(Analysis by Nick Wu)

We'll start by focusing on the subtask where $N = 1$. Analyzing some small cases, we see that Farmer John appears to win if and only if $a_1$ is not divisible by $4$. To prove this, we note that if $a_1$ is divisible by $4$, then Farmer John must change the value of $a_1$ by some value that is not divisible by $4$. Farmer Nhoj can then decrease it by either $1$, $2$, or $3$ to return it to a smaller value that is divisible by $4$. Eventually, Farmer Nhoj will decrease $a_1$ to zero with this strategy.

This therefore defines the winning strategy for both players - pick some valid value that is equivalent to $a_1 \pmod{4}$ and decrease $a_1$ accordingly. If $a_1$ is divisible by $4$, you have lost.

Before moving on to the case where $N > 1$, we ask the following question - in how many turns will the game end if $N = 1$, the loser wishes to maximize the number of turns the game takes, while the winner wishes to minimize the number of turns the game takes? When $a_1$ is even, we can show that the answer is exactly $\frac{a_1}{2}$ by induction. Otherwise, the winning player wants to pick the largest prime that is equivalent to $a_1 \pmod{4}$. If that prime is $p$, then it will take $1 + \frac{a_1 - p}{2}$ turns.

We are now prepared to solve the problem in full. We start by precomputing all primes less than $5 \cdot 10^6$ using the sieve of Eratosthenes. For each value, we keep track of how many turns it would take for the value to get to zero. Per the formula in the previous paragraph, we track how many turns it would take to get a given room to become empty. Among all rooms that share the minimum number of turns to become empty divided by two, the first one determines which farmer wins.

Yuval Vaknin's C++ code:

#include <iostream>
using namespace std;

const int mx = 5000005;
int min_turns[mx] = {0, 1};
bool composite[mx] = {false};
int max_mod4[4] = {2, 1, 2, 3};

int main() {
for(int i = 2; i < mx; i++) {
if(!composite[i]) {
for(int j = i; j < mx; j += i) {
composite[j] = true;
}
max_mod4[i % 4] = i;
}
min_turns[i] = (i - max_mod4[i % 4]) / 2 + 1;
}

int t; cin >> t;
while(t--) {
int n; cin >> n;
int ans = mx;
for(int i = 0; i < n; i++) {
int a_i; cin >> a_i;
if(min_turns[a_i] / 2 < ans / 2) ans = min_turns[a_i];
}
if(ans & 1) cout << "Farmer John" << endl;
else cout << "Farmer Nhoj" << endl;
}
}


Danny Mittal's Java code:

import java.io.BufferedReader;
import java.io.IOException;
import java.util.StringTokenizer;

public class CircularBarn {
public static final int MAXVAL = 5000000;

public static void main(String[] args) throws IOException {
boolean[] isPrime = new boolean[MAXVAL + 1];
for (int p = 1; p <= MAXVAL; p++) {
isPrime[p] = true;
}
for (int p = 2; p <= MAXVAL; p++) {
if (isPrime[p]) {
for (int k = 2 * p; k <= MAXVAL; k += p) {
isPrime[k] = false;
}
}
}
int[] lastPrimes = new int[4];
int[] steps = new int[MAXVAL + 1];
for (int k = 1; k <= MAXVAL; k++) {
if (k % 2 == 0) {
steps[k] = k / 2;
} else {
if (isPrime[k]) {
lastPrimes[k % 4] = k;
}
steps[k] = 1 + steps[k - lastPrimes[k % 4]];
}
}

StringBuilder out = new StringBuilder();
for (int t = Integer.parseInt(in.readLine()); t > 0; t--) {
int[] rooms = new int[n];
for (int j = 0; j < n; j++) {
rooms[j] = Integer.parseInt(tokenizer.nextToken());
}

boolean johnWins = true;
int minSteps = MAXVAL + 1;
for (int j = 0; j < n; j++) {
int stepsHere = steps[rooms[j]] / 2;
if (stepsHere < minSteps) {
minSteps = stepsHere;
johnWins = steps[rooms[j]] % 2 == 1;
}
}
out.append("Farmer ").append(johnWins ? "John" : "Nhoj").append('\n');
}
System.out.print(out);
}
}


My Python 3 code:

MAXV = 5000000
sieve = [False] * (MAXV + 1)
i = 2
while i * i < len(sieve):
if not sieve[i]:
for j in range(i*i, len(sieve), i):
sieve[j] = True
i += 1

def solve():
n = int(input())
l = [int(x) for x in input().split()]
moves = 1e9
for i, x in enumerate(l):
if x % 2 == 0:
numMoves = x // 2
else:
cand = x
while sieve[cand]:
cand -= 4
numMoves = (x - cand) // 2 + 1
if numMoves // 2 < moves // 2:
moves = numMoves
if moves == 1:
break
if moves%2:
print("Farmer John")
else:
print("Farmer Nhoj")

t = int(input())
for _ in range(t):
solve()