C 语言程序：检查一个数字是否为氖数

编写一个C语言程序，使用 for 循环检查一个数字是否为氖数。如果一个数字等于其平方的数字之和，则它是一个氖数。

1. 我们使用 math pow 函数来计算数字的平方。
2. 将输出的平方值分解为单个数字，并计算它们的总和。
3. 使用 if else 语句检查平方的数字之和是否等于原始数字。如果为真，则该数字是氖数。

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

int main()
{
  int Number, squr, rem, Sum;
  
  printf("Enter Number to Check = ");
  scanf("%d", &Number);

  squr = pow(Number, 2);

  for (Sum = 0; squr > 0; squr = squr / 10)
  {
    rem = squr % 10;
    Sum = Sum + rem;
  }

  if (Number == Sum)
    printf("\n%d is a Neon Number.\n", Number);
  else
    printf("\n%d is not.\n", Number);
}

这个C语言程序使用 while 循环检查给定的数字是否为氖数。

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

int main()
{
  int Number, squr, rem, Sum = 0;
  
  printf("Enter Number to Check = ");
  scanf("%d", &Number);

  squr = pow(Number, 2);
  printf("The Square of a Number %d = %d\n", Number, squr);

  while(squr > 0)
  {
    rem = squr % 10;
    Sum = Sum + rem;
    squr = squr / 10;
  }
  printf("The Sum of Digits in a Square = %d\n", Sum);
  if (Number == Sum)
    printf("\n%d is a Neon Number.\n", Number);
  else
    printf("\n%d is not.\n", Number);
}

Enter Number to Check = 9
The Square of a Number 9 = 81
The Sum of Digits in a Square = 9

9 is a Neon Number.

在这个检查氖数 示例 中，squareDigitsSum 递归函数通过递归调用来分解并计算总和。

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

int squareDigitsSum(int num)
{
  static int rem, sum = 0;

  if(num > 0)
  {
    rem = num % 10;
    sum = sum + rem;
    squareDigitsSum(num / 10);
  }
  return sum;
}

int main()
{
  int Number;
  
  printf("Enter Number to Check = ");
  scanf("%d", &Number);

  int squr = pow(Number, 2);

  int Sum = squareDigitsSum(squr);

  printf("The Square of %d Number = %d\n", Number, squr);

  printf("The Sum of Digits in %d = %d\n", squr, Sum);

  if (Number == Sum)
    printf("\n%d is a Neon Number.\n", Number);
  else
    printf("\n%d is not.\n", Number);
}

Enter Number to Check = 5
The Square of 5 Number = 25
The Sum of Digits in 25 = 7

5 is not.