📜  阿姆斯特朗编号的C / C++程序

📅  最后修改于: 2021-04-29 04:34:41             🧑  作者: Mango

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”

例子:

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)

想要从精选的最佳视频中学习和练习问题,请查看《基础到高级C的C基础课程》。