Armstrong Numbers Between Two Integers using c program
C Programming Language / Loop control in C Language
163Tip: Before trying this program, learn what is armstrong-number
Given Input:
Enter two numbers(intervals): 200
2000
Expected Output:
Armstrong numbers between 200 and 2000 are: 370 371 407 1634
Program:
#include
#include
int main() {
int low, high, number, originalNumber, rem, count = 0;
double result = 0.0;
printf("Enter two numbers(intervals): ");
scanf("%d %d", &low, &high);
printf("Armstrong numbers between %d and %d are: ", low, high);
// swap numbers if high < low
if (high < low) {
high += low;
low = high - low;
high -= low;
}
// iterate number from (low + 1) to (high - 1)
// In each iteration, check if number is Armstrong
for (number = low + 1; number < high; ++number) {
originalNumber = number;
// number of digits calculation
while (originalNumber != 0) {
originalNumber /= 10;
++count;
}
originalNumber = number;
// result contains sum of nth power of individual digits
while (originalNumber != 0) {
rem = originalNumber % 10;
result += pow(rem, count);
originalNumber /= 10;
}
// check if number is equal to the sum of nth power of individual digits
if ((int)result == number) {
printf("%d ", number);
}
// resetting the values
count = 0;
result = 0;
}
return 0;
}
Output:
Enter two numbers(intervals): 200
2000
Armstrong numbers between 200 and 2000 are: 370 371 407 1634
Explanation:
In the program, the outer loop is iterated from (low+ 1) to (high - 1). In each iteration, it's checked whether number is an Armstrong number or not.
Inside the outer loop, the number of digits of an integer is calculated first and stored in count
. And, the sum of the power of individual digits is stored in the result variable.
If number is equal to result
, the number is an Armstrong number.
Notes:
- You need to swap low and high if the user input for high is less than that of low. To learn more, check our example on swapping two numbers.
- You need to reset count and result to 0 in each iteration of the outer loop.
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Armstrong numbers between 200 and 2000 are: 370 371 407 1634
Program:
#include #include int main() { int low, high, number, originalNumber, rem, count = 0; double result = 0.0; printf("Enter two numbers(intervals): "); scanf("%d %d", &low, &high); printf("Armstrong numbers between %d and %d are: ", low, high); // swap numbers if high < low if (high < low) { high += low; low = high - low; high -= low; } // iterate number from (low + 1) to (high - 1) // In each iteration, check if number is Armstrong for (number = low + 1; number < high; ++number) { originalNumber = number; // number of digits calculation while (originalNumber != 0) { originalNumber /= 10; ++count; } originalNumber = number; // result contains sum of nth power of individual digits while (originalNumber != 0) { rem = originalNumber % 10; result += pow(rem, count); originalNumber /= 10; } // check if number is equal to the sum of nth power of individual digits if ((int)result == number) { printf("%d ", number); } // resetting the values count = 0; result = 0; } return 0; }
Output:
Enter two numbers(intervals): 200 2000 Armstrong numbers between 200 and 2000 are: 370 371 407 1634
Explanation:
In the program, the outer loop is iterated from (low+ 1) to (high - 1). In each iteration, it's checked whether number is an Armstrong number or not.
Inside the outer loop, the number of digits of an integer is calculated first and stored in count
. And, the sum of the power of individual digits is stored in the result variable.
If number is equal to result
, the number is an Armstrong number.
Notes:
- You need to swap low and high if the user input for high is less than that of low. To learn more, check our example on swapping two numbers.
- You need to reset count and result to 0 in each iteration of the outer loop.
This Particular section is dedicated to Programs only. If you want learn more about C Programming Language. Then you can visit below links to get more depth on this subject.
# C Tutorials
# JAVA Tutorials
# HTML Tutorials
# Computer Fundamental
# Data Structure
# DBMS Tutorials
SQL
# C# Language
# R Language
# PHP
# Python
# Vue JS