Acowdemia I Bronze

Link to Bronze Acowdemia I:

I tried solving this problem using binary search.

#include <bits/stdc++.h>

using namespace std;

int32_t main()
    int N, L;
    cin >> N >> L;
    vector<int> c(N);
    for(int i=0; i<N; i++){
        cin >> c[i];
    L = min(L, N);
    sort(begin(c), end(c), greater<int>());
    int l = 0, h = N;
    while(l < h) {
        int m = (l + h + 1)/2;
        int cit = 0;
        for(int i=0; i<m; i++){
            cit += max(0, m-c[i]);
            if(cit > L) break;
        if(cit <= L)
            l = m;
            h = m-1;
    cout << l << '\n';

I check a particular m-index by going through the m most-cited papers and calculating the additional citations I need to get every c[i] up to m. I then check if this number of citations is at most L to see if the m-index is valid.

My code gives WA on test cases 9, 11, 12, 13. I think the error in my code might be that the survey can only cite each of the N papers at most once. However, I set L = min(L, N), and I don’t see how exactly my code overlooks the case when we double cite some paper. Can you please help me spot the error in my logic and its implications?

I understood my problem. I am citing the i-th paper \max(0, m-c[i]) times. Therefore, I need to include if(m-c[i] > 1){ good = false; break; } in the for loop as well.

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