Answer:
Explanation:
Since all of the items in the array would be integers sorting them would not be a problem regardless of the difference in integers. O(n) time would be impossible unless the array is already sorted, otherwise, the best runtime we can hope for would be such a method like the one below with a runtime of O(n^2)
static void sortingMethod(int arr[], int n)
{
int x, y, temp;
boolean swapped;
for (x = 0; x < n - 1; x++)
{
swapped = false;
for (y = 0; y < n - x - 1; y++)
{
if (arr[y] > arr[y + 1])
{
temp = arr[y];
arr[y] = arr[y + 1];
arr[y + 1] = temp;
swapped = true;
}
}
if (swapped == false)
break;
}
}
There are al types of dogs what specific breed are you looking for
Answer:
Following are the code in the C Programming Language.
//set integer datatype variable
int score;
//check condition is the score is in the range of 0 to 100
if(score > 0 && score < 100){
//print if condition is true
printf("Valid test scores");
}else{
//otherwise print the following string.
printf("test scores are Invalid");
}
Explanation:
<u>Following are the description of the code.</u>
In the following code that is written in the C Programming Language.
- Set an integer data type variable i.e., score.
- Then, set the if conditional statement to check the condition is the variable "score" is greater than 0 and less the 100.
- If the following statement is true then print "Valid test scores".
- Otherwise, it print "test scores are Invalid".
<u>Explanation:</u>
Hey there! you need not to panic about it ,your program didn't have Driver program i.e main program! the correct & working code is given below:
// C++ program to count even digits in a given number .
#include <iostream>
using namespace std;
// Function to count digits
int countEven(int n)
{
int even_count = 0;
while (n > 0)
{
int rem = n % 10;
if (rem % 2 == 0)
even_count++;
n = n / 10;
}
cout << "Even count : "
<< even_count;
if (even_count % 2 == 0 )
return 1;
else
return 0;
}
// Driver Code
int main()
{
int n;
std::cin >>n;
int t = countEven(n);
return 0;
}
There are two ways to convert from hexadecimal to denary gcse method. They are:
- Conversion from hex to denary via binary.
- The use of base 16 place-value columns.
<h3>How is the conversion done?</h3>
In Conversion from hex to denary via binary:
One has to Separate the hex digits to be able to know or find its equivalent in binary, and then the person will then put them back together.
Example - Find out the denary value of hex value 2D.
It will be:
2 = 0010
D = 1101
Put them them together and then you will have:
00101101
Which is known to be:
0 *128 + 0 * 64 + 1 *32 + 0 * 16 + 1 *8 + 1 *4 + 0 *2 + 1 *1
= 45 in denary form.
Learn more about hexadecimal from
brainly.com/question/11109762
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