检查是否有友好的一对-yiteyi-C++库

友好的数字两个不同的数字是如此相关以至于真因子之和每一个数等于另一个数。（一个数的适当除数是该数的正因子，而不是该数本身。）。 例如：

null

Input : x = 220, y = 284
Output : Yes 
Proper divisors of 220 are 1, 2, 4, 5,
10, 11, 20, 22, 44, 55 and 110. Sum of 
these is 284. Proper divisors of 284 
are 1, 2, 4, 71 and 142 with sum 220.

Input : 1 2
Output :No

建议：请在“上解决” 实践 “首先，在讨论解决方案之前。

逻辑很简单。我们比较真因子之和并将一个数字的和与另一个数字的和进行比较。

C++

                         // CPP program to check if two numbers are                       
                         // Amicable or not.                       
                         #include <bits/stdc++.h>                       
                         using                                     namespace                                     std;                       
                                   
                         // Function to calculate sum of all                       
                         // proper divisors of a given number                       
                         int                                     divSum(                                     int                                     n)                       
                         {                       
                                                 // Sum of divisors                       
                                                 int                                     result = 0;                       
                                   
                                                 // find all divisors which divides 'num'                       
                                                 for                                     (                                     int                                     i = 2; i <=                                     sqrt                                     (n); i++)                       
                                                 {                       
                                                 // if 'i' is divisor of 'n'                       
                                                 if                                     (n % i == 0)                       
                                                 {                       
                                                 // if both divisors are same                       
                                                 // then add it once else add                       
                                                 // both                       
                                                 if                                     (i == (n / i))                       
                                                 result += i;                       
                                                 else                       
                                                 result += (i + n/i);                       
                                                 }                       
                                                 }                       
                                   
                                                 // Add 1 and n to result as above loop                       
                                                 // considers proper divisors greater                       
                                                 // than 1.                       
                                                 return                                     (result + 1);                       
                         }                       
                                   
                         // Returns true if x and y are Amicable                       
                         // else false.                       
                         bool                                     areAmicable(                                     int                                     x,                                     int                                     y)                       
                         {                       
                                                 if                                     (divSum(x) != y)                       
                                                 return                                     false                                     ;                       
                                   
                                                 return                                     (divSum(y) == x);                       
                         }                       
                                   
                         int                                     main() {                       
                                                 int                                     x = 220, y = 284;                       
                                                 if                                     (areAmicable(x, y))                       
                                                 cout <<                                     "Yes"                                     ;                       
                                                 else                       
                                                 cout <<                                     "No"                                     ;                       
                                                 return                                     0;                       
                         }                       

JAVA

                         // JAVA program to check if two numbers are                       
                         // Amicable or not.                       
                         import                                     java.io.*;                       
                                   
                         class                                     GFG                       
                         {                       
                                                 // Function to calculate sum of all                       
                                                 // proper divisors of a given number                       
                                                 static                                     int                                     divSum(                                     int                                     n)                       
                                                 {                       
                                                 // Sum of divisors                       
                                                 int                                     result =                                     0                                     ;                       
                                   
                                                 // find all divisors which divides 'num'                       
                                                 for                                     (                                     int                                     i =                                     2                                     ; i <= Math.sqrt(n); i++)                       
                                                 {                       
                                                 // if 'i' is divisor of 'n'                       
                                                 if                                     (n % i ==                                     0                                     )                       
                                                 {                       
                                                 // if both divisors are same                       
                                                 // then add it once else add                       
                                                 // both                       
                                                 if                                     (i == (n / i))                       
                                                 result += i;                       
                                                 else                       
                                                 result += (i + n / i);                       
                                                 }                       
                                                 }                       
                                   
                                                 // Add 1 and n to result as above loop                       
                                                 // considers proper divisors greater                       
                                                 // than 1.                       
                                                 return                                     (result +                                     1                                     );                       
                                                 }                       
                                   
                                                 // Returns true if x and y are Amicable                       
                                                 // else false.                       
                                                 static                                     boolean                                     areAmicable(                                     int                                     x,                                     int                                     y)                       
                                                 {                       
                                                 if                                     (divSum(x) != y)                       
                                                 return                                     false                                     ;                       
                                   
                                                 return                                     (divSum(y) == x);                       
                                                 }                       
                                   
                                                 public                                     static                                     void                                     main (String[] args)                       
                                                 {                       
                                                 int                                     x =                                     220                                     , y =                                     284                                     ;                       
                                                 if                                     (areAmicable(x, y))                       
                                                 System.out.println(                                     "Yes"                                     );                       
                                                 else                       
                                                 System.out.println(                                     "No"                                     );                       
                                   
                                                 }                       
                         }                       
                                   
                         // This code is contributed by vt_m.                       

