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3 years ago

Write an algorithm which gets a number A, if it is even, prints even, and if it is odd prints odd.

Computers and Technology

1 answer:

juin [17]3 years ago

7 0

Answer:

The algorithm to find A is even or odd.

input A.
Check the remainder on diving by 2 by A%2.
If remainder is 1 then A is odd Print(Odd).
If remainder is 0 print(Even).

Explanation:

To check if the number is even or odd we use modulo operator(%).Which gives the remainder on dividing.So if we do this A%2 it will give the remainder that will come out on dividing the value of A by 2.

So if the remainder comes out is 1 then the number is odd and if the remainder is 0 then the number is odd.

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The sum of all the minterms of a boolean function of n variables is equal to 1.

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The answers are as follows:

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= A'(B'C' + B'C + BC' + BC) + A((B'C' + B'C + BC' + BC)

= (A' + A)(B'C' + B'C + BC' + BC) = B'C' + B'C + BC' + BC

= B'(C' + C) + B(C' + C) = B' + B = 1

b) F(x1, x2, x3, ..., xn) = ∑mi has 2n/2 minterms with x1 and 2n/2 minterms

with x'1, which can be factored and removed as in (a). The remaining 2n1

product terms will have 2n-1/2 minterms with x2 and 2n-1/2 minterms

with x'2, which and be factored to remove x2 and x'2, continue this

process until the last term is left and xn + x'n = 1

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Answer:

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2 years ago

Array Challenge Have the function ArrayChallenge (arr) take the array of numbers stored in arr and return the string true if any

vampirchik [111]

Answer:

The code is given as follows,

Explanation:

Code:

#include <stdio.h>

#include <string.h>

int n; //to store size of array

char* ArrayChallenge(int arr[]) //function returns string true or false

{

int i, j;

int sum = 0;

for(i = 0; i < n; i++)

{

sum = sum + arr[i]; //count sum

}

for(i = 0; i < n; i++)

{

for(j = i+1; j < n; j++) //loop for every two elements in array

{

if(arr[i]*arr[j] > 2*sum) //check if proudct of two elements > 2 times sum

{

printf("\n%d x %d = %d, 2xSum = %d\n", arr[i], arr[j], arr[i]*arr[j], 2*sum);

return "true"; //If proudct of two elements > 2 times sum. return true

}

return "false"; // If proudct of two elements < 2 times sum. return false

}

int main()

{

printf("\nEnter size of array: ");

scanf("%d", &n); //read size of array

int A[n]; //array of size n

printf("\nEnter array elements: ");

int i;

for(i = 0; i < n; i++)

{

scanf("%d", &A[i]); //read array from stdin

}

printf("%s\n",ArrayChallenge(A)); //ccall function and print answer

return 0;

}

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3 years ago