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📜  C程序检查阿姆斯壮编号

📅  最后修改于: 2020-10-04 11:20:04             🧑  作者: Mango

在此示例中，您将学习检查用户输入的整数是否是Armstrong数字。

如果满足以下条件，则正整数称为阿姆斯壮数( n阶) 

abcd... = an + bn + cn + dn +

对于3位数的阿姆斯特朗数，每个数位的立方和等于该数字本身。例如，153是Armstrong的数字，因为

153 = 1*1*1 + 5*5*5 + 3*3*3
检查阿姆斯壮三位数
#include 
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 three-digit integer: 371
371 is an Armstrong number.
检查阿姆斯壮n位数字
#include 
#include 

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.

在此程序中，首先计算一个整数的位数，并将其存储在n 。并且， pow() 函数用于计算第二个for循环的每次迭代中各个数字的幂。