求一个数的奇数因子和的程序-yiteyi-C++库

给定一个数字n，任务是找到奇数因子和。 例如：

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Input : n = 30Output : 24Odd dividers sum 1 + 3 + 5 + 15 = 24 Input : 18Output : 13Odd dividers sum 1 + 3 + 9 = 13

让p _1. P _2. ，…p _K 是主要因素让我们 _1. A. _2. , .. A. _K 是p的最高权力 _1. P _2. , .. P _K 分别除以n，也就是说，我们可以把n写成 n=（p _1. ^A. _^1. )*（p _2. ^A. _^2. )*…（p _K ^A. _^K ) .

Sum of divisors = (1 + p₁ + p₁² ... p₁^a_¹) *                   (1 + p₂ + p₂² ... p₂^a_²) *                  .............................................                  (1 + p_k + p_k² ... p_k^a_^k)

为了求奇数因子之和，我们只需要忽略偶数因子及其幂。例如，考虑n＝18。它可以写成2 ^1. 3. ^2. 所有因素中的太阳是（1）*（1+2）*（1+3+3） ^2. ).奇数因子之和（1）*（1+3+3） ^2. ) = 13. 为了去除所有偶数因子，我们反复地将n除以2。在这一步之后，我们只得到奇数因子。请注意，2是唯一的偶数素数。

C++

                         // Formula based Java program                       
                         // to find sum of all divisors                       
                         // of n.                       
                         #include <iostream>                       
                         #include <cmath>                       
                         using                                     namespace                                     std;                       
           
                         // Returns sum of all                       
                         // factors of n.                       
                         int                                     sumofoddFactors(                                     int                                     n)                       
                         {                       
                                                 // Traversing through                       
                                                 // all prime factors.                       
                                                 int                                     res = 1;                       
           
                                                 // ignore even factors by                       
                                                 // removing all powers                       
                                                 // of 2                       
                                                 while                                     (n % 2 == 0)                       
                                                 n = n / 2;                       
           
                                                 for                                     (                                     int                                     i = 3;                       
                                                 i <=                                     sqrt                                     (n); i++)                       
                                                 {                       
           
                                                 // While i divides n,                       
                                                 // print i and divide n                       
                                                 int                                     count = 0, curr_sum = 1;                       
                                                 int                                     curr_term = 1;                       
                                                 while                                     (n % i == 0)                       
                                                 {                       
                                                 count++;                       
                                                 n = n / i;                       
                                                 curr_term *= i;                       
                                                 curr_sum += curr_term;                       
                                                 }                       
                                                 res *= curr_sum;                       
                                                 }                       
           
                                                 // This condition is to handle                       
                                                 // the case when n is a                       
                                                 // prime number.                       
                                                 if                                     (n >= 2)                       
                                                 res *= (1 + n);                       
           
                                                 return                                     res;                       
                         }                       
           
                         // Driver code                       
                         int                                     main()                       
                         {                       
                                                 int                                     n = 30;                       
                                                 cout << sumofoddFactors(n);                       
                                                 return                                     0;                       
                         }                       
                         // This code is contributed by                       
                         // Manish Shaw(manishshaw1)                       

JAVA

                         // Formula based Java program                       
                         // to find sum of all divisors                       
                         // of n.                       
                         import                                     java.io.*;                       
                         import                                     java.math.*;                       
           
                         class                                     GFG {                       
                                   
                                                 // Returns sum of all                       
                                                 // factors of n.                       
                                                 static                                     int                                     sumofoddFactors(                                     int                                     n)                       
                                                 {                       
                                                 // Traversing through                       
                                                 // all prime factors.                       
                                                 int                                     res =                                     1                                     ;                       
                                   
                                                 // ignore even factors by                       
                                                 // removing all powers                       
                                                 // of 2                       
                                                 while                                     (n %                                     2                                     ==                                     0                                     )                       
                                                 n = n /                                     2                                     ;                       
                                   
                                                 for                                     (                                     int                                     i =                                     3                                     ; i <= Math.sqrt(n); i++)                       
                                                 {                       
                                   
                                                 // While i divides n, print i                       
                                                 // and divide n                       
                                                 int                                     count =                                     0                                     , curr_sum =                                     1                                     ;                       
                                                 int                                     curr_term =                                     1                                     ;                       
                                                 while                                     (n % i ==                                     0                                     )                       
                                                 {                       
                                                 count++;                       
                                   
