Official Analysis (C++)

For every $15$ numbers, there are only $8$ numbers where the cows moo. (see below)

$[\underline{1}, \underline{2}, 3, \underline{4}, 5, 6, \underline{7}, \underline{8}, 9, 10, \underline{11}, 12, \underline{13}, \underline{14}, 15]$

$[\underline{16}, \underline{17}, 18, \underline{19}, 20, 21, \underline{22}, \underline{23}, 24, 25, \underline{26}, 27, \underline{28}, \underline{29}, 30]$

Therefore, we can just get the distance away $N$ is from the first $8$ by calculating $\lfloor \frac{N}{8} \rfloor \times 15$, since we have to add $15$ for every grouping of $15$.

Now it's just a matter of adding the corresponding first $8$ to the answer.

Implementation

Time Complexity: $\mathcal{O}(1)$

Copy
import sys

sys.stdin = open("moobuzz.in", "r")
sys.stdout = open("moobuzz.out", "w")

N = int(input())

# Lists the first 8.
# 14 is in the zeroth position since if were trying to find the
# eighth number, 8 % 8 = 0.

Copy
#include <bits/stdc++.h>
using namespace std;

int main() {
	int n;
	cin >> n;

	/*
	 * Lists the first 8.
	 * 14 is in the zeroth position since if were trying to find the

Copy
import java.io.*;
import java.util.*;

public class MooBuzz {
	public static void main(String[] args) throws IOException {
		BufferedReader br = new BufferedReader(new FileReader("moobuzz.in"));
		int N = Integer.parseInt(br.readLine());
		br.close();

		int[] first8 = {14, 1, 2, 4, 7, 8, 11, 13};