阿姆斯特朗编号的C / C++程序

给定数字N ，任务是检查给定数字是否为阿姆斯特朗数字。如果给定的号码是阿姆斯壮号码，则打印“是”，否则打印“否” 。

A positive integer of D digits is called an armstrong-numbers of order D (order is the number of digits) if
$N_{1}N_{2}N_{3}N_{4}... = N_{1}^{D} + N_{2}^{D} + N_{3}^{D} + N_{4}^{D}...$
Where D is number of digit in number N
and N(1), N(2), N(3)… are digit of number N.

例子：

Input: N = 153
Output: Yes
Explanation:
153 is an Armstrong number.
1*1*1 + 5*5*5 + 3*3*3 = 153

Input: 120
Output: No
Explanation:
120 is not an Armstrong number.
1*1*1 + 2*2*2 + 0*0*0 = 9

方法：想法是计算给定数字N中的位数(例如d )。对于给定数字N中的每个数字(例如r )，找到r ^d的值，如果所有值的总和为N，则打印“是”，否则打印“否” 。

下面是上述方法的实现：

C

// C program to find Armstrong number
#include 
  
// Function to calculate N raised
// to the power D
int power(int N, unsigned int D)
{
    if (D == 0)
        return 1;
  
    if (D % 2 == 0)
        return power(N, D / 2)
               * power(N, D / 2);
  
    return N * power(N, D / 2)
           * power(N, D / 2);
}
  
// Function to calculate the order of
// the number
int order(int N)
{
    int r = 0;
  
    // For each digit
    while (N) {
        r++;
        N = N / 10;
    }
    return r;
}
  
// Function to check whether the given
// number is Armstrong number or not
int isArmstrong(int N)
{
    // Calling order function
    int D = order(N);
  
    int temp = N, sum = 0;
  
    // For each digit
    while (temp) {
        int Ni = temp % 10;
        sum += power(Ni, D);
        temp = temp / 10;
    }
  
    // If satisfies Armstrong condition
    if (sum == N)
        return 1;
    else
        return 0;
}
  
// Driver Code
int main()
{
    // Given Number N
    int N = 153;
  
    // Function Call
    if (isArmstrong(N) == 1)
        printf("True\n");
    else
        printf("False\n");
    return 0;
}

C++

// C++ program to find Armstrong number
#include 
using namespace std;
  
// Function to calculate N raised
// to the power D
int power(int N, unsigned int D)
{
    if (D == 0)
        return 1;
  
    if (D % 2 == 0)
        return power(N, D / 2)
               * power(N, D / 2);
  
    return N * power(N, D / 2)
           * power(N, D / 2);
}
  
// Function to calculate the order of
// the number
int order(int N)
{
    int r = 0;
  
    // For each digit
    while (N) {
        r++;
        N = N / 10;
    }
    return r;
}
  
// Function to check whether the given
// number is Armstrong number or not
int isArmstrong(int N)
{
    // To find order of N
    int D = order(N);
  
    int temp = N, sum = 0;
  
    // Traverse each digit
    while (temp) {
        int Ni = temp % 10;
        sum += power(Ni, D);
        temp = temp / 10;
    }
  
    // If satisfies Armstrong condition
    if (sum == N)
        return 1;
    else
        return 0;
}
  
// Driver Code
int main()
{
    // Given Number N
    int N = 153;
  
    // Function Call
    if (isArmstrong(N) == 1)
        cout << "True";
    else
        cout << "False";
    return 0;
}

输出：

True

时间复杂度： O(log ₁₀ N)
辅助空间： O(1)

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