C Programming

You've answered your own question, I'm afraid. A 16-bit memory address requires 2 bytes of storage (8 bits * 2 bytes = 16 bits). Note that, without using XMS or EMS, you can only address 2^20 bytes of memory, with the last 384k or so being restricted (BIOS, video, etc) unless you enable the A20 line (it actually wraps from FFFF:0010 back to 0000:0000). Also, you should note that 16-bit pointers work because of a "segment:offset" feature; your current "data segment" determines where in memory your pointer actually points to. If you need a 4-byte pointer, use a far pointerinstead; these pointers can reference any point in memory (up to the aforementioned 1MB limit).

#include<stdio.h>

main()

{

int n,m,i,max;

printf("How many numbers(n) you going to enter:");

scanf("%d",&n);

printf("Enter the numbers:");

scanf("%d",&m);

max=m;

for(i=2;i<=n;i++)

{

scanf("%d",&m);

if(m>max)

max=m;

}

printf("The Largest Number is %d",max);

}

Here is a code with explanation and dry run hope it can be helpful

http://fahad-cprogramming.blogspot.com/2014/04/find-maximum-and-minimum-number-in.html

To become a Chief Operator, typically one must first gain experience in the relevant industry, often starting in entry-level positions and progressing through roles such as operator or supervisor. Pursuing a degree in a related field, such as engineering or operations management, can enhance qualifications. Additionally, developing strong leadership, problem-solving, and communication skills is essential, as Chief Operators oversee operations and lead teams. Networking and seeking mentorship within the industry can also facilitate career advancement.

The simple (brute-force) way to do it would be something like this:

For every integer i from 2 to n-1 do:

If n modulo i equals 0, output i.

This would be very slow for large n (linear in the size of n, in the best case).

Picot

increment the loop control variable

The letter "D."

Define normalization explain the conditions under which a relation need to be normalized to 2nf and 3nf with the help of an example ?

That is false; the very definition of object-oriented programming is to create objects that model real objects, which places an emphasis on data encapsulation, polymorphic objects, and so on to reduce code complexity, common programming errors, and other problems associated with a non-object-oriented language. The procedures are not nearly as important as the objects that are designed.

20 Nickels in a Dollar

broad side array very important array in this attenna communication

210=1024

int complement (int n) {

return -n;

}

or int complement (int n) {

return ~n+1;

}

both does the same thing.

#include <stdio.h>

The function getchar() returns an int corresponding to the next character in standard input. The value EOF indicates error or end-of-file.

Communication, A designer's goal is to communicate something to all the pople who will see the design.

Or even three. Actually, pseudo-code has no rules.

You mean SQL?

NULL = anything IS NULL

NULL <> anything IS NULL

...

NULL IS NULL = TRUE

NULL IS NOT NULL = FALSE

yes

it is fifth root of 2 = exp (log 2)/5) = 1.1487 = 14.87%

The above applies to Compound Interest:

At Simple Interest I = PTR/100. In this case I = P, T = 5 and R is unknown.

100 x I = I x 5 x R ie 100 = 5 x R so R = 20%

#include<iostream.h>

#include<conio.h>

void main()

{

clrscr();

int n,x,y=1;

cout<<"Please enter x: ";

cin>>x;

cout<<"Please enter n: ";

cin>>n;

//Coding by: Snehil Khanor

//http://WapCPP.blogspot.com

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

y=y*x;

cout<<y;

getch();

}

Anything that is used to determine values during the exchange of goods and services called a unit of account. This field of study is called macroeconomics.

In a "language" containing just 10 letters, there are 10,000 four-letter permutations. It's easy to work out if you simply replace the letters with decimal digits 0-9. The first permutation is 0000, followed by 0001, 0002, 0003, ..., 9997, 9998 and 9999.

int max = a>b?a:b; // set max to the larger of a and b

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