C Programming

Mainly, we store values in them.

If there are two digits than the first one is ten times its value.

So as an algebraic equation we get

10A + B - 3 = 7(A x B)

This can be simplified to

10A + B - 3 = 7A + 7B

and this reduces to

10A - 7A = 7B - B +3

or

3A =6B +3

A solution to this is

3(3) = 6(2) +3

Which would make the original equation

32 = 7(3+2)-3

#include<stdio.h>

#define size 100

int main(){

int x[size], sum=0;

float average;

for(int i=0;i<size;i++){

scanf("%d",x[i]);

sum+=x[i];

}

average=sum/size;

printf("The Average is :\t%d",average);

return 0;

}

even

A prime number is a number that has no factors other than 1 and itself. For instance, 1, 2 and 3 are prime numbers. However, 4 isn't because it can also be divided evenly by 2.

Although a complicated floating-point function can be written to test if the division of a number is even, the C modulus operator (%) is far quicker. If 'n' modulus 'm' is zero, then the division is even - there is no fractional component.

Using this rule, the pseudo-code for testing if a number is prime is as follows:

- if number is zero, return false

- if number is less than 4, return true since 1, 2 and 3 are primes

- iterate a variable 'c' from 2 until the number 'n' minus 1

- if n modulus c is zero, return false, since 'n' has 'c' as a factor

- return true

All that remains is converting the above code to C (or any other programming language) using inline code or a function.

I don't know about C but in C++ I would have the compiler do a bool statement for return type (int, float, long, short, signed short, signed long) and have make them true and have it output. I'm still learning myself so this might not be accurate but it's what I would do.

If you wanted to ask 'Which is the only numeric value that can be assigned to a pointer?', then the answer would be: 0.

Okay, I'm pretty sure that 864 binary is 30 hexadecimal. - RG

11100011

int is_prime(int n){

int flag=1;

if(n==1)

return 0;

int root=(int)sqrt(n);

while(root!=1){

if(n%root==0){

flag=0;

break;

}

root--;

}

return flag;

}

#include<conio.h>

#include<stdio.h>

void main (void)

{

int a,b,c,d;

printf("enter the max limit : ");

scanf("%d",&a);

for(b=1;b<=a;b++)

{

for (c=2;c<b;c++)

{d=b%c;

if (d==0) break;}

if (d!=0) printf("%d\n",b);}

getch();

}

by WAli Ahsan

Ch-007

yes

#include
#include
void main()
{
int n,i;
clrscr();
printf("Enter a Number:");
scanf("%d",&n);
printf("\n\nPrime Factors of %d is: ",n);
for(i=2;i<=n;i++)
{
if(n%i==0)
{
printf("%d,",i);
n=n/i;
i--;
if(n==1)
break;
}
}
getche();
}




this program will find the prime factors of a given number


any assistance


mail at :- devilllcreature@yahoo.com


thank you

This is typically done by importing math.h, and calling the sqrt function.

I'll assume the list is supplied as arguments to the program:

#include <stdlib.h>

#include <stdio.h>

main (argc, argv)

int argc;

char **argv;

{

int c = 0, mn, mx, z;

while (--argc)

{

z = atoi(*(++argv));

if (!c++) mx = mn = z;

else if (z > mx) mx = z;

else if (z < mn) mn = z;

}

if (c) printf("%d numbers, min = %d, max = %d\n", c, mn, mx);

else printf("No arguments given: please supply a list of numbers as arguments\n");

}

Any error trapping is left as an exercise.

pongada punda vayanungala ..................

Not tested:

void printPrimes(int n) {

int a = 0, b = 0;

float c = 0.0f;

bool isp = true;

int primes = 0;

while(true) {

a++;

do {

b++;

if(b > 1) {

c = (float)a/(float)b;

if(c = (int)c)

if(b != a) {

isp = false;

break;

}

else

isp=true;

}

} while(b<a);

if(isp) {

primes++;

std::cout << a << endl;

}

if(primes > n)

break;

b = 0;

c = 0;

isp = true;

}

}

void main() {

printPrimes(10);

}

import java.io.*; public class p_factor

{

public static void main(String args[])throws IOException

{

int arr[]=new int[50];

int n,i,j=0,k,f=0;

BufferedReader br=new BufferedReader(new InputStreamReader(System.in));

System.out.println("Enter a number");

n=Integer.parseInt(br.readLine());

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

{

if(n%i==0)

{

arr[j]=i;

j++;

}

}

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

{

for(k=2;k<arr[i];k++)

{

if(arr[i]%k==0)

{

f=1;

}

}

if(f==0)

{

System.out.println(arr[i]);

}

}

}

}

8

Multiply the digit to the left of the "decimal" point by 2^0 = 1.

Multiply the digit to the left of it by 2^1 = 2

Multiply the digit to the left of that by 2^2 = 4 and so on.

Also

Multiply the digit to the right of the "decimal" point by 2^(-1) = 1/2.

Multiply the digit to the right of that by 2^(-2) = 1/4 and so on.

Add all these together.

Example:

Binary 1101.1011

1*2^3 = 1*8 = 8

1*2^2 = 1*4 = 4

0*2^1 = 0*2 =0

1*2^0 = 1*1 = 1

1*2^-1 = 1*1/2 = 0.5

0*2^-2 = 0*1/4 = 0

1*2^-3 = 1*1/8 = 0.125

1*2^-4 = 1*1/16 = 0.0625

Sum = 13.6875

Happiness

You write the number and then follow it with the digits in reverse order.

Please post a new question with the bill's denomination and condition. Note that the serial number is unimportant to its value.

int LCM3 (int a, int b, int c)

{

return LCM2 (a, LCM2 (b, c));

}

int LCM2 (int a, int b)

{

return a*b/GCD2(a, b);

}