This problem immediately seems to be looking for a greedy solution. There are two obvious greedy orderings: sort the cows by increasing $w_i$, or sort the cows by decreasing $w_i$.

Trying both orderings, it turns out that sorting the cows from largest $w_i$ to smallest $w_i$ is provably optimal. There are several ways to show this. One way is to consider any ordering in which the $w_i$ are not ordered largest-to-smallest, and show that swapping two out-of-order cows will improve the solution or leave it the same.

Here is a slightly cleaner proof. We do not explicitly show that the above ordering is optimal, but this can be deduced from the properties we do prove about the ordering.

Given any ordering, we can shuffle the cows who do not join the line to the end of the ordering -- and they will still not join the line in this new ordering. So there is an optimal ordering in which all cows who join the line precede all cows who do not join the line. Now consider any cow $i$ who does join the line, and any cow $j$ who does not join the line. If $w_i < w_j$, we can swap cows $i$ and $j$, and cow $j$ will join the line but cow $i$ will not. So in some optimal ordering, the cows who join the line have the $k$ largest $w_i$ for some $k \leq N$.

Now, the optimum is at most $k$ if and only if $w_{k+1} < k$, where $w_{k+1}$ is the waiting time of the $(k+1)^\text{st}$ cow after sorting largest-to-smallest. This gives our algorithm: after sorting the cows from largest $w_i$ to smallest $w_i$, we can simply scan through the cows for the first $k$ such that $w_{k+1} < k$.

#include <iostream> #include <fstream> #include <algorithm> using namespace std; int N, W[100000]; int main(void) { ifstream fin("lemonade.in"); ofstream fout("lemonade.out"); fin >> N; for (int i=0; i<N; i++) fin >> W[i]; sort(W,W+N); int i, num_in_line=0; for (i=N-1; i>=0; i--) { if (W[i] < num_in_line) break; num_in_line++; } fout << num_in_line << "\n"; return 0; }