# Acowdemia I Bronze

Link to Bronze Acowdemia I: http://www.usaco.org/index.php?page=viewproblem2&cpid=1131.

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;
}
else
{
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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