 2 4
Sample Output:
10 2 1 2 2 1
______________________________________________________________________________________________ Shahriar Manzoor 16-12-2000
Source
Root::AOAPC I: Beginning Algorithm Contests (Rujia Liu)::  Volume 4. Algorithm Design
Root::Competitive Programming 3: The New Lower Bound of Programming Contests (Steven & Felix Halim)::Data Structures and Libraries::Linear Data Structures with Built-in Libraries::C++ STL algorithm (Java Collections)
作り方:一組の数を入力して、並べ替えて、真ん中の(すなわち中位数)を探して、それから奇偶数に分けて討論すればいいです.最後の数は第1中位数と第2中位数の間の数を意味することに注意してください.
ACコード:

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
using namespace std;

const int maxn = 1000005;
int num[maxn]; 

int main()
{
	int n;
	while(scanf("%d", &n)!=EOF)
	{
		for(int i=0; i<n; i++)
		{
			scanf("%d", &num[i]);
		}
		sort(num, num+n);
		int k = (n - 1) / 2;
		int n1 = 0, n2 = 1;
		if(n%2 == 1)
		{
			for(int i=k; i>=0 && num[k] == num[i]; i--) n1++;
			for(int i=k+1; i<n && num[k] == num[i]; i++) n1++;
		}
		else 
		{
			for(int i=k; i>=0 && num[k] == num[i]; i--) n1++;
			for(int i=k+1; i<n && num[k+1] == num[i]; i++) n1++;
			n2 = num[k+1] - num[k] +1;
		}
		printf("%d %d %d
", num[k], n1, n2);
	}
	return 0;
}