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: Accepted)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>


void inp_str(int instr[]){
char *ptr;
int z,i,count;
char str[100]={'\0'};
char *ln;

fgets(str, 100, stdin); // 入力："123456\n"
ln = strchr(str, '\n'); /* 改行文字を検索 */
if (ln != NULL) { /* 改行が読み取られていたかどうか */
   *ln = '\0';    /* 改行文字を終端文字に置き換える */
}
else {            /* 入力ストリーム上に文字が残ってる場合 */
   while (1) {    /* 改行文字が読み取られるまで空読みする */
       z = getchar();
      if (z == '\n' || z == EOF) break;
   }
}
	count=0;
    ptr = strtok(str, " ");
    i = atoi(ptr);
    instr[count]=i;
    count=count+1;
    // 2回目以降
    while(ptr != NULL) {
        // strtok関数により変更されたNULLのポインタが先頭
        ptr = strtok(NULL, " ");
        // ptrがNULLの場合エラーが発生するので対処
        if(ptr != NULL) {
		    i = atoi(ptr);
    instr[count]=i;
    		count=count+1;
        }
    }
}

int main(void) {

//int input[120];
//char input1[120];
//int j,count;
int a,b,c,d,e,k;

int num[5];

//inp_str(input);
//N=input[0];

	//printf("%d\n",N);

scanf("%d",&num[0]);
scanf("%d",&num[1]);
scanf("%d",&num[2]);
scanf("%d",&num[3]);
scanf("%d",&num[4]);
scanf("%d",&k);

/*printf("%d\n",a);
printf("%d\n",b);
printf("%d\n",c);
printf("%d\n",d);
printf("%d\n",e);
printf("%d\n",k);*/

int B;
int C;
int ans;
int flag=0;


for(B=0;B<5;B++){
for(C=0;C<5;C++){

ans=num[B]-num[C];

if(ans>k) {
	flag=1;
}

}
}

if(flag==1) {
	printf(":(\n");
}else{
printf("Yay!\n");
}                        
    return 0;
}