(Analysis by Nick Wu)

For this problem, we can try unlocking all possible starting doors and seeing how far each cow travels.

One tricky implementation detail here is that the barn is circular, so if we store the data in an array and we assume that the index to inspect is simply the sum of the unlocked door index and the distance to travel, we might exceed the bounds of the array. Because the barn is circular, the barn that would be at location N is actually the barn at location 0, so we can just take the remainder of the sum modulo N.

Here is my Java code.

import java.io.*;
import java.util.*;
public class cbarn {
public static void main(String[] args) throws IOException {
// initialize file I/O
PrintWriter pw = new PrintWriter(new BufferedWriter(new FileWriter("cbarn.out")));
int[] cows = new int[n];
// read in how many cows need to be in each room
for(int i = 0; i < n; i++) {
}
// the answer cannot exceed N * N * 100, since there are at most 100N cows, each of which can move at most N
int ans = n * n * 100;
for(int unlock = 0; unlock < n; unlock++) {
// assume we unlock the door at index "unlock", compute the distance all cows travel
int currentDistance = 0;
for(int offset = 0; offset < n; offset++) {
// count how many cows have to walk a distance of "offset"
currentDistance += offset * cows[(unlock+offset)%n];
}