C Programming

Non-parametric statistical methods.

largestNum = 0;

for ( i = 0; i < 10 ; i++)

{

if ( currentNum[i] > largestNum)

largestNum = currentNum[i];

}

recu

First, review the definition of an Armstrong, or narcissistic, number:

"...is a number that is the sum of its own digits each raised to the power of the number of digits."

So, you need to count the digits (to know what power to use), and extract the individual digits. This can be done in several ways; for example, you might convert the number to a string. In Java:

String numberAsString = "" + number;

Now it should be easy to figure out the length of the String (use the .length() method), and to extract the individual digits - check the methods available for strings. Then you need to convert the digits back to numeric data.

Another way is to get one digit at a time, starting from the right, using the "%" operator.

For example, 153 % 10 is equal to 3. Divide the number, 153, by 10 (integer division), then repeat to get the remaining digits. You might store the digits to an array.

First, review the definition of an Armstrong, or narcissistic, number:

"...is a number that is the sum of its own digits each raised to the power of the number of digits."

So, you need to count the digits (to know what power to use), and extract the individual digits. This can be done in several ways; for example, you might convert the number to a string. In Java:

String numberAsString = "" + number;

Now it should be easy to figure out the length of the String (use the .length() method), and to extract the individual digits - check the methods available for strings. Then you need to convert the digits back to numeric data.

Another way is to get one digit at a time, starting from the right, using the "%" operator.

For example, 153 % 10 is equal to 3. Divide the number, 153, by 10 (integer division), then repeat to get the remaining digits. You might store the digits to an array.

First, review the definition of an Armstrong, or narcissistic, number:

"...is a number that is the sum of its own digits each raised to the power of the number of digits."

So, you need to count the digits (to know what power to use), and extract the individual digits. This can be done in several ways; for example, you might convert the number to a string. In Java:

String numberAsString = "" + number;

Now it should be easy to figure out the length of the String (use the .length() method), and to extract the individual digits - check the methods available for strings. Then you need to convert the digits back to numeric data.

Another way is to get one digit at a time, starting from the right, using the "%" operator.

For example, 153 % 10 is equal to 3. Divide the number, 153, by 10 (integer division), then repeat to get the remaining digits. You might store the digits to an array.

First, review the definition of an Armstrong, or narcissistic, number:

"...is a number that is the sum of its own digits each raised to the power of the number of digits."

So, you need to count the digits (to know what power to use), and extract the individual digits. This can be done in several ways; for example, you might convert the number to a string. In Java:

String numberAsString = "" + number;

Now it should be easy to figure out the length of the String (use the .length() method), and to extract the individual digits - check the methods available for strings. Then you need to convert the digits back to numeric data.

Another way is to get one digit at a time, starting from the right, using the "%" operator.

For example, 153 % 10 is equal to 3. Divide the number, 153, by 10 (integer division), then repeat to get the remaining digits. You might store the digits to an array.

printf ("3*19 is %d\n", sizeof (char [19][3]));

Warning: never do this in an actual program!

Mainly, we store values in them.

-- First of all, if the number is a fraction or a decimal, then don't worry about it.

Only whole numbers are considered odd or even.

-- All multiples of 2 are even numbers, and that's easy to recognize. It doesn't

matter how big the number is, or how many digits it has. You only have to look

at the last digit (the one on the right-hand end). If that digit is 2, 4, 6, 8, or 0,

then the whole big number is even. Otherwise it's odd.

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;

}

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.

even

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.

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

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.