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solmaris [256]
3 years ago
12

Create a static method called fillArray, which takes an integer array as an input parameter, along with an integer initial value

. It should then fill the array with values starting from the initial value, and counting up by 1s. For example, if the data array has length 5, and if the initialValue is 3, then after execution the array should hold this sequence: 3 4 5 6 7. public static void fillArray (int[] data, int initialValue) {
Computers and Technology
1 answer:
JulijaS [17]3 years ago
3 0

Answer:

public class print{

   

   public static void fillArray(int[] arr, int initialValue){

       

       int n = arr.length;

      for(int i=0;i<n;i++){

          arr[i] = initialValue++;

      }

       for(int i=0;i<n;i++){

           System.out.print(arr[i]+" ");

      }

       

   }

    public static void main(String []args){

       

        int[] array = new int[5];

        int initialValue =3;

       

        fillArray(array,initialValue);

       

    }

}

Explanation:

Create the function with two parameter first is array and second is integer.

Then, declare the variable and store the size of array.

Take the for and fill the array from the incremented value of initialValue  by 1 at every run of loop.

After loop, print the element of the array.

Create the main function which is used for calling the function and also declare the array with size 5 and initialValue with 3. After that, call the function with this argument.

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Write a program that prompts the user to enter in a postive number. Only accept positive numbers - if the user supplies a negati
marusya05 [52]

Answer:

The c++ program to check prime numbers is shown below.

#include <iostream>

using namespace std;

int main() {

   int num, prime=0;

   do

   {

       cout<<"Enter a positive number."<<endl;

       cin>>num;

       if(num<1)

       {

           cout<<"Invalid number. Enter a positive number"<<endl;

           cin>>num;

       }

   }while(num<1);

   

   if(num==1 || num==2 || num==3)

       cout<<num<<" is a prime number."<<endl;

   else if(num%2 == 0)

       cout<<num<<" is not a prime number."<<endl;

   else

   {

       for(int k=3; k<num/2; k++)

       {

           if(num%k == 0)

               prime++;

       }    

   if(prime>1)

       cout<<num<<" is not a prime number."<<endl;

   else

       cout<<num<<" is a prime number."<<endl;

   }

}

OUTPUT

Enter a positive number.

-7

Invalid number. Enter a positive number

0

Enter a positive number.

79

79 is a prime number.

Explanation:

The user input is validated for positivity. A do while loop along with an if statement is implemented for verification.

do

   {

       cout<<"Enter a positive number."<<endl;

       cin>>num;

       if(num<1)

       {

           cout<<"Invalid number. Enter a positive number"<<endl;

           cin>>num;

       }

   }while(num<1);

The test for prime number is done by using multiple if else statements.

If user inputs 1, 2, or 3, message is displayed.

Else If user inputs an even number, message is displayed for not prime. This is done by taking modulo of the number upon division by 2.

Else if user inputs neither an even number nor a number less than 3, the modulus of the number is taken with divisors beginning from 3 up to half of the input number.

For this, an integer variable prime is initialized to 0. A number can be completely divisible by itself or by its factors.

If the number is divisible by any of the divisors, value of variable prime is increased by 1. If value of prime is greater than 1, this means that the user input is divisible by more than one divisor. Hence, the given number is not a prime number.

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