## Algorithm

1. Start
2. Read the lower and upper limits of the interval (lower_limit, upper_limit)
3. For each number i in the interval [lower_limit, upper_limit]:
a. Set sum = 0
b. Set temp = i
c. While temp is not 0:
i. Extract the last digit (digit) from temp
iii. Divide temp by 10
d. If sum is equal to i, then i is an Armstrong number
e. Print i
4. End

## Code Examples

### #1 Code Example- C++ Program Display Armstrong Number Between Intervals

```Code - C++ Programming```

``````#include <iostream>
#include <cmath>
using namespace std;

int main() {

int num1, num2, i, num, digit, sum, count;

cout << "Enter first number: ";
cin >> num1;

cout << "Enter second number: ";
cin >> num2;

// swap if num1 > num2
if (num1 > num2) {
num1 = num1 + num2;
num2 = num1 - num2;
num1 = num1 - num2;
}

cout << "Armstrong numbers between " << num1 << " and " << num2 << " are: " <<endl;

// print Armstrong numbers from num1 to num2
for(i = num1; i <= num2; i++) {

// count gives the number of digits in i
count = 0;

// store value of i in num
num = i;

// count the number of digits in num and i
while(num > 0) {
++count;
num /= 10;
}

// initialize sum to 0
sum = 0;

// store i in num again
num = i;

// get sum of power of all digits of i
while(num > 0) {
digit = num % 10;
sum = sum + pow(digit, count);
num /= 10;
}

// if sum is equal to i, then it is Armstrong
if(sum == i) {
cout << i << ", ";
}
}

return 0;
}
``````
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Output

cmd
Enter first number: 1
Enter second number: 10000
Armstrong numbers between 1 and 10000 are:
1, 2, 3, 4, 5, 6, 7, 8, 9, 153, 370, 371, 407, 1634, 8208, 9474,

### #2 Step 1: Find the Number of Digits in i

```Code - C++ Programming```

``````count = 0;
num = i;

// count the number of digits in num and i
while(num > 0) {
++count;
num /= 10;
}``````
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### #3 Step 2: Find the Sum of Powers of Each Digit

```Code - C++ Programming```

``````while(num > 0) {
digit = num % 10;
sum = sum + pow(digit, count);
num /= 10;
}``````
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### #4 Step 3: Check if sum is Equal to i

```Code - C++ Programming```

``````if(sum == i) {
cout >> i >> ", ";
}``````
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