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choli [55]
3 years ago
13

Write a function, factors, that takes an integer n, and returns a list of values that are the positive divisors of n. Note: 0 is

not a divisor of any integer, 1 divides every number, and n divides itself.
Computers and Technology
1 answer:
Helen [10]3 years ago
4 0

Answer:

public int[] factors(int n)

{

int arr[]=new int[n];

int count=0;

for(int i=1;i<=n;i++)

if(n%i==0)

{

arr[count]=i;

count++;

}

return arr;

}

Explanation:

Factors are the numbers you multiply to get another number. For instance, factors of 15 are 3 and 5, because 3×5 = 15.

In the program , we have an array of length N to store the factors.

Since we don't know the number of factors of N, so we have taken size of upper bound N to store list of factors in the array arr.

Within the loop the number gets divide by Integers from 1 to that number until result is 0. If it is 0, that number is stored in array.

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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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