1037B--Reach Median(中位数)

Posted 2021-01-09 lllaih

median 中位数 odd 奇数 even 奇数

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 $\mathrm{ss.}$

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 $\mathrm{6,5,86,5,8 is\; equal\; to 66,\; since\; if\; we\; sort\; this\; array\; we\; will\; get 5,6,85,6,8,\; and 66 is\; located\; on\; the\; middle\; position.}$

Input

The first line contains two integers $\mathrm{nn and ss (1\le n\le 2?105?11\le n\le 2?105?1, 1\le s\le 1091\le s\le 109) —\; the\; length\; of\; the\; array\; and\; the\; required\; value\; of\; median.}$

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 $\mathrm{nn is\; odd.}$

Output

In a single line output the minimum number of operations to make the median being equal to $\mathrm{ss.}$

Examples

input

Copy

3 8
6 5 8

output

Copy

2

input

Copy

7 20
21 15 12 11 20 19 12

output

Copy

6

Note

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

$\mathrm{题意：给你一个含有n个数字的数组，问改动多少次才能使这个数组的中位数为s，每次改动只能使一个数字加1或减1}$

$\mathrm{分析：先给这个数组从低到高排序，找出其中的中位数，从后面到中位数的位置遍历，即（n\; to\; n/2+1）,判断，如果其中有数字小于s,则改动次数加上s-a[i],当i等于n/2+1时，判断，如果大于了s，改动次数加上a[i]-s;再从（1\; to\; n/2）遍历，如果有a[i]>s,改动次数加上a[i]-s.}$

$\mathrm{总之，要使中位数之前的都小于等于中位数，中位数之后的都大于等于中位数。}$

 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
",sum);
26     }
27     return 0;
28 }