自然数的所有真因子之和-yiteyi-C++库

给定一个自然数，计算其所有适当因子之和。自然数的适当除数是严格小于该数的除数。例如，数字20有5个适当的除数：1，2，4，5，10，除数之和为：1+2+4+5+10=22。 例如：

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Input : num = 10Output: 8// proper divisors 1 + 2 + 5 = 8 Input : num = 36Output: 55// proper divisors 1 + 2 + 3 + 4 + 6 + 9 + 12 + 18 = 55

建议：请在“上解决” 实践 “首先，在讨论解决方案之前。

这个问题有着非常重要的意义 简单解决方案 ，我们都知道，对于任何一个数’num’，它的所有除数总是小于等于’num/2’，所有素因子总是小于等于 sqrt（数量） .所以我们迭代’i’直到i<=sqrt（num），对于任何’i’如果它除以’num’，那么我们得到两个除数’i’和’num/i’，不断地添加这些除数，但是对于一些数字，除数’i’和’num/i’在这种情况下只需添加一个除数，例如；num=36，所以对于i=6，我们将得到（num/i）=6，这就是为什么我们在求和中只得到一次6。最后我们加一，因为一是所有自然数的除数。

C++

                         // C++ program to find sum of all divisors of                       
                         // a natural number                       
                         #include<bits/stdc++.h>                       
                         using                                     namespace                                     std;                       
           
                         // Function to calculate sum of all proper divisors                       
                         // num --> given natural number                       
                         int                                     divSum(                                     int                                     num)                       
                         {                       
                                                 // Final result of summation of divisors                       
                                                 int                                     result = 0;                       
                                                 if                                     (num == 1)                                     // there will be no proper divisor                       
                                                 return                                     result;                       
                                                 // find all divisors which divides 'num'                       
                                                 for                                     (                                     int                                     i=2; i<=                                     sqrt                                     (num); i++)                       
                                                 {                       
                                                 // if 'i' is divisor of 'num'                       
                                                 if                                     (num%i==0)                       
                                                 {                       
                                                 // if both divisors are same then add                       
                                                 // it only once else add both                       
                                                 if                                     (i==(num/i))                       
                                                 result += i;                       
                                                 else                       
                                                 result += (i + num/i);                       
                                                 }                       
                                                 }                       
           
                                                 // Add 1 to the result as 1 is also a divisor                       
                                                 return                                     (result + 1);                       
                         }                       
           
                         // Driver program to run the case                       
                         int                                     main()                       
                         {                       
                                                 int                                     num = 36;                       
                                                 cout << divSum(num);                       
                                                 return                                     0;                       
                         }                       

JAVA

                         // JAVA program to find sum of all divisors                       
                         // of a natural number                       
                         import                                     java.math.*;                       
           
                         class                                     GFG {                       
                                   
                                                 // Function to calculate sum of all proper                       
                                                 // divisors num --> given natural number                       
                                                 static                                     int                                     divSum(                                     int                                     num)                       
                                                 {                       
                                                 // Final result of summation of divisors                       
                                                 int                                     result =                                     0                                     ;                       
                                   
                                                 // find all divisors which divides 'num'                       
                                                 for                                     (                                     int                                     i =                                     2                                     ; i <= Math.sqrt(num); i++)                       
                                                 {                       
                                                 // if 'i' is divisor of 'num'                       
                                                 if                                     (num % i ==                                     0                                     )                       
                                                 {                       
                                                 // if both divisors are same then                       
                                                 // add it only once else add both                       
                                                 if                                     (i == (num / i))                       
                                                 result += i;                       
                                                 else                       
                                                 result += (i + num / i);                       
                                                 }                       
                                                 }                       
                                   
                                                 // Add 1 to the result as 1 is also                       
                                                 // a divisor                       
                                                 return                                     (result +                                     1                                     );                       
                                                 }                       
                                   
                                                 // Driver program to run the case                       
                                                 public                                     static                                     void                                     main(String[] args)                       
                                                 {                       
                                                 int                                     num =                                     36                                     ;                       
                                                 System.out.println(divSum(num));                       
                                                 }                       
                         }                       
           
                         /*This code is contributed by Nikita Tiwari*/                       

Python3

                         # PYTHON program to find sum of all                       
                         # divisors of a natural number                       
                         import                                     math                       
                                   
                         # Function to calculate sum of all proper                       
                         # divisors num --> given natural number                       
                         def                                     divSum(num) :                       
                                   
                                                 # Final result of summation of divisors                       
                                                 result                                     =                                     0                       
                                   
                                                 # find all divisors which divides 'num'                       
                                                 i                                     =                                     2                       
                                                 while                                     i<                                     =                                     (math.sqrt(num)) :                       
                                   
                                                 # if 'i' is divisor of 'num'                       
                                                 if                                     (num                                     %                                     i                                     =                                     =                                     0                                     ) :                       
                                   
