阿姆斯壮数是指一个n位数（n≥3），它的每个数位的n次方的和等于它本身。例如：371=3^3+7^3+1^3，因此371就是一个三位的阿姆斯壮数。

一个常见的阿姆斯壮数判断程序如下：

#include #include

int main() { int number, originalNumber, remainder, result = 0, n = 0 ;

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

originalNumber = number;

while (originalNumber != 0) { originalNumber /= 10; ++n; }

originalNumber = number;

while (originalNumber != 0) { remainder = originalNumber % 10; result += pow(remainder, n); originalNumber /= 10; }

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

return 0; }

问题在于，在计算n位数的指数之和时，如果使用pow函数对每个位上的数字求n次方，可能会产生精度损失或溢出，导致运行时错误。解决这个问题的方法是改用一个自定义的幂函数，通过循环依次计算每个数字的n次方并累计。

一个修改后的程序如下：

#include

int power(int base, int exponent);

int main() { int number, originalNumber, remainder, result = 0, n = 0 ;

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

originalNumber = number;

while (originalNumber != 0) { originalNumber /= 10; ++n; }

originalNumber = number;

while (originalNumber != 0) { remainder = originalNumber % 10; result += power(remainder, n); originalNumber /= 10; }

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

return 0; }

int power(int base, int exponent) { int result = 1; for (int i = 0; i < exponent; ++i) result *= base; return result; }

通过使用自定义的