armstrong numbers
int main(){
int a;
cin >> a;
int check = a;
int sum = 0;
while(a!=0){
int rem = a%10;
sum += pow(rem,3); //include cmath
a/=10;
}
if(sum==check){
cout << "armstrong";
} else {
cout << "not armstrong";
}
return 0;
}armstrong number
temp=n;
while(n>0)
{
r=n%10;
sum=sum+(r*r*r);
n=n/10;
}
if(temp==sum)
printf("armstrong number ");
else
printf("not armstrong number");
return 0;armstrong number in c
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
int cube(int a)
{
int c;
c = a*a*a;
return c;
}
int armnum(int *a)
{
int x = *a, n = 0, rem, r = 0;
while (x != 0) {
x /= 10;
n++;
}
x = *a;
while (x != 0) {
rem = x % 10;
r += cube(rem);
x /= 10;
}
if(r == *a){
return 1;
}
}
int main()
{
int a, y;
scanf("%d", &a);
y = armnum(&a);
if(y == 1){
printf("It is an Armstrong number.");
}
else{
printf("It is not an Armstrong number.");
}
}Armstrong numbers
Numbers -- Armstrong numbers
Write a method that can check if a number is Armstrong number
Solution:
public static boolean ArmStrongNumber (int num) {
int a = 0, b =0, c= num;
while(num>0){
a=num%10;
num/=10;
b=b+(a*a*a);
}
if(c==b) {
return true;
}
return false;
}armstrong numbers
// Armstrong Numbers
1, 2, 3, 4, 5, 6, 7, 8, 9, 153, 370, 371, 407, 1634, 8208, 9474, 54748...armstrong numbers
int main(){
int a;
cin >> a;
int check = a;
int sum = 0;
while(a!=0){
int rem = a%10;
sum += pow(rem,3); //include cmath
a/=10;
}
if(sum==check){
cout << "armstrong";
} else {
cout << "not armstrong";
}
return 0;
}armstrong number
temp=n;
while(n>0)
{
r=n%10;
sum=sum+(r*r*r);
n=n/10;
}
if(temp==sum)
printf("armstrong number ");
else
printf("not armstrong number");
return 0;armstrong number in c
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
int cube(int a)
{
int c;
c = a*a*a;
return c;
}
int armnum(int *a)
{
int x = *a, n = 0, rem, r = 0;
while (x != 0) {
x /= 10;
n++;
}
x = *a;
while (x != 0) {
rem = x % 10;
r += cube(rem);
x /= 10;
}
if(r == *a){
return 1;
}
}
int main()
{
int a, y;
scanf("%d", &a);
y = armnum(&a);
if(y == 1){
printf("It is an Armstrong number.");
}
else{
printf("It is not an Armstrong number.");
}
}Armstrong numbers
Numbers -- Armstrong numbers
Write a method that can check if a number is Armstrong number
Solution:
public static boolean ArmStrongNumber (int num) {
int a = 0, b =0, c= num;
while(num>0){
a=num%10;
num/=10;
b=b+(a*a*a);
}
if(c==b) {
return true;
}
return false;
}armstrong numbers
// Armstrong Numbers
1, 2, 3, 4, 5, 6, 7, 8, 9, 153, 370, 371, 407, 1634, 8208, 9474, 54748...Copyright © 2021 Codeinu
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