                                                 # if both divisors are same then                       
                                                 # add it only once else add both                       
                                                 if                                     (i                                     =                                     =                                     (num                                     /                                     i)) :                       
                                                 result                                     =                                     result                                     +                                     i;                       
                                                 else                                     :                       
                                                 result                                     =                                     result                                     +                                     (i                                     +                                     num                                     /                                     i);                       
                                                 i                                     =                                     i                                     +                                     1                       
                                   
                                                 # Add 1 to the result as 1 is also                       
                                                 # a divisor                       
                                                 return                                     (result                                     +                                     1                                     );                       
                                   
                         # Driver program to run the case                       
                         num                                     =                                     36                       
                         print                                     (divSum(num))                       
           
                         # This code is contributed by Nikita Tiwari                       

C#

                         // C# program to find sum of all                       
                         // divisorsof a natural number                       
                         using                                     System;                       
           
                         class                                     GFG {                       
                                   
                                                 // Function to calculate sum of all proper                       
                                                 // divisors num --> given natural number                       
                                                 static                                     int                                     divSum(                                     int                                     num)                       
                                                 {                       
                                   
                                                 // Final result of summation of divisors                       
                                                 int                                     result = 0;                       
                                   
                                                 // find all divisors which divides 'num'                       
                                                 for                                     (                                     int                                     i = 2; i <= Math.Sqrt(num); i++)                       
                                                 {                       
                                   
                                                 // if 'i' is divisor of 'num'                       
                                                 if                                     (num % i == 0)                       
                                                 {                       
                                   
                                                 // if both divisors are same then                       
                                                 // add it only once else add both                       
                                                 if                                     (i == (num / i))                       
                                                 result += i;                       
                                                 else                       
                                                 result += (i + num / i);                       
                                                 }                       
                                                 }                       
                                   
                                                 // Add 1 to the result as 1                       
                                                 // is also a divisor                       
                                                 return                                     (result + 1);                       
                                                 }                       
                                   
                                                 // Driver Code                       
                                                 public                                     static                                     void                                     Main()                       
                                                 {                       
                                                 int                                     num = 36;                       
                                                 Console.Write(divSum(num));                       
                                                 }                       
                         }                       
           
                         // This code is contributed by Nitin Mittal.                       

PHP

                         <?php                       
                         // PHP program to find sum of                       
                         // all divisors of a natural number                       
           
                         // Function to calculate sum of                       
                         // all proper divisors                       
                         // num --> given natural number                       
                         function                                     divSum(                                     $num                                     )                       
                         {                       
                                                 // Final result of                       
                                                 // summation of divisors                       
                                                 $result                                     = 0;                       
           
                                                 // find all divisors                       
                                                 // which divides 'num'                       
                                                 for                                     (                                     $i                                     = 2;                                     $i                                     <= sqrt(                                     $num                                     );                       
                                                 $i                                     ++)                       
                                                 {                       
                                                 // if 'i' is divisor of 'num'                       
                                                 if                                     (                                     $num                                     %                                     $i                                     == 0)                       
                                                 {                       
                                                 // if both divisors are                       
                                                 // same then add it only                       
                                                 // once else add both                       
                                                 if                                     (                                     $i                                     == (                                     $num                                     /                                     $i                                     ))                       
                                                 $result                                     +=                                     $i                                     ;                       
                                                 else                       
                                                 $result                                     += (                                     $i                                     +                                     $num                                     /                                     $i                                     );                       
                                                 }                       
                                                 }                       
           
                                                 // Add 1 to the result as                       
                                                 // 1 is also a divisor                       
                                                 return                                     (                                     $result                                     + 1);                       
                         }                       
           
                         // Driver Code                       
                         $num                                     = 36;                       
                         echo                                     (divSum(                                     $num                                     ));                       
           
                         // This code is contributed by Ajit.                       
                         ?>                       

Javascript

                         <script>                       
           
                         // Javascript program to find sum of all divisors of                       
                         // a natural number                       
           
                         // Function to calculate sum of all proper divisors                       
                         // num --> given natural number                       
                         function                                     divSum(num)                       
                         {                       
                                                 // Final result of summation of divisors                       
                                                 let result = 0;                       
           
                                                 // find all divisors which divides 'num'                       
                                                 for                                     (let i=2; i<=Math.sqrt(num); i++)                       
                                                 {                       
                                                 // if 'i' is divisor of 'num'                       
                                                 if                                     (num%i==0)                       
                                                 {                       
                                                 // if both divisors are same then add                       
                                                 // it only once else add both                       
                                                 if                                     (i==(num/i))                       
                                                 result += i;                       
                                                 else                       
                                                 result += (i + num/i);                       
                                                 }                       
                                                 }                       
           
                                                 // Add 1 to the result as 1 is also a divisor                       
                                                 return                                     (result + 1);                       
                         }                       
           
                         // Driver program to run the case                       
           
                                                 let num = 36;                       
                                                 document.write(divSum(num));                       
                                   
                         // This code is contributed by Mayank Tyagi                       
                                   
                         </script>                       

输出：

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