UVa 10057 - A mid-summer night's dream.

链接：

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=113&page=show_problem&problem=998

原题：

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<n<=1000000) indicating how many numbers he saw in the dream. Next there will be n numbers. All the numbers will be less than 65536. The input will be terminated by end of file.

 

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

分析与总结：

求出A，使得(|X1-A| + |X2-A| + … … + |Xn-A|)的值最小。

 那么这个A其实就是中位数。那么在排序之后，

如果数字个数为奇数，那么中间那个就是唯一的中位数，如果是偶数，则会有中间的两位数，而且在这两位数之间的数也是。

再求数列中有几个数时需要注意的是，奇数是求出中间两个数之后，这两个数的旁边也可能会有相等的数

/*
 * UVa: 10057 - A mid-summer night's dream.
 * Time: 0.236s
 * Result: Accept
 * Author: D_Double 
 * 
 */
#include<iostream>
#include<cstdio>
#include<algorithm>
#define MAXN 1000010
using namespace std;
int arr[MAXN];

int main(){
    int n;
    while(~scanf("%d",&n)){
        for(int i=0; i<n; ++i)
            scanf("%d",&arr[i]);

        sort(arr, arr+n);
        
        int mid, left, right;
        int ans_min, ans_num, ans_val;

        if(n%2==0){ // 如果有偶数个数字
            left=(n-2)/2;  right=(n)/2;
            
            ans_min=arr[left];
            
            ans_num=0;
            for(int i=left; arr[i]==arr[left]&&i>=0; --i) ++ans_num;
            for(int i=right; arr[i]==arr[right]&&i<n; ++i) ++ans_num;
            
            ans_val = arr[right]-arr[left]+1;
        }
        else{
            mid = (n-1)/2;

            ans_min = arr[mid];

            ans_num=0;
            for(int i=mid; arr[i]==arr[mid]&&i>=0; --i) ++ans_num;
            for(int i=mid+1; arr[i]==arr[mid]&&i<n; ++i) ++ans_num;

            ans_val = 1;
        }
        printf("%d %d %d\n", ans_min, ans_num, ans_val);
    }
    return 0;
}

——  生命的意义，在于赋予它意义。
  
   
  
  
  

    
   
   
             
   
  
  
  
   
  
  
  

    
   
   
        原创 http://blog.csdn.net/shuangde800 ， By   D_Double  (转载请标明)