https://oj.turingedu.cn/class/40/lesson/1/problem/2求调

#include <iostream>
#include <algorithm>
#include <cmath>
#define ll long long

using namespace std;
const int N = 200000 + 10;

int n, q;
int a[N], sum[N]; // sum[i] 表示a[1] - a[i] 的和(前缀和)

int main() {
	cin >> n >> q;
	for (int i = 1; i <= n; i++) {
		cin >> a[i];
		sum[i] = sum[i - 1] + a[i];
	}
	sort(a + 1, a + n + 1);
	for (int i = 1; i <= q; i++) {
		int x;
		cin >> x;
		int pos = lower_bound(a + 1, a + n + 1, x) - a;
		ll ans = 0;
		ans = (n - pos) * x - sum[pos - 1];// 小于x
		ans += (sum[n] - sum[pos - 1]) * x - pos * x; // 大于等于x
		cout << ans << endl;
	}
	return 0;
}