检查阿姆斯特朗数的 C 程序

要理解此示例，您应该具备以下 C 语言编程主题的知识：

如果一个 n 位正整数等于其各位数字的 n 次方之和,则称该数为阿姆斯特朗数。

如果是 3 位数的阿姆斯特朗数，则每位数的立方的和等于该数本身。例如 153 是一个阿姆斯特朗数。因为： 153 = 1*1*1 + 5*5*5 + 3*3*3

检查阿姆斯壮的三位数字

#include <stdio.h>
int main() {
    int num, originalNum, remainder, result = 0;
    printf("Enter a three-digit integer: ");
    scanf("%d", &num);
    originalNum = num;

    while (originalNum != 0) {
       // remainder contains the last digit
        remainder = originalNum % 10;

       result += remainder * remainder * remainder;

       // removing last digit from the orignal number
       originalNum /= 10;
    }

    if (result == num)
        printf("%d is an Armstrong number.", num);
    else
        printf("%d is not an Armstrong number.", num);

    return 0;
}

输出

Enter a positive integer: 371
371 is an Armstrong number.

检查阿姆斯壮 n 位数

#include <math.h>
#include <stdio.h>

int main() {
   int num, originalNum, remainder, n = 0;
   float result = 0.0;

   printf("Enter an integer: ");
   scanf("%d", &num);

   originalNum = num;

   // store the number of digits of num in n
   for (originalNum = num; originalNum != 0; ++n) {
       originalNum /= 10;
   }

   for (originalNum = num; originalNum != 0; originalNum /= 10) {
       remainder = originalNum % 10;

      // store the sum of the power of individual digits in result
      result += pow(remainder, n);
   }

   // if num is equal to result, the number is an Armstrong number
   if ((int)result == num)
    printf("%d is an Armstrong number.", num);
   else
    printf("%d is not an Armstrong number.", num);
   return 0;
}

输出

Enter an integer: 1634
1634 is an Armstrong number.

在这个程序中，首先计算一个整数的位数并存储在变量 n 中. 然后，使用 pow() 函数计算第二个 for 循环的每次迭代中各个数字的幂。