參考了佬的c題題解思路，感覺很巧妙，記錄一下
https://zhuanlan.zhihu.com/p/618685370

#include <bits/stdc++.h>
using namespace std;
#define int long long
const int N= 2 * 100010;
int a[N];
void solve()
{
	int n,c,d;
	cin >> n >> c >> d;
	set<int> se;
	int sh = 0;//去重花費
	for(int i=1; i <= n;i ++) {
		cin >> a[i];
	}
	int tp = c*n + d;//特判花費
	sort(a + 1,a + 1 + n);
	int n1 = unique(a + 1, a + 1 + n) - (a + 1);//返回值是去重之后的尾地址
	sh = c*(n-n1);//重復的個數就是n-n1，去重代價*c
	
	int cnt = 1e18;
	n = n1;
	for(int i = 1; i <= n; i ++){
		cnt =min(cnt, (n-i)*c + (a[i]-i)*d);
	}
	tp = min(sh+cnt,tp);
	cout << tp << '\n';
}

signed main() {
	int T;
	cin >> T;
	while(T --)
		solve();
	return 0;
}