AtCoder Regular Contest 095 C - Many Medians(去掉一个数求中位数)-优快云博客

该博客介绍了AtCoder Regular Contest 095的一道题目，涉及计算在一组偶数个整数中，去掉其中一个数后新的中位数。问题的核心是确定在排除某个特定元素后，序列的中位数如何变化。解决方案指出，根据原始中位数和被移除数字的位置，中位数的变化分为两种情况：如果移除的数字小于等于原中位数，中位数位置将加1；否则，中位数不变。

Problem Statement

When l is an odd number, the median of l numbers a1,a2,…,al is the (

l+1

)-th largest value among a1,a2,…,al.

You are given N numbers X1,X2,…,XN, where N is an even number. For each i=1,2,…,N, let the median of X1,X2,…,XN excluding Xi, that is, the median of X1,X2,…,Xi−1,Xi+1,…,XN be Bi.

Find Bi for each i=1,2,…,N.

Constraints

2≤N≤200000
N is even.
1≤Xi≤109
All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
X1 X2 ... XN

Output

Print N lines. The i-th line should contain Bi.

Sample Input 1

Copy

4
2 4 4 3

Sample Output 1

Copy

Since the median of X2,X3,X4 is 4, B1=4.
Since the median of X1,X3,X4 is 3, B2=3.
Since the median of X1,X2,X4 is 3, B3=3.
Since the median of X1,X2,X3 is 4, B4=4.

Sample Input 2

Copy

2
1 2

Sample Output 2

Copy

2
1

Sample Input 3

Copy

6
5 5 4 4 3 3

Sample Output 3

Copy

题意：去掉第i个数之后的中位数

思路：很容易分析得到，1-n的中位数去掉第i个数之后只有2种情况，原本数字中排序后的中位数和中位数的位数+1，如果当前数字小于等于中位数，则会对结果造成影响，所以最后结果为中位数的位数+1，否则无影响。

#include <bits/stdc++.h>
 
using namespace std;
const int N = 200005;
typedef long long ll;
int a[N];
int b[N];
int main()
{
    int n;
    cin>>n;
    for(int i=1;i<=n;i++){
        cin>>a[i];
        b[i]=a[i];
   }
   sort(a+1,a+1+n);
    int mid=n/2;
    for(int i=1;i<=n;i++){
        if(b[i]<=a[mid])
        printf("%d\n",a[mid+1]);
        else
        printf("%d\n",a[mid]);
    }
}