Problem Statement
In this program we will code a basic program to find armstrong
numbers within a given range(3 digits only).
An Armstrong number is an integer such that the sum of the cubes
of its digits is equal to the number itself. For example, 371 is an Armstrong
number since 3*3*3 + 7*7*7 + 1*1*1 = 371.
#include
void armstronginrange(int a,int b)
{
int i,j;
int num=0;
int n=0;
int rem=0,sum=0;
for(i=a;i<=b;i++)
{
rem=0,sum=0;
n=i;
while(n>0)
{
rem=n%10;
sum=sum+rem*rem*rem;
n=n/10;
}
if(sum==i)
{
printf("%d ",sum);
}
}
}
int main()
{
int a,b;
printf("ENTER THE INNER RANGE : \n");
scanf("%d",&a);
printf("ENTER THE OUTER RANGE : \n");
scanf("%d",&b);
armstronginrange(a,b);
return 0;
}
Output
ENTER
THE INNER RANGE : 10
ENTER
THE OUTER RANGE : 1000
153 370 371 407
Explanation
In
this problem we have to find all the armstrong number under a specified range
To do so we declare a variable ‘sum’,which will store the result value (don’t
forget to initialize it with zero). Then,we update our ‘sum’ value by adding
cube of remainder value (rem*rem*rem) to it in each iteration (until ‘n’
becomes 0).
After the last iteration,we compare the value of given number and our sum, if
both values are equal then given number is armstrong else it isn’t. Repeat the
process for every element in the range.
For n digits :
#include
#include
int main() {
int a, b, i, num,
rem, count = 0;
int sum = 0;
printf("Enter
two is(intervals): ");
scanf("%d
%d", &a, &b);
for (i = a+1; i
< b; i++)
{
num = i;
while (num != 0)
{
num /= 10;
count++;
}
num = i;
while (num != 0)
{
rem = num %
10;
sum +=
pow(rem, count);
num /= 10;
}
if(sum == i)
{
printf("%d ", i);
}
count = 0;
sum = 0;
}
return 0;
}
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