Sort the citation counts largest-to-smallest, and consider a bar graph where bar $i$ has width $1$ and height $c_i$. The $h$-index of the papers before the surveys is simply the dimension of the largest square that fits under this bar graph. We want to determine how much we can increase the $h$-index with $K$ surveys each citing $L$ distinct papers.

Let's binary search on the final $h$-index. Then we just need a way to check whether it's possible to achieve a given $h$-index $h$.

It's clearly optimal to work only with the $h$ papers that start off with the largest citation counts. Note that we have $KL$ total citations to allocate to these papers. If the total citation count of these papers is less than $h^2 - KL$, then we cannot hope to achieve $h$. This is one failure mode.

Unfortunately, it's not the only failure mode. That is, the converse of the above statement is not true, because we have an added restriction that no survey can cite a paper twice. So for example if $h=3$, $K = 1$, and $L = 2$, and the top three papers initially have citation counts $(3, 3, 1)$, then we cannot raise the third paper to citation count $3$ since we only have one survey. This illuminates another possible failure mode: if there is a paper with less than $h-K$ citations initially, then we cannot achieve $h$.

It turns out that these are the only two failure modes. We can prove this by induction on $K$. If $K=0$ then every paper already has at least $h$ citations, so we're certainly happy. Suppose $K>0$. Every paper needs at most $K$ more citations. There is some set of papers which need exactly $K$ more citations. By the assumption on the sum of citation counts, this set has size at most $L$, so with one survey we can cite all these papers (plus some others, until we've hit $L$ citations or this survey has cited every paper). Now we're in a situation with $K-1$ remaining surveys, but every paper needs at most $K-1$ more citations. Moreover, it can be checked that the total number of needed citations is now at most $(K-1)L$. So we can indeed induct, completing the proof.

Below is Danny Mittal's code, implementing the above idea.

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.Comparator;
import java.util.StringTokenizer;
 
public class AcowdemiaSilver {
 
    public static void main(String[] args) throws IOException {
        BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer tokenizer = new StringTokenizer(in.readLine());
        int n = Integer.parseInt(tokenizer.nextToken());
        int k = Integer.parseInt(tokenizer.nextToken());
        int l = Integer.parseInt(tokenizer.nextToken());
        Integer[] publications = new Integer[n];
        tokenizer = new StringTokenizer(in.readLine());
        for (int j = 0; j < n; j++) {
            publications[j] = Integer.parseInt(tokenizer.nextToken());
        }
        Arrays.sort(publications, Comparator.reverseOrder());
        int upper = n;
        int lower = 0;
        while (upper > lower) {
            int mid = (upper + lower + 1) / 2;
            long needed = 0;
            for (int j = 0; j < mid; j++) {
                needed += (long) Math.max(0, mid - publications[j]);
            }
            if (needed <= ((long) k) * ((long) l) && publications[mid - 1] + k >= mid) {
                lower = mid;
            } else {
                upper = mid - 1;
            }
        }
        System.out.println(upper);
    }
}