C Language Examples - Check Armstrong Number
An Armstrong number is one where, for an n-digit number, the sum of each digit raised to the nth power equals the number itself, such as:
153=1^3+5^3+3^3 1634=1^4+6^4+3^4+4^4
Example
#include <stdio.h>
int main()
{
int number, originalNumber, remainder, result = 0;
printf("Enter a three-digit number:");
scanf("%d", &number);
originalNumber = number;
while (originalNumber != 0)
{
remainder = originalNumber%10;
result += remainder*remainder*remainder;
originalNumber /= 10;
}
if(result == number)
printf("%d is an Armstrong number",number);
else
printf("%d is not an Armstrong number",number);
return 0;
}
Run result:
输入三位数: 371 371 是 Armstrong
Example - Armstrong Numbers Between Two Numbers
#include <stdio.h>
#include <math.h>
int main()
{
int low, high, i, temp1, temp2, remainder, n = 0, result = 0;
printf("Enter two integers:");
scanf("%d %d", &low, &high);
printf("The Armstrong numbers between %d and %d are:", low, high);
for(i = low + 1; i < high; ++i)
{
temp2 = i;
temp1 = i;
// Calculate
while (temp1 != 0)
{
temp1 /= 10;
++n;
}
while (temp2 != 0)
{
remainder = temp2 % 10;
result += pow(remainder, n);
temp2 /= 10;
}
if (result == i) {
printf("%d ", i);
}
n = 0;
result = 0;
}
return 0;
}
Run result:
输入两个整数: 100 1000 100 和 1000 之间的 Armstrong 数为: 153 370 371 407
Example - Check Armstrong Number Using a Function
#include <stdio.h>
#include <math.h>
int checkPrimeNumber(int n);
int checkArmstrongNumber(int n);
int main()
{
int n, flag;
printf("Enter a positive integer:");
scanf("%d", &n);
// Check prime number
flag = checkPrimeNumber(n);
if (flag == 1)
printf("%d is a prime number.\n", n);
else
printf("%d is not a prime number\n", n);
// Check for Armstrong number
flag = checkArmstrongNumber(n);
if (flag == 1)
printf("%d is an Armstrong number.", n);
else
printf("%d is not an Armstrong number.",n);
return 0;
}
int checkPrimeNumber(int n)
{
int i, flag = 1;
for(i=2; i<=n/2; ++i)
{
// Non-prime condition
if(n%i == 0)
{
flag = 0;
break;
}
}
return flag;
}
int checkArmstrongNumber(int number)
{
int originalNumber, remainder, result = 0, n = 0, flag;
originalNumber = number;
while (originalNumber != 0)
{
originalNumber /= 10;
++n;
}
originalNumber = number;
while (originalNumber != 0)
{
remainder = originalNumber%10;
result += pow(remainder, n);
originalNumber /= 10;
}
// Determine condition
if(result == number)
flag = 1;
else
flag = 0;
return flag;
}
The output result is:
输入正整数: 371 371 不是素数 371 是 Armstrong 数。other extensions