这题貌似只能用朴素做法
#include <iostream>
#include <algorithm>
using namespace std;
const int N = 1e5 + 10;
int n;
int s[N], a[N], b[N];
int main()
{
cin >> n;
for (int i = 0; i < n; i ++ ) cin >> s[i]; //输入数据
sort(s, s + n);
int mid = n >> 1;
for (int i = 0; i < mid; i ++ ) a[i] = s[i]; //排序后按中位数分隔数组
for (int i = mid; i < n; i ++ ) b[i - mid] = s[i];
int sum1 = 0, sum2 = 0;
for (int i = 0; i < mid; i ++ ) sum1 += a[i]; //分别求和
for (int i = 0; i < n - mid; i ++ ) sum2 += b[i];
cout << n - 2 * mid << ' ' << sum2 - sum1 << endl; //(n - mid) - mid可替换为n % 2
return 0;
}