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nikklg [1K]
3 years ago
10

The elements of an integer-valued array can be initialized so that a[i] == i in a recursive fashion as follows: An array of size

0 is already initialized; Otherwise set the last element of the array to n-1 (where n is the number of elements in the array, for example, an array of size 3 will have its last element -- index 2-- set to 2; and initialize the portion of the array consisting of the first n-1 elements (i.e., the other elements of the array) Write a void method named init that accepts an integer array, and the number of elements in the array and recursively initializes the array so that a[i] == i.
Computers and Technology
1 answer:
blondinia [14]3 years ago
4 0

Answer:

public static void init(int[] arr, int n) {

    if (n==0)

        arr[0] = 0;

    else {

        arr[n-1] = n - 1;

        init(arr, n-1);

    }

}

Explanation:

Create a method called init that takes two parameters, an array and the number of elements in the array

When n reaches 0, set the first element to 0 (This is a base for our recursive method)

Otherwise, set the element in index i to i

Call the init inside the init, this is the recursion part, with same array but decrease the number of elements by 1 (We decrease the number of element by 1 in each time so that it goes through all the elements in the array)

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Consider two vectors that are NOT sorted, each containing n comparable items. How long would it take to display all items (in an
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". The further explanation is given below.

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7 0
3 years ago
Create a 4x5 matrix with ones everywhere and zeros on the last row.
Stells [14]

Answer:

#include <iostream>

using namespace std;

int main() {

   int a[4][5];//declaring a matrix of 4 rows and 5 columns.

   for(int i=0;i<4;i++)

   {

       for(int j=0;j<5;j++)

       {

           if(i==3)//initializing last row as 0.

           {

               a[i][j]=0;

           }

           else//initializing last row as 1.

           {

               a[i][j]=1;

           }

       }

   }

   for(int i=0;i<4;i++)

   {

       for(int j=0;j<5;j++)

       cout<<a[i][j]<<" ";//printing the matrix.

       cout<<endl;

   }

return 0;

}

Output:-

1 1 1 1 1  

1 1 1 1 1  

1 1 1 1 1  

0 0 0 0 0

Explanation:

I have created a matrix of size 4 rows and 5 columns.I have used for loops to fill the array.To fill the last row with 0 i have used if statement.else we are filling it with 1.

7 0
3 years ago
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