判断给定数字是否为迪萨里奥数的程序

2025年1月8日 | 阅读 4 分钟

说明

在此程序中，我们需要检查给定的数字是否为迪萨里奥数。

迪萨里奥数

当一个数字的各位数字的幂次之和等于该数字本身时，该数字就被称为迪萨里奥数。

例如，175 是一个 Disarium 数，如下所示

1¹ + 7² + 5³ = 1 + 49 + 125 = 175

迪萨里奥数的一些其他示例包括 89、135、518 等。

要找出给定的数字是否为迪萨里奥数，请计算各位数字的幂次之和。如果和等于原始数字，则给定的数字是迪萨里奥数。

算法

calculateLength() 函数计算数字中的位数。
1. 使用 while 循环检查数字是否不等于 0。
2. 将该数字除以 10，并将变量 length 递增 1。
3. 返回 length。
定义并初始化变量num。
通过将 num 的值存储在 n 中来复制 num。
使用 while 循环，通过将 num 除以 10 来重复计算余数 rem。
计算 rem 的幂次（即 remlen），并将计算出的值存储在变量 sum 中。
检查 sum 是否等于 number。如果是，则给定的数字是迪萨里奥数。否则，它不是迪萨里奥数。

解决方案

Python

#calculateLength() will count the digits present in a number
def calculateLength(n):
    length = 0;
    while(n != 0):
        length = length + 1;
        n = n//10;
    return length;
 
num = 175;
rem = sum = 0;
len = calculateLength(num);
 
#Makes a copy of the original number num
n = num;
 
#Calculates the sum of digits powered with their respective position
while(num > 0):
    rem = num%10;
    sum = sum + int(rem**len);
    num = num//10;
    len = len - 1;
 
#Checks whether the sum is equal to the number itself
if(sum == n):
    print(str(n) + " is a disarium number");
else:
    print(str(n) + " is not a disarium number");

输出

 175 is a disarium number

C

#include <stdio.h>
 
//calculateLength() will count the digits present in a number
int calculateLength(int n){
    int length = 0;
    while(n != 0){
        length = length + 1;
        n = n/10;
    }
    return length;
}
 
int main()
{
    int num = 175, sum = 0, rem = 0, n;
    int len = calculateLength(num);
    
    //Makes a copy of the original number num
    n = num;
    
    //Calculates the sum of digits powered with their respective position
    while(num > 0){
        rem = num%10;
        sum = sum + (int)pow(rem,len);
        num = num/10;
        len--;
    }
    
    //Checks whether the sum is equal to the number itself
    if(sum == n)
        printf("%d is a disarium number", n);
    else
        printf("%d is not a disarium number", n);
 
    return 0;
}

输出

175 is a disarium number

JAVA

public class DisariumNumber
{
    //calculateLength() will count the digits present in a number
    public static int calculateLength(int n){
        int length = 0;
        while(n != 0){
            length = length + 1;
            n = n/10;
        }
        return length;
    }
    
    public static void main(String[] args) {
        int num = 175, sum = 0, rem = 0, n;
        int len = calculateLength(num);
        
        //Makes a copy of the original number num
        n = num;
        
        //Calculates the sum of digits powered with their respective position
        while(num > 0){
            rem = num%10;
            sum = sum + (int)Math.pow(rem,len);
            num = num/10;
            len--;
        }
        
        //Checks whether the sum is equal to the number itself
        if(sum == n)
            System.out.println(n + " is a disarium number");
        else
            System.out.println(n + " is not a disarium number");
    }
}

输出

175 is a disarium number

C#

using System;                    
public class DisariumNumber
{
    //calculateLength() will count the digits present in a number
    public static int calculateLength(int n){
        int length = 0;
        while(n != 0){
            length = length + 1;
            n = n/10;
        }
        return length;
    }
    
    public static void Main()
    {
        int num = 175, sum = 0, rem = 0, n;
        int len = calculateLength(num);
        
        //Makes a copy of the original number num
        n = num;
        
        //Calculates the sum of digits powered with their respective position
        while(num > 0){
            rem = num%10;
            sum = sum + (int)Math.Pow(rem,len);
            num = num/10;
            len--;
        }
        
        //Checks whether the sum is equal to the number itself
        if(sum == n)
            Console.WriteLine(n + " is a disarium number");
        else
            Console.WriteLine(n + " is not a disarium number");
    }
}

输出

175 is a disarium number

PHP

<!DOCTYPE html>
<html>
<body>
<?php
//calculateLength() will count the digits present in a number
function calculateLength($n){
    $length = 0;
    while($n != 0){
        $length = $length + 1;
        $n = intval($n/10);
    }
    return $length;
}
    
$num = 175;
$rem = $sum = 0;
$len = calculateLength($num);
    
//Makes a copy of the original number num
$n = $num;
 
//Calculates the sum of digits powered with their respective position
while($num > 0){
    $rem = $num%10;
    $sum = $sum + pow($rem,$len);
    $num = intval($num/10);
    $len--;
}
 
//Checks whether sum is equal to the number itself
if($sum == $n)
    print($n . " is a disarium number");
else
    print($n . " is not a disarium number");
?>
</body>
</html>

输出

175 is a disarium number

下一主题#

判断给定数字是否为迪萨里奥数的程序

说明

算法

解决方案

Python

C

JAVA

C#

PHP

联系信息

关注我们

教程

面试题

在线编译器

Python

Java

.Net Framework

AI, ML and Data Science

Cloud Technology

B.Tech and MCA

Web Technology

PHP

Software Testing

Technical Interview

Java Interview

Python

Web Interview

Database Interview

B.Tech / MCA

Important Interview

Software Testing Interview

Company Interviews

Online Compilers

Multiple Choice Questions

程序

判断给定数字是否为迪萨里奥数的程序

说明

算法

解决方案

Python

C

JAVA

C#

PHP

相关帖子

查找给定矩阵中奇数和偶数出现频率的程序

计算字符串中所有标点符号字符数量的程序

将厘米转换为英尺和英寸的程序

将千米转换为厘米的程序

检查快乐数（Happy number）的程序

查找给定文本文件中单词数量的程序

单向链表示例

右旋数组元素的程序

打印前 10 个素数的程序

创建包含 n 个节点的单向链表并计算节点数量的程序

订阅 Tpoint Tech

联系信息

关注我们

教程

面试题

在线编译器