Consider this data sequence: "3 11 5 5 5 2 4 6 6 7 3 -8". any value that is the same as the immediately preceding value is considered a consecutive duplicate. in this example, there are three such consecutive duplicates: the 2nd and 3rd 5s and the second 6. note that the last 3 is not a consecutive duplicate because it was preceded by a 7.

write some code that uses a loop to read such a sequence of non-negative integers , terminated by a negative number. when the code exits the loop it should print the number of consecutive duplicates encountered. in the above case, that value would be 3.

Following is the source code required:

int entry_1 ,entry_2 = -1, consecutive_duplicates = 0;

do {

cin >> entry_1;

if ( entry_2 == -1)

{

entry_2 = entry_1;

}else

{

if ( entry_2  == entry_1 )

consecutive_duplicates++;

else

entry_2  = entry_1;

}

}

while(entry_1 > 0 );

cout << consecutive_duplicates;

Explanation:

Following is the explanation for given code:

3 integers are declared as entry_1, entry_2  and consecutive_duplicates.entry_2 = -1 (it will tell that the integers are non-zero)user will enter the value for entry_1The loop will be applied which will check for the value of entry_2. If entry_2 = -1, the value of entry_1 will be stored in entry_2 else it the value of entry_2 is already equal to entry_1, the integer consecutive_duplicate in increased by one.Now the value of entry_1 is stored in entry_2 , so that next entity might be checked.In the end, while the entry_1 is greater than zero (unless the last (negative)element of array is reached), print the integer consecutive_duplicates.

i hope it will help you!

The answer is b; as a lab tech, you’re more behind the scenes and don’t interact with the patients.