Disk Scheduling（第二届中国计量大学ACM程序设计竞赛个人赛）

链接：https://ac.nowcoder.com/acm/contest/3190/G
来源：牛客网

示例1
输入
9 100
55 58 39 18 90 160 150 38 184

输出
248

本题其实就是先将n个数据排序后，将m插入进去，左右判断，选取较小的差值累加，具体思路见代码。

#include <iostream>
#include <cstdio>
#include <cstring>
#include<algorithm>
#include<cmath>
using namespace std
long long int a[1000005]
int main()
{
    long long int n, m, left, right, s = 0
    scanf("%lld %lld", &n,&m)
    for (int i = 0 i < n i++)
        scanf("%lld", &a[i])
    sort(a, a + n)				//对n个数据进行排序
    for (int i = 1 i < n i++)  		//将m值插入进去后，获取m左边的值left与右边的值right
    {
        if (m <= a[0])			//特判m是否比第一个值还小
        {
            left = -1
            s = s + a[0] - m
            m = a[0]
            right = 1
            break
        }
        if (m > a[n - 1])		//特判m是否比最后一个值还大
        {
            right = n
            s = s + m - a[n - 1]
            m = a[n - 1]
            left = n - 2
            break
        }
        if (m >= a[i - 1] && m <= a[i])		//遍历搜索m在n个数据中，排到哪个位置
        {
            long long int sl,sr
            sl = m - a[i-1]
            sr = a[i] - m
            if (sl <= sr)
            {
                s = s + sl
                m = a[i - 1]
                left = i - 2
                right = i
                break
            }
            else
            {
                s = s + sr
                m = a[i]
                left = i - 1
                right = i+1
                break
            }
        }
    }
    while (left != -1 && right!=n)	//开始左右比较，并且累加s，替换m值
    {
        long long int sl, sr
        sl =m-a[left]
        sr = a[right]-m
        if (sl <= sr)
        {
            s = s + sl
            m = a[left]
            left--
        }
        else
        {
            s = s + sr
            m = a[right]
            right++
        }
    }
    if (left == -1)		//若m到达最左边后，只需不断计算右边没计算的值
    {
        while (right != n)
        {
            long long int sr
            sr = a[right]-m
            s = s + sr
            m = a[right]
            right++
        }
    }
    if (right == n)		//若m到达最右边后，只需不断计算左边没计算的值
    {
        while (left != -1)
        {
            long long int sl
            sl = m - a[left]
            s = s + sl
            m = a[left]
            left--
        }
    }
    printf("%lldn", s)
 
    return 0
}