Write a C program to print Armstrong numbers in a given range?

Answer 1

Sorry. I had to delete the older version, for it was written in C++, while the questioner had asked one in C. So here it is. In fact, I will give the questioner a bonus, it will print not only Armstrong numbers but all Narcissistic numbers (here the power to which the individual digits should be raised before being added to each other can be any integer. Clearly Armstrong numbers form a subset of these numbers, with the power equal to three), and that too not only upto 10000, but upto the maximum number a C compiler can process. Enjoy!

#include

#include

main()

{

int m=1,b;

int a[200],i,j,k,l;

while (m<=2147483647)

{

b=m;

i=0;

while (b>=1)

{

a[i]=(b%10);

b=floor(b/10);

++i;

}

for (j=1;j<=9;++j)

{

k=0;

for (l=(i-1);l>=0;--l)

k+= pow(a[l],j);

if (k==m)

goto sexy;

}

goto label;

sexy: printf("%d (raise every digit to power of %d)\n",m,j);

label: ++m;

}

}

Answer 2

#include
#include
using namespace std;
int main(void)
{
int upper,lower;
cout<<"Enter the limits"<cin>>lower>>upper;
for(int i=lower;i<=upper;i++)
{
int temp=i;
int sum=0;
int count=0;
while(temp>0)
{
temp=temp/10;
count++;
}
temp=i;
while(temp>0)
{
n=temp%10;
temp=temp/10;
int cal=pow(n,count);
sum=sum+cal;
}
if(sum==i)
cout<<"Armstrong Numbers are"<}
return 0;
}

Answer 3

//This program checks whether or not a number is palindrome #include #include


void main()
{
int rem,sum=0,n,num;


clrscr();
printf("Enter a number\n");
scanf("%d",&num);
n=num;
while(n!=0)
{
rem=n%10;
n=n/10;
sum+=rem*rem*rem;
}
if(sum==num)
printf("%d is an armstrong",num);
else
printf("%d is not armstrong",num);
getch();
}