/*Program to find Armstrong number between 1 to N*/
int main() {
int n = 0, remainder, sum = 0, i = 0, noDigits = 0, isArm = 0;
char ch[60] = {0};
printf("Find the Arm Strong Numbers between 1 to N");
scanf("%d", &n);
for(i = 1; i<n; i++)
{
isArm = i;
itoa(isArm, ch, 10);
noDigits = strlen(ch);
while(isArm)
{
remainder = isArm%10;
isArm=isArm/10;
sum= sum+pow(remainder, noDigits);
}
if(sum == i)
printf("\nArm Strong Nos are %d\n", i);
sum = noDigits = 0;
}
}
if the sum of the cubic value of each digit of a number is equal to the original number , then that particular number is an Armstrong number :-)
Actually, Armstrong numbers are the sum of their own digits to the power of the number of digits.
#include(stdio.h)
#include(conio.h)
int main( )
{
int no, temp, rem, sum;
clrscr( );
printf("Armstrong numbers between 1 and 1000 are:\n");
for(no=1; no<=1000; no++)
{
temp=no;
sum=0;
while(temp>0)
{
rem=temp%10;
sum=sum+(rem*rem*rem);
temp=temp/10;
}
if(no==sum)
printf("\n%d", no);
}
getch( );
return 0;
}
#include <iostream>
#include <math.h> // for std::pow()
unsigned int get_length(unsigned int num,const unsigned int base=10)
{
unsigned int len=1;
while(num && (num/=base))
++len;
return( len );
}
bool is_armstrong(const unsigned int num,const unsigned int base=10)
{
unsigned int len=get_length(num,base);
unsigned int sum=0;
unsigned int tmp=num;
while(tmp)
{
sum+=(unsigned int)std::pow((double)(tmp%base),(double)len);
tmp/=base;
}
return(num==sum);
}
int main()
{
std::cout << "Armstrong series (base 10):";
for(unsigned int num=0; num<=0xffffffff; ++num)
if(is_armstrong(num))
std::cout << " " << num;
std::cout << std::endl;
return(0);
}
You will need:
class armstrong
{
public void m()
{
int s,d,i,n;
for(i=1;i<=500;i++)
{
s=0;
n=i;
while(n>0)
{
d=n%10;
s=s+(d*d*d);
n=n/10;
}
if(s==i)
{
System.out.println(s+" ");
}
}
}
}
/*Program to check for armstrong number */
#include<stdio.h>
#include<math.h>
main()
{
int a, b, z,x, sum = 0;
printf("enter the number to check if it is armstrong number\n");
scanf("%d", &x);
z=x;
if ((z >= 100) && (z <= 900))
{
a = z % 10;
z = z / 10;
b = z % 10;
z = z / 10;
sum = pow(a, 3) + pow(b, 3) + pow(z, 3);
if (x== sum)
printf("a is an armstrong number.\n");
else
printf("not a armstrong number\n");
}
else
printf("armstrong number must be of three digit\n");
}
/*posted by prabhat5052@gmail.com*/
class ArmstrongNumbers{
public static void main(String[] args){
int num, temp, digit1, digit2, digit3;
System.out.println("Armstrong numbers between 1 and 500: ");
num = 1;
while (num <= 1000){
digit1 = num - ((number / 10) * 10);
digit2 = (num / 10) - ((number / 100) * 10);
digit3 = (num / 100) - ((number / 1000) * 10);
temp = (digit1*digit1*digit1) + (digit2*digit2*digit2) + (digit3*digit3*digit3);
if (temp == num){
System.out.println("Armstrong Number:"+ temp);
}
num++;
}
}
}
m,kj,kh,kj
Q.1 Write a program to print first ten odd natural numbers. Q.2 Write a program to input a number. Print their table. Q.3 Write a function to print a factorial value.
#include
PRINT 2,3,5,7,11,13,17,19,23,29,31,37
find even number in array
you do this 10 print "0112358132134" use the whole of the thing
You can use int i; for (i = 10; i <= 50; i += 2) {//print i} as a program to print even numbers between 10 and 50.
Prime numbers are numbers that are only divisible by themselves and the number 1. You can write a program to print all prime numbers from 1 to 100 in FoxPro.
Q.1 Write a program to print first ten odd natural numbers. Q.2 Write a program to input a number. Print their table. Q.3 Write a function to print a factorial value.
Write a c program to print the 100 to 1 nos
for (int i = 2; i < 10; i ++) printf("%d\n", i); You did say even and odd numbers between 1 and 10. That's allnumbers between 1 and 10.
how do we use loops in c plus plus programing and what are basic differences between do,for and while loop
#include
PRINT 2,3,5,7,11,13,17,19,23,29,31,37
Install your phone software program and download the list to your PC then print. Your phone program can be downloaded from your phone maker web site if it did not come with your phone package.
Try the triangle program on a search engine. Replace numbers with stars and that should do the trick
find even number in array
you do this 10 print "0112358132134" use the whole of the thing