Problem:
Score : 200 points

Problem Statement

La Confiserie d'ABC sells cakes at 4 dollars each and doughnuts at 7 dollars each.
Determine if there is a way to buy some of them for exactly N dollars. You can buy two or more doughnuts and two or more cakes, and you can also choose to buy zero doughnuts or zero cakes.

Constraints

N is an integer between 1 and 100, inclusive.

Input

Input is given from Standard Input in the following format:

N

Output

If there is a way to buy some cakes and some doughnuts for exactly N dollars, print Yes; otherwise, print No.

Sample Input 1

11

Sample Output 1

Yes

If you buy one cake and one doughnut, the total will be 4 + 7 = 11 dollars.

Sample Input 2

40

Sample Output 2

Yes

If you buy ten cakes, the total will be 4 \times 10 = 40 dollars.

Sample Input 3

3

Sample Output 3

No

The prices of cakes (4 dollars) and doughnuts (7 dollars) are both higher than 3 dollars, so there is no such way.

Solution in Kotlin (1.0.0) (judge verdict: Accepted)

fun main(args: Array<String>) {
  val N = readInt()
  for (i in 0..20) {
    val d = (N - 4 * i)
    if (d < 0) break
    if (d % 7 == 0) {
      println("Yes")
      return
    }
  }
  println("No")
}

fun gcd(a: Int, b: Int): Int = if (a < b) gcd(b, a) else if (b == 0) a else gcd(b, a % b)
fun gcd(a: Long, b: Long): Long = if (a < b) gcd(b, a) else if (b == 0L) a else gcd(b, a % b)
fun lcm(a: Int, b: Int): Int = (a * b) / gcd(a, b)
fun lcm(a: Long, b: Long): Long = (a * b) / gcd(a, b)

fun readString() = readLine()!!
fun readInt() = readLine()!!.toInt()
fun readInts() = readLine()!!.split(" ").map { it.toInt() }
fun readLongs() = readLine()!!.split(" ").map { it.toLong() }