                                                 n = n / i;                       
                                   
                                                 curr_term *= i;                       
                                                 curr_sum += curr_term;                       
                                                 }                       
                                   
                                                 res *= curr_sum;                       
                                   
                                                 }                       
                                   
                                                 // This condition is to handle                       
                                                 // the case when n is a                       
                                                 // prime number.                       
                                                 if                                     (n >=                                     2                                     )                       
                                                 res *= (                                     1                                     + n);                       
                                   
                                                 return                                     res;                       
                                                 }                       
                                   
                                                 // Driver code                       
                                                 public                                     static                                     void                                     main(String args[])                       
                                                 throws                                     IOException                       
                                                 {                       
                                                 int                                     n =                                     30                                     ;                       
                                                 System.out.println(sumofoddFactors(n));                       
                                                 }                       
                         }                       
           
                         /* This code is contributed by Nikita Tiwari.*/                       

Python3

                         # Formula based Python program                       
                         # to find sum of all divisors                       
                         # of n.                       
                         import                                     math                       
           
                         # Returns sum of all                       
                         # factors of n.                       
                         def                                     sumofoddFactors(n) :                       
           
                                                 # Traversing through                       
                                                 # all prime factors.                       
                                                 res                                     =                                     1                       
           
                                                 # ignore even factors by                       
                                                 # removing all powers                       
                                                 # of 2                       
                                                 while                                     (n                                     %                                     2                                     =                                     =                                     0                                     ) :                       
                                                 n                                     =                                     int                                     (n                                     /                                     2                                     )                       
           
                                                 for                                     i                                     in                                     range                                     (                                     3                                     ,                                     int                                     (math.sqrt(n))                                     +                                     1                                     ) :                       
                                   
                                                 # While i divides n,                       
                                                 # pri and divide n                       
                                                 count                                     =                                     0                       
                                                 curr_sum                                     =                                     1                       
                                                 curr_term                                     =                                     1                       
                                                 while                                     (n                                     %                                     i                                     =                                     =                                     0                                     ) :                       
                                   
                                                 count                                     =                                     count                                     +                                     1                       
                                                 n                                     =                                     int                                     (n                                     /                                     i)                       
                                                 curr_term                                     *                                     =                                     i                       
                                                 curr_sum                                     =                                     curr_sum                                     +                                     curr_term                       
                                   
                                                 res                                     =                                     res                                     *                                     curr_sum                       
                                   
                                                 # This condition is to                       
                                                 # handle the case when                       
                                                 # n is a prime number.                       
                                                 if                                     (n >                                     =                                     2                                     ) :                       
                                                 res                                     =                                     res                                     *                                     (                                     1                                     +                                     n)                       
           
                                                 return                                     res                       
           
                         # Driver code                       
                         n                                     =                                     30                       
                         print                                     (sumofoddFactors(n))                       
           
                         # This code is contributed by                       
                         # Manish Shaw(manishshaw1)                       

C#

                         // Formula based C# program to find sum                       
                         // of all divisors of n.                       
                         using                                     System;                       
           
                         class                                     GFG {                       
                                   
                                                 // Returns sum of all factors of n.                       
                                                 static                                     int                                     sumofoddFactors(                                     int                                     n)                       
                                                 {                       
                                   
                                                 // Traversing through                       
                                                 // all prime factors.                       
                                                 int                                     res = 1;                       
                                   
                                                 // ignore even factors by                       
                                                 // removing all powers                       
                                                 // of 2                       
                                                 while                                     (n % 2 == 0)                       
                                                 n = n / 2;                       
                                   
                                                 for                                     (                                     int                                     i = 3; i <= Math.Sqrt(n); i++)                       
                                                 {                       
                                   
                                                 // While i divides n, print i                       
                                                 // and divide n                       
                                                 int                                     count = 0, curr_sum = 1;                       
                                                 int                                     curr_term = 1;                       
                                                 while                                     (n % i == 0)                       
                                                 {                       
                                                 count++;                       
                                   
                                                 n = n / i;                       
                                   
                                                 curr_term *= i;                       
                                                 curr_sum += curr_term;                       
                                                 }                       
                                   
