失序数-yiteyi-C++库

给定一个数字“n”，找出它是否是精神错乱。如果一个数字加上其各自位置的数字之和等于该数字本身，则该数字称为“失序”。

null

例如：

Input   : n = 135Output  : Yes 1^1 + 3^2 + 5^3 = 135Therefore, 135 is a Disarium numberInput   : n = 89Output  : Yes 8^1+9^2 = 89Therefore, 89 is a Disarium numberInput   : n = 80Output  : No8^1 + 0^2 = 8

建议：请在“上解决” 实践 “首先，在讨论解决方案之前。

这个想法是先计算给定数字中的数字。计数完成后，我们从最右边遍历所有数字（使用%运算符），将其幂提高到数字计数，并减少数字计数。

下面是上述想法的实现。

C++

                         // C++ program to check whether a number is Desoriam                       
                         // or not                       
                         #include<bits/stdc++.h>                       
                         using                                     namespace                                     std;                       
           
                         // Finds count of digits in n                       
                         int                                     countDigits(                                     int                                     n)                       
                         {                       
                                                 int                                     count_digits = 0;                       
           
                                                 // Count number of digits in n                       
                                                 int                                     x = n;                       
                                                 while                                     (x)                       
                                                 {                       
                                                 x = x/10;                       
           
                                                 // Count the no. of digits                       
                                                 count_digits++;                       
                                                 }                       
                                                 return                                     count_digits;                       
                         }                       
           
                         // Function to check whether a number is disarium or not                       
                         bool                                     check(                                     int                                     n)                       
                         {                       
                                                 // Count digits in n.                       
                                                 int                                     count_digits = countDigits(n);                       
           
                                                 // Compute sum of terms like digit multiplied by                       
                                                 // power of position                       
                                                 int                                     sum = 0;                                     // Initialize sum of terms                       
                                                 int                                     x = n;                       
                                                 while                                     (x)                       
                                                 {                       
                                                 // Get the rightmost digit                       
                                                 int                                     r = x%10;                       
           
                                                 // Sum the digits by powering according to                       
                                                 // the positions                       
                                                 sum = sum +                                     pow                                     (r, count_digits--);                       
                                                 x = x/10;                       
                                                 }                       
           
                                                 // If sum is same as number, then number is                       
                                                 return                                     (sum == n);                       
                         }                       
           
                         //Driver code to check if number is disarium or not                       
                         int                                     main()                       
                         {                       
                                                 int                                     n = 135;                       
                                                 if                                     ( check(n))                       
                                                 cout <<                                     "Disarium Number"                                     ;                       
                                                 else                       
                                                 cout <<                                     "Not a Disarium Number"                                     ;                       
                                                 return                                     0;                       
                         }                       

JAVA

                         // Java program to check whether a number is Desoriam                       
                         // or not                       
           
                         class                                     Test                       
                         {                       
                                                 // Method to check whether a number is disarium or not                       
                                                 static                                     boolean                                     check(                                     int                                     n)                       
                                                 {                       
                                                 // Count digits in n.                       
                                                 int                                     count_digits = Integer.toString(n).length();                       
                                   
                                                 // Compute sum of terms like digit multiplied by                       
                                                 // power of position                       
                                                 int                                     sum =                                     0                                     ;                                     // Initialize sum of terms                       
                                                 int                                     x = n;                       
                                                 while                                     (x!=                                     0                                     )                       
                                                 {                       
                                                 // Get the rightmost digit                       
                                                 int                                     r = x%                                     10                                     ;                       
                                   
                                                 // Sum the digits by powering according to                       
                                                 // the positions                       
                                                 sum = (                                     int                                     ) (sum + Math.pow(r, count_digits--));                       
                                                 x = x/                                     10                                     ;                       
                                                 }                       
                                   
                                                 // If sum is same as number, then number is                       
                                                 return                                     (sum == n);                       
                                                 }                       
                                   
                                                 // Driver method                       
                                                 public                                     static                                     void                                     main(String[] args)                       
                                                 {                       
                                                 int                                     n =                                     135                                     ;                       
                                   
                                                 System.out.println(check(n) ?                                     "Disarium Number"                                     :                                     "Not a Disarium Number"                                     );                       
                                                 }                       
                         }                       

Python3

                         # Python program to check whether a number is Disarium                       
                         # or not                       
                         import                                     math                       
           
                         # Method to check whether a number is disarium or not                       
                         def                                     check(n) :                       
           
                                                 # Count digits in n.                       
                                                 count_digits                                     =                                     len                                     (                                     str                                     (n))                       
                                   
