You are given an array $\mathrm{aa\; of\; nn\; integers\; and\; an\; integer\; ss.\; It\; is\; guaranteed\; that\; nn\; is\; odd.}$

In one operation you can either increase or decrease any single element by one. Calculate the minimum number of operations required to make the median of the array being equal to

The median of the array with odd length is the value of the element which is located on the middle position after the array is sorted. For example, the median of the array $\mathrm{6,5,86,5,8\; is\; equal\; to\; 66,\; since\; if\; we\; sort\; this\; array\; we\; will\; get\; 5,6,85,6,8,\; and\; 66\; is\; located\; on\; the\; middle\; position.}$

The first line contains two integers

The second line contains $\mathrm{nn\; integers\; a1,a2,\dots ,ana1,a2,\dots ,an\; (1\le ai\le 1091\le ai\le 109)\; —\; the\; elements\; of\; the\; array\; aa.}$

It is guaranteed that $\mathrm{nn\; is\; odd.}$

In a single line output the minimum number of operations to make the median being equal to

3 8
6 5 8

2

7 20
21 15 12 11 20 19 12

6

In the first sample, $\mathrm{66\; can\; be\; increased\; twice.\; The\; array\; will\; transform\; to\; 8,5,88,5,8,\; which\; becomes\; 5,8,85,8,8\; after\; sorting,\; hence\; the\; median\; is\; equal\; to\; 88.}$

In the second sample, $\mathrm{1919\; can\; be\; increased\; once\; and\; 1515\; can\; be\; increased\; five\; times.\; The\; array\; will\; become\; equal\; to\; 21,20,12,11,20,20,1221,20,12,11,20,20,12.\; If\; we\; sort\; this\; array\; we\; get\; 11,12,12,20,20,20,2111,12,12,20,20,20,21,\; this\; way\; the\; median\; is\; 2020.}$

$\mathrm{題意：給你一個含有n個數字的數組，問改動多少次才能使這個數組的中位數為s，每次改動只能使一個數字加1或減1}$

$\mathrm{分析：先給這個數組從低到高排序，找出其中的中位數，從後面到中位數的位置遍歷，即（n\; to\; n/2+1）,判斷，如果其中有數字小於s,則改動次數加上s-a[i],當i等於n/2+1時，判斷，如果大於了s，改動次數加上a[i]-s;再從（1\; to\; n/2）遍歷，如果有a[i]>s,改動次數加上a[i]-s.}$

$\mathrm{總之，要使中位數之前的都小於等於中位數，中位數之後的都大於等於中位數。}$

 1 #include<cstdio>
 2 #include<algorithm>
 3 using namespace std;
 4 int main()
 5 {
 6     long long n,s,a[200005];
 7     while(~scanf("%lld %lld",&n,&s))
 8     {
 9         for(int i=1;i<=n;i++)
10             scanf("%lld",&a[i]);
11         sort(a+1,a+1+n);
12         long long sum=0;
13         for(int i=n;i>=n/2+1;i--)
14         {
15             if(a[i]<s)
16                 sum+=s-a[i];
17             if(i==n/2+1&&a[i]>s)
18                 sum+=a[i]-s;
19         }
20         for(int i=n/2;i>=1;i--)
21         {
22             if(a[i]>s)
23                 sum+=a[i]-s;
24         }
25         printf("%lld\n",sum);
26     }
27     return 0;
28 }

1037B--Reach Median(中位數)