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Law Incorporation [45]
3 years ago
8

Write a function SwapArrayEnds() that swaps the first and last elements of the function's array parameter. Ex: sortArray = {10,

20, 30, 40} becomes {40, 20, 30, 10)
#include
/* Your solution goes here
*/ int main(void) {
const int SORT_ARR_SIZE = 4;
int sortArray[SORT_ARR_SIZE];
int i;
int userNum;
for (i = 0; i < SORT_ARR_SIZE; ++i) {
scanf("%d", &sortArray[i]);
}
SwapArrayEnds(sortArray, SORT_ARR_SIZE);
for (i = 0; i < SORT_ARR_SIZE; ++i) {
printf("%d ", sortArray[i]);
}

Computers and Technology
1 answer:
sergey [27]3 years ago
4 0

Answer:

Here is the SwapArrayEnds()

//function with two parameters; the array and array size

void SwapArrayEnds(int sortArray[],int SORT_ARR_SIZE){  

int temp;   //temporary variable to hold the first element of sortArray

if(SORT_ARR_SIZE > 1){   //if sortArray size is greater than 1

temp = sortArray[0];   // set value of temp to the first element of sortArrray

sortArray[0] = sortArray[SORT_ARR_SIZE-1];   // set the value of first element of sortArray to the last element of sortArray

sortArray[SORT_ARR_SIZE-1] = temp;  }  } // set last element  value to temp

Explanation:

Lets understand the working of the above function with the help of an example. Suppose the sortArray has the following elements: {10,20,30,40}. So the size of this array is 4.

if(SORT_ARR_SIZE > 1) has an if statement which checks if the array size is greater than 1. This condition evaluates to true as 4 > 1. So the body of if condition executes.

Next the statement temp = sortArray[0];   sets the value of temp variable to the value of first element of sortArray. As sortArray[0] means the element at 0-th index position of the sortArray. So using the above example the first element of the array is 10. So temp = 10.

Next the statement sortArray[0] = sortArray[SORT_ARR_SIZE-1];   sets the last element of the sortArray to the position of first element of the sortArray.

SORT_ARR_SIZE-1 is the position of last element of the array. As the size of sortArray is 4, so 4-1 = 3 which points to the last element. Now this element is moved to the first position of the array. This means that the value of first element of array becomes 40 as the element at 3-th index of the sortArray is 40.

Next the statement sortArray[SORT_ARR_SIZE-1] = temp moves the value in temp variable to the last position of the sortArray. As we know that the value of temp= 10. So this element 10 of the sortArray is positioned at SORT_ARR_SIZE-1 position of sortArray which is 4-1 = 3-th index of sortArray. This simply means that 10 is moved to the last position of sortArray.

So now the final sortArray becomes: 40,20,30,10 as the first and last element of the array are swapped using above function SwapArrayEnds().

The program along with the output is attached in screenshot.

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2 years ago
in java how do i Write a program that reads a set of integers, and then prints the sum of the even and odd integers.
hammer [34]

Answer:

Here is the JAVA program:

import java.util.Scanner; //to take input from user

public class Main{

public static void main(String[] args) { //start of main function

Scanner input = new Scanner(System.in); // creates Scanner class object  

       int num, integer, odd = 0, even = 0; //declare variables

       System.out.print("Enter the number of integers: "); //prompts user to enter number of integers

       num = input.nextInt(); //reads value of num from user

       System.out.print("Enter the integers:\n"); //prompts user to enter the integers

       for (int i = 0; i < num; i++) { //iterates through each input integer

           integer = input.nextInt(); // reads each integer value

               if (integer % 2 == 0) //if integer value is completely divisible by 2

                   even += integer; //adds even integers

               else //if integer value is not completely divisible by 2

                   odd += integer;  } //adds odd integers

           System.out.print("Sum of Even Numbers: " + even); //prints the sum of even integers

           System.out.print("\nSum of Odd Numbers: " + odd);//prints the sum of odd integers

           }}

Explanation:

The program is explained in the comments mentioned with each line of the code. I will explain the logic of the program with the help of an example.

Suppose user wants to input 5 integers. So,

num = 5

Suppose the input integers are:

1, 2 , 3, 4, 5

for (int i = 0; i < num; i++)  is a for loop that has a variable i initialized to 0. The condition i<num is true because i=0 and num=5 so 0<5. Hence the statements inside body of loop execute.

At first iteration:

integer = input.nextInt(); statement reads the value of input integer. The first integer is 1

if (integer % 2 == 0) checks if integer is completely divisible by 2. The modulo operator is used which returns the remainder of the division and if this remainder is equal to 0 then it means that the integer is completely divisible by 2 and if the integer is completely divisible by 2 then this means the integer is even. 1%2 returns 1 so this means this condition evaluates to false and the integer is not even. So the else part executes:

odd += integer;   this becomes:

odd = odd + integer

odd = 1

Now the value of i is incremented to 1 so i=1

At second iteration:

integer = input.nextInt(); statement reads the value of input integer. The first integer is 2

if (integer % 2 == 0) checks if integer is completely divisible by 2. 2%2 returns 0 so this means this condition evaluates to true and the integer is even. So

even += integer;   this becomes:

even= even+ integer

even = 2

Now the value of i is incremented to 1 so i=2

At third iteration:

integer = input.nextInt(); statement reads the value of input integer. The first integer is 3

if (integer % 2 == 0) checks if integer is completely divisible by 2. 3%2 returns 1 so this means this condition evaluates to false and the integer is odd. So

odd += integer;   this becomes:

odd= odd + integer

odd= 1 + 3

odd = 4

Now the value of i is incremented to 1 so i=3

At fourth iteration:

integer = input.nextInt(); statement reads the value of input integer. The first integer is 4

if (integer % 2 == 0) checks if integer is completely divisible by 2. 4%2 returns 0 so this means this condition evaluates to true and the integer is even. So

even+= integer;   this becomes:

even = even + integer

even = 2 + 4

even = 6

Now the value of i is incremented to 1 so i=4

At fifth iteration:

integer = input.nextInt(); statement reads the value of input integer. The first integer is 5

if (integer % 2 == 0) checks if integer is completely divisible by 2. 5%2 returns 1 so this means this condition evaluates to false and the integer is odd. So

odd+= integer;   this becomes:

odd= odd+ integer

odd= 4 + 5

odd = 9

Now the value of i is incremented to 1 so i=5

When i=5 then i<num condition evaluate to false so the loop breaks. The next two print statement prints the values of even and odd. So

even = 6

odd = 9

Hence the output is:

Sum of Even Numbers: 6                                                                                                           Sum of Odd Numbers: 9  

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Answer:

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using namespace std;

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int main() {

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// print the output

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return 0;

}

Explanation:

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Output:

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enter birth year:2000                                                                                                      

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Answer:

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