Problem:
﻿Score: 100 points

Problem Statement

In AtCoder city, there are five antennas standing in a straight line. They are called Antenna A, B, C, D and E from west to east, and their coordinates are a, b, c, d and e, respectively.

Two antennas can communicate directly if the distance between them is k or less, and they cannot if the distance is greater than k.

Determine if there exists a pair of antennas that cannot communicate directly.

Here, assume that the distance between two antennas at coordinates p and q (p < q) is q - p.

Constraints

a, b, c, d, e and k are integers between 0 and 123 (inclusive).

a < b < c < d < e

Input

Input is given from Standard Input in the following format:

a
b
c
d
e
k

Output

Print :( if there exists a pair of antennas that cannot communicate directly, and print Yay! if there is no such pair.

Sample Input 1

1
2
4
8
9
15

Sample Output 1

Yay!

In this case, there is no pair of antennas that cannot communicate directly, because:

the distance between A and B is 2 - 1 = 1

the distance between A and C is 4 - 1 = 3

the distance between A and D is 8 - 1 = 7

the distance between A and E is 9 - 1 = 8

the distance between B and C is 4 - 2 = 2

the distance between B and D is 8 - 2 = 6

the distance between B and E is 9 - 2 = 7

the distance between C and D is 8 - 4 = 4

the distance between C and E is 9 - 4 = 5

the distance between D and E is 9 - 8 = 1

and none of them is greater than 15. Thus, the correct output is Yay!.

Sample Input 2

15
18
26
35
36
18

Sample Output 2

:(

In this case, the distance between antennas A and D is 35 - 15 = 20 and exceeds 18, so they cannot communicate directly.
Thus, the correct output is :(.

Solution in C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <math.h>

int main() {
  int t[5];
  int max = 0;
  scanf("%d\n", &t[0]);
  scanf("%d\n", &t[1]);
  scanf("%d\n", &t[2]);
  scanf("%d\n", &t[3]);
  scanf("%d\n", &t[4]);
  double d[5];
  double sum = 0;
  for(int i; i < 5; i++) {
    d[i] = ceil((double)(t[i])/10);
    sum += d[i];
  }
  int sum2 = (int)sum;

  for(int j = 0; j < 5; j++) {
    int a = (10 - (t[j] % 10)) % 10;
    if (a > max) {
     max = a;
   }
  }
  int ans = (sum2*10)-max;
  printf("%d\n", ans);
  return 0;
}