Problem:
Score : 700 points

Problem Statement

N problems are proposed for an upcoming contest. Problem i has an initial integer score of A_i points.

M judges are about to vote for problems they like. Each judge will choose exactly V problems, independently from the other judges,
and increase the score of each chosen problem by 1.

After all M judges cast their vote, the problems will be sorted in non-increasing order of score, and the first P problems will be chosen for the problemset.
Problems with the same score can be ordered arbitrarily, this order is decided by the chief judge.

How many problems out of the given N have a chance to be chosen for the problemset?

Constraints

2 \le N \le 10^5

1 \le M \le 10^9

1 \le V \le N - 1

1 \le P \le N - 1

0 \le A_i \le 10^9

Input

Input is given from Standard Input in the following format:

N M V P
A_1 A_2 ... A_N

Output

Print the number of problems that have a chance to be chosen for the problemset.

Sample Input 1

6 1 2 2
2 1 1 3 0 2

Sample Output 1

5

If the only judge votes for problems 2 and 5, the scores will be 2 2 1 3 1 2.
The problemset will consist of problem 4 and one of problems 1, 2, or 6.

If the only judge votes for problems 3 and 4, the scores will be 2 1 2 4 0 2.
The problemset will consist of problem 4 and one of problems 1, 3, or 6.

Thus, problems 1, 2, 3, 4, and 6 have a chance to be chosen for the problemset. On the contrary, there is no way for problem 5 to be chosen.

Sample Input 2

6 1 5 2
2 1 1 3 0 2

Sample Output 2

3

Only problems 1, 4, and 6 have a chance to be chosen.

Sample Input 3

10 4 8 5
7 2 3 6 1 6 5 4 6 5

Sample Output 3

8

Solution in C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
#define int long long
using namespace std;
const int N=1e5+5;
int n,m,v,p,cnt;
int a[N];
signed main() 
{
	cin>>n>>m>>v>>p;//m个评委,每个评委给v个题目加分，选前p名 
	for (int i=1;i<=n;i++) cin>>a[i];
	sort(a+1,a+n+1);
	if (v<=p)
	{
		for (int i=1;i<=n;i++) if (a[i]+m>=a[n-p+1]) cnt++;
		cout<<cnt<<endl;
	} else
	{
		for (int i=1;i<=n;i++) 
		{
			int j=n-i+1;//正数第j名 
			if (j<=p) 
			{
				cout<<n-i+1<<endl;
				break;
			}
			if (i+p-1>=v && a[i]+m>=a[n-p+1]) //取比自己低的人和最高的人加满 
			{
//				cout<<"CASE2"<<endl;
				cout<<n-i+1<<endl;
				break;
			}
			int tmp=v*m-(i+p-1)*m;//还需要给这么多
			int temp=0;
			for (int k=i+1;a[k]<a[i]+m,k<=n-p+1;k++)
				temp+=a[i]+m-a[k];
			if (temp>=tmp)
			{
				cout<<n-i+1<<endl;
				break;
			}
		}
	}
	return 0;
}