                                                 # Compute sum of terms like digit multiplied by                       
                                                 # power of position                       
                                                 sum                                     =                                     0                                     # Initialize sum of terms                       
                                                 x                                     =                                     n                       
                                                 while                                     (x!                                     =                                     0                                     ) :                       
           
                                                 # Get the rightmost digit                       
                                                 r                                     =                                     x                                     %                                     10                       
                                   
                                                 # Sum the digits by powering according to                       
                                                 # the positions                       
                                                 sum                                     =                                     (                                     int                                     ) (                                     sum                                     +                                     math.                                     pow                                     (r, count_digits))                       
                                                 count_digits                                     =                                     count_digits                                     -                                     1                       
                                                 x                                     =                                     x                                     /                                     /                                     10                       
                                   
                                                 # If sum is same as number, then number is                       
                                                 if                                     sum                                     =                                     =                                     n :                       
                                                 return                                     1                       
                                                 else                                     :                       
                                                 return                                     0                       
                                   
                         # Driver method                       
                         n                                     =                                     135                       
                         if                                     (check(n)                                     =                                     =                                     1                                     ) :                       
                                                 print                                     (                                     "Disarium Number"                                     )                       
                         else                                     :                       
                                                 print                                     (                                     "Not a Disarium Number"                                     )                       
                                   
                         # This code is contributed by Nikita Tiwari.                       

C#

                         // C# program to check whether a number                       
                         // is Desoriam or not                       
                         using                                     System;                       
           
                         class                                     GFG{                       
                                   
                         // Method to check whether a number                       
                         // is disarium or not                       
                         static                                     bool                                     check(                                     int                                     n)                       
                         {                       
                                   
                                                 // Count digits in n.                       
                                                 int                                     count_digits = n.ToString().Length;                       
                                   
                                                 // Compute sum of terms like digit                       
                                                 // multiplied by power of position                       
                                                 // Initialize sum of terms                       
                                                 int                                     sum = 0;                       
                                                 int                                     x = n;                       
                                   
                                                 while                                     (x != 0)                       
                                                 {                       
                                   
                                                 // Get the rightmost digit                       
                                                 int                                     r = x % 10;                       
                                   
                                                 // Sum the digits by powering according                       
                                                 // to the positions                       
                                                 sum = (                                     int                                     )(sum + Math.Pow(                       
                                                 r, count_digits--));                       
                                                 x = x / 10;                       
                                                 }                       
                                   
                                                 // If sum is same as number,                       
                                                 // then number is                       
                                                 return                                     (sum == n);                       
                         }                       
           
                         // Driver code                       
                         public                                     static                                     void                                     Main(                                     string                                     [] args)                       
                         {                       
                                                 int                                     n = 135;                       
                                   
                                                 Console.Write(check(n) ?                                     "Disarium Number"                                     :                       
                                                 "Not a Disarium Number"                                     );                       
                         }                       
                         }                       
           
                         // This code is contributed by rutvik_56                       

Javascript

                         <script>                       
           
                         // JavaScript program to check whether a number is Desoriam                       
                         // or not                       
                         // Method to check whether a number is disarium or not                       
                         function                                     check(n)                       
                                                 {                       
                                   
                                                 // Count digits in n.                       
                                                 var                                     count_digits = Number.toString();                       
                                   
                                                 // Compute sum of terms like digit multiplied by                       
                                                 // power of position                       
                                                 var                                     sum = 0;                                     // Initialize sum of terms                       
                                                 var                                     x = n;                       
                                                 while                                     (x!=0)                       
                                                 {                       
                                                 // Get the rightmost digit                       
                                                 var                                     r = x%10;                       
                                   
                                                 // Sum the digits by powering according to                       
                                                 // the positions                       
                                                 sum = (sum + Math.pow(r, count_digits--));                       
                                                 x = x/10;                       
                                                 }                       
                                   
                                                 // If sum is same as number, then number is                       
                                                 return                                     (sum = n);                       
                                                 }                       
                                   
                                                 // Driver method                       
                                                 var                                     n = 135;                       
                                   
                                                 document.write(check(n) ?                                     "Disarium Number"                                     :                                     "Not a Disarium Number"                                     );                       
                                   
                         // This code is contributed by shivanisinghss2110                       
                         </script>                       

输出：

Disarium Number

本文由 萨希尔·查布拉（杀手） .如果你喜欢GeekSforgek，并想贡献自己的力量，你也可以使用写极客。组织或者把你的文章寄去评论-team@geeksforgeeks.org.看到你的文章出现在Geeksforgeks主页上，并帮助其他极客。如果您发现任何不正确的地方，或者您想分享有关上述主题的更多信息，请写下评论。

文章版权归作者所有，未经允许请勿转载。

THE END

技术文章