阿姆斯特朗数是一个等于其立方数字总和的数字。它由 N 位数字组成，相当于每位数字的 N 次方之和。例如，包括 0、1、153、370、371 和 407、1634、8208、9474 在内的数字被视为阿姆斯特朗数。在本主题中，我们将学习 C# 中的阿姆斯特朗数。
加密和解密的过程使用阿姆斯特朗号码，该号码将被指定为密钥。为了在两个目标用户之间进行身份验证和保护，需要使用服务器。发送者和接收者将在服务器的帮助下得到验证。
逻辑
为了了解阿姆斯特朗数背后的逻辑，我们将用其中一个数字来简单解释一下阿姆斯特朗数法。考虑数字 407 来解释逻辑。
407 = (4*4*4) + (0*0*0) + (7*7*7)
在这里，将每个数字乘以三，如下所示：
(4*4*4) = 64
(0*0*0) = 0
(7*7*7) = 343
因此，将上面相乘的数字相加，您将得到如下结果：
64+0+343 = 407
C# 中阿姆斯特朗数的示例
以下是下面提到的以下示例。
示例#1
代码：
using System;
using System.Collections.Generic;
using System.Linq;
public class Program
{
public static void Main()
{
int i = 0;
int num = 0;
int[] numArray = new int[10];
double res = 0;
//Step 1 : Provide the number in this step
Console.Write("Please enter the number : ");
int number = int.Parse(Console.ReadLine());
//Step 2 : Now store the number in a temporary variable
int tmpryNum = number;
//Step 3 : Calculate the total number of integers in number and store each integer in the digit array
while (number > 0)
{
numArray[i++] = number % 10;
number = number / 10;
num++;
}
//Step 4 : Calculate the number
for (i = 0; i < num; i++)
{
res += Math.Pow(numArray[i], num);
}
//Step 5 : Check whether number is prime or not
if (res == tmpryNum)
{
Console.WriteLine($"The number {tmpryNum} is armstrong");
}
else
{
Console.WriteLine($"The number {tmpryNum} is not armstrong");
}
Console.ReadLine();
}
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输出：
示例#2
代码：
using System;
using System.Collections.Generic;
using System.Linq;
public class Program
{
public static void Main()
{
int mynum, remainder, res = 0;
Console.Write("Please enter an integer number: ");
mynum = int.Parse(Console.ReadLine());
for (int m = mynum; m > 0; m = m / 10)
{
remainder = m % 10;
res = res + remainder*remainder*remainder;
}
if (res == mynum)
{
Console.Write("The number entered by you is an Armstrong number...");
}
else
Console.Write("The number entered by you is not an Armstrong number...");
Console.ReadLine();
}
}登录后复制
输出：
示例#3
代码：
using System;
using System.Collections.Generic;
using System.Linq;
public class Program
{
public static void Main()
{
int mynum,m, res=0,r;
Console.Write("Please enter the number :");
mynum = Convert.ToInt32(Console.ReadLine());
m = mynum;
while(m!=0)
{
r=m%10;
res =res+(r*r*r);
m = m / 10;
}
if(res==mynum)
Console.WriteLine("The number which you have entered is an Armstrong Number...");
else
Console.WriteLine("The number which you have entered is not an Armstrong Number...");
Console.ReadKey();
}
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输出
示例#4
代码：
using System;
using System.Collections.Generic;
using System.Linq;
public class Program
{
public static void Main()
{
int mynum,val,res,temp_var;
int start_no,end_no;
Console.Write("\n\n");
Console.Write("Searching for the Armstrong number in a given range of range of numbers:\n");
Console.Write("--------------------------------------------------------");
Console.Write("\n\n");
Console.Write("Enter the starting number... ");
start_no= Convert.ToInt32(Console.ReadLine());
Console.Write("Enter the ending number... ");
end_no= Convert.ToInt32(Console.ReadLine());
Console.Write("The list of Armstrong numbers in given above range are: ");
for(mynum=start_no;mynum<=end_no;mynum++){
temp_var=mynum;
res = 0;
while(temp_var!=0){
val=temp_var % 10;
temp_var=temp_var/10;
res=res+(val*val*val);
}
if(res==mynum)
Console.Write("{0} ",mynum);
}
Console.Write("\n");
}
}登录后复制
输出：
示例#5
代码：
using System;
using System.Collections.Generic;
using System.Linq;
public class Program
{
public static void Main()
{
Console.Write("Enter the Start Number : ");
int StartNumber = int.Parse(Console.ReadLine());
Console.Write("Enter the End Number : ");
int EndNumber = int.Parse(Console.ReadLine());
Console.WriteLine($"The Armstrong Numbers between {StartNumber} and {EndNumber} are : ");
for (int i = StartNumber; i <= EndNumber; i++)
{
if (IsArmstrongNumber(i))
Console.WriteLine(i);
}
Console.ReadLine();
}
static bool IsArmstrongNumber(int number)
{
int sum = 0;
int temporaryNumber = number;
int temp = 0;
int length = number.ToString().Length;
while (number != 0)
{
temp = number % 10;
number = number / 10;
sum += (int)Math.Pow(temp, length);
}
if (sum == temporaryNumber)
{
return true;
}
else
{
return false;
}
}
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输出：
示例#6
代码：
using System;
public class Program
{
// This function is used to find Nth Armstrong Number
static int ArmstrongNum(int n)
{
int cnt = 0;
// specifying the upper limit from integer
for(int m = 1; m <= int.MaxValue; m++)
{
int num = m, rem, digit = 0, res = 0;
num = m;
// Calculate the total digits in num
digit = (int) Math.Log10(num) + 1;
// Calculate sum of power of digits
while(num > 0)
{
rem = num % 10;
res = res + (int)Math.Pow(rem, digit);
num = num / 10;
}
// Check the Armstrong number
if(m == res)
cnt++;
if(cnt == n)
return m;
}
return n;
}
public static void Main()
{
int n = 15;
Console.WriteLine(ArmstrongNum(n));
}
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输出：
示例#7
代码：
using System;
public class Program
{
int power(int a, long b)
{
if( b == 0)
return 1;
if (b % 2 == 0)
return power(a, b / 2) * power(a, b / 2);
return a * power(a, b / 2) * power(a, b / 2);
}
int myfunc(int a)
{
int n = 0;
while (a != 0)
{
n++;
a = a / 10;
}
return n;
}
bool ArmstrongNum (int a)
{
int n = myfunc(a);
int tmp_var = a, res = 0;
while (tmp_var != 0)
{
int rs = tmp_var % 10;
res = res + power(rs, n);
tmp_var = tmp_var / 10;
}
return (res == a);
}
public static void Main()
{
Program pgm = new Program();
int a = 8208;
Console.WriteLine(pgm.ArmstrongNum(a));
a = 1423;
Console.WriteLine(pgm.ArmstrongNum(a));
}
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输出：
结论
到目前为止，我们已经研究了如何在 C# 中使用阿姆斯特朗数。在数论中，它被称为自恋数。通过本文，我希望上面解释的内容能够为您的 C# 知识增添价值。尝试使用上述不同方法中所示的不同类型的数字来玩阿姆斯特朗数字技术。
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