What is the value of n after this code runs?
1 Point
int n = 0;...
What is the value of n after this code runs?
1 Point
int n = 0;
int j = 7;
if ((j!=0) && (n < 25))
{
n = 1;
if (j > 4)
{
n = 2;
}
else
{
n = 3;
}
}
else
{
n = 4;
if (j%2 >= 2)
{
n = 5;
}
else
{
n = 6;
}
}
1. What will be the value of numbers[1] after the following code
is executed?
int[] numbers = {22, 33, 44};
for(int k = 0; k < 3; k++)
{
numbers[k] = numbers[k] + 5;
}
}
2. What will be the results of the following code?
final int ARRAY_SIZE = 5;
double[] x = new double[ARRAY_SIZE];
for(int i = 1; i <= ARRAY_SIZE; i++)
{
x[i] = 10.0;
}
3.What is the value of numbers [3] after the following line...
n following code if variable n=0, a=5 and b=10 then what will be
the value of variable ‘n’ after while loop executed. Show your
working as well.
while(n<=(a^b)) { n++; } printf("%d",n)
int value = 1;
do
{
if (value % 2 == 0)
cout << value << " ";
value = value + 1;
}
while (value % 7 != 0);
cout << "\n" << value;
What will be displayed?
Consider the following function:
int counter( int n)
{ int s = 0;
for ( int i = 0; i < n; i++)
for
( int j = i; j > 0; j--)
s
= s + 1;
return s;
}
Part (a): How many times "s=s+1;" will be executed? Determine a
precise count. Show all your work.
Part (b): What is the time complexity of the "counter" function
in terms of Big-O notation? Justify and show all
your work.
Consider the following C code that outlines Fibonacci
function
int fib (int n) {
if (n == 0) return 0;
else if (n==1) return 1;
else return fib(n-1) + fib (n-2);
}
For this programming assignment, write and test an ARMv8 program
to find Fibonacci (n).
You need to write a main function that calls the recursive fib
function and passes an argument n.
The function fib calls itself (recursively) twice to compute
fib(n-1) and fib (n-2). The input to...
n = int(input("Enter any number: "))
sum1 = 0
for i in range(1, n):
if(n % i == 0):
sum1 = sum1 + i
if (sum1 == n):
print("The number is a Perfect number!")
else:
print("The number is not a Perfect number!")
taking the code above, how would i create a new function that
calls the program above and finds all perfect numbers within a
certain range? the function would accept two parameters, the two
ranges, and use a for...
// 1.
System.out.println("1.");
int max = 5;
for (int n = 1; n <= max; n++) {
System.out.println(n);
}
System.out.println();
// 2.
System.out.println("2.");
int total = 25;
for (int number = 1; number <= (total / 2); number++) {
total = total - number;
System.out.println(total + " " + number);
}
System.out.println();
// 3.
System.out.println("3.");
for (int i = 1; i <= 2; i++) {
for (int j = 1; j <= 3; j++) {
for (int k = 1;...
Write a full MIPS assembly code for: Function Compare(int A[][],
int B[][], int M, int N, int i, int j, int k, int l). This function
compares the elements of the array A placed between (i,j) and (k,l)
with the corresponding elements of the array B placed between the
same indexes. The function returns the number of elements of both
matrices that coincide in the same position (c,d). That is, it is
necessary to check each element A[c,d] with the...