                                                 res *= curr_sum;                       
                                   
                                                 }                       
                                   
                                                 // This condition is to handle                       
                                                 // the case when n is a                       
                                                 // prime number.                       
                                                 if                                     (n >= 2)                       
                                                 res *= (1 + n);                       
                                   
                                                 return                                     res;                       
                                                 }                       
                                   
                                                 // Driver code                       
                                                 public                                     static                                     void                                     Main()                       
                                   
                                                 {                       
                                                 int                                     n = 30;                       
                                   
                                                 Console.Write(sumofoddFactors(n));                       
                                                 }                       
           
                         }                       
           
                         // This code is contributed by nitin mittal.                       

PHP

                         <?php                       
                         // Formula based PHP program                       
                         // to find sum of all divisors                       
                         // of n.                       
           
                         // Returns sum of all                       
                         // factors of n.                       
                         function                                     sumofoddFactors(                                     $n                                     )                       
                         {                       
                                                 // Traversing through                       
                                                 // all prime factors.                       
                                                 $res                                     = 1;                       
           
                                                 // ignore even factors by                       
                                                 // removing all powers                       
                                                 // of 2                       
                                                 while                                     (                                     $n                                     % 2 == 0)                       
                                                 $n                                     =                                     intval                                     (                                     $n                                     / 2);                       
           
                                                 for                                     (                                     $i                                     = 3;                       
                                                 $i                                     <= sqrt(                                     $n                                     );                                     $i                                     ++)                       
                                                 {                       
           
                                                 // While i divides n,                       
                                                 // pr$i and divide n                       
                                                 $count                                     = 0;                       
                                                 $curr_sum                                     = 1;                       
                                                 $curr_term                                     = 1;                       
                                                 while                                     (                                     $n                                     %                                     $i                                     == 0)                       
                                                 {                       
                                                 $count                                     ++;                       
                                                 $n                                     =                                     intval                                     (                                     $n                                     /                                     $i                                     );                       
                                                 $curr_term                                     *=                                     $i                                     ;                       
                                                 $curr_sum                                     +=                                     $curr_term                                     ;                       
                                                 }                       
                                   
                                                 $res                                     *=                                     $curr_sum                                     ;                       
                                                 }                       
           
                                                 // This condition is to                       
                                                 // handle the case when                       
                                                 // n is a prime number.                       
                                                 if                                     (                                     $n                                     >= 2)                       
                                                 $res                                     *= (1 +                                     $n                                     );                       
           
                                                 return                                     $res                                     ;                       
                         }                       
           
                         // Driver code                       
                         $n                                     = 30;                       
                         echo                                     (sumofoddFactors(                                     $n                                     ));                       
           
                         // This code is contributed by                       
                         // Manish Shaw(manishshaw1)                       
                         ?>                       

Javascript

                         <script>                       
           
                                                 // Formula based Javascript program                       
                                                 // to find sum of all divisors                       
                                                 // of n.                       
           
                                                 // Returns sum of all                       
                                                 // factors of n.                       
                                                 function                                     sumofoddFactors(n)                       
                                                 {                       
                                                 // Traversing through                       
                                                 // all prime factors.                       
                                                 let res = 1;                       
           
                                                 // ignore even factors by                       
                                                 // removing all powers                       
                                                 // of 2                       
                                                 while                                     (n % 2 == 0)                       
                                                 n = n / 2;                       
           
                                                 for                                     (let i = 3; i <= n; i++)                       
                                                 {                       
           
                                                 // While i divides n,                       
                                                 // print i and divide n                       
                                                 let count = 0, curr_sum = 1;                       
                                                 let curr_term = 1;                       
                                                 while                                     (n % i == 0)                       
                                                 {                       
                                                 count++;                       
                                                 n = n / i;                       
                                                 curr_term *= i;                       
                                                 curr_sum += curr_term;                       
                                                 }                       
                                                 res *= curr_sum;                       
                                                 }                       
           
                                                 // This condition is to handle                       
                                                 // the case when n is a                       
                                                 // prime number.                       
                                                 if                                     (n >= 2)                       
                                                 res *= (1 + n);                       
           
                                                 return                                     res;                       
                                                 }                       
           
                                                 let n = 30;                       
                                                 document.write(sumofoddFactors(n));                       
           
                         </script>                       

输出：

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