Python3

                         # Python program to check                       
                         # if two numbers are                       
                         # Amicable or not.                       
                         import                                     math                       
                                   
                         # def to calculate sum                       
                         # of all proper divisors                       
                         # of a given number                       
                         def                                     divSum(n) :                       
                                   
                                                 # Sum of divisors                       
                                                 result                                     =                                     0                       
                                   
                                                 # find all divisors                       
                                                 # which divides 'num'                       
                                                 for                                     i                                     in                                     range                                     (                                     2                                     ,                                     int                                     (math.sqrt(n))                                     +                                     1                                     ) :                       
                                   
                                                 # if 'i' is                       
                                                 # divisor of 'n'                       
                                                 if                                     (n                                     %                                     i                                     =                                     =                                     0                                     ) :                       
                                   
                                                 # if both divisors are same                       
                                                 # then add it once else add                       
                                                 # both                       
                                                 if                                     (i                                     =                                     =                                     int                                     (n                                     /                                     i)) :                       
                                                 result                                     =                                     result                                     +                                     i                       
                                                 else                                     :                       
                                                 result                                     =                                     result                                     +                       
                                                 (i                                     +                                     int                                     (n                                     /                                     i))                       
                                   
                                                 # Add 1 and n to result                       
                                                 # as above loop considers                       
                                                 # proper divisors greater                       
                                                 # than 1.                       
                                                 return                                     (result                                     +                                     1                                     )                       
                                   
                         # Returns true if x and y                       
                         # are Amicable else false.                       
                         def                                     areAmicable(x, y) :                       
                                   
                                                 if                                     (divSum(x) !                                     =                                     y) :                       
                                                 return                                     False                       
                                   
                                                 return                                     (divSum(y)                                     =                                     =                                     x)                       
                                   
                         # Driver Code                       
                         x                                     =                                     220                       
                         y                                     =                                     284                       
                         if                                     (areAmicable(x, y)) :                       
                                                 print                                     (                                     "Yes"                                     )                       
                         else                                     :                       
                                                 print                                     (                                     "No"                                     )                       
                                   
                         # This code is contributed by                       
                         # Manish Shaw(manishshaw1)                       

C#

                         // C# program to check if two numbers are                       
                         // Amicable or not.                       
                         using                                     System;                       
                                   
                         class                                     GFG                       
                         {                       
                                                 // Function to calculate sum of all                       
                                                 // proper divisors of a given number                       
                                                 static                                     int                                     divSum(                                     int                                     n)                       
                                                 {                       
                                                 // Sum of divisors                       
                                                 int                                     result = 0;                       
                                   
                                                 // find all divisors which divides 'num'                       
                                                 for                                     (                                     int                                     i = 2; i <= Math.Sqrt(n); i++)                       
                                                 {                       
                                                 // if 'i' is divisor of 'n'                       
                                                 if                                     (n % i == 0)                       
                                                 {                       
                                                 // if both divisors are same                       
                                                 // then add it once else add                       
                                                 // both                       
                                                 if                                     (i == (n / i))                       
                                                 result += i;                       
                                                 else                       
                                                 result += (i + n / i);                       
                                                 }                       
                                                 }                       
                                   
                                                 // Add 1 and n to result as above loop                       
                                                 // considers proper divisors greater                       
                                                 // than 1.                       
                                                 return                                     (result + 1);                       
                                                 }                       
                                   
                                                 // Returns true if x and y are Amicable                       
                                                 // else false.                       
                                                 static                                     bool                                     areAmicable(                                     int                                     x,                                     int                                     y)                       
                                                 {                       
                                                 if                                     (divSum(x) != y)                       
                                                 return                                     false                                     ;                       
                                   
                                                 return                                     (divSum(y) == x);                       
                                                 }                       
                                   
                                                 public                                     static                                     void                                     Main ()                       
                                                 {                       
                                                 int                                     x = 220, y = 284;                       
                                   
                                                 if                                     (areAmicable(x, y))                       
                                                 Console.WriteLine(                                     "Yes"                                     );                       
                                                 else                       
                                                 Console.WriteLine(                                     "No"                                     );                       
                                   
                                                 }                       
                         }                       
                                   
                         // This code is contributed by vt_m.                       

PHP

                         <?php                       
                         // PHP program to check if two                       
                         // numbers are Amicable or not.                       
                                   
                         // Function to calculate sum of all                       
                         // proper divisors of a given number                       
                         function                                     divSum(                                     $n                                     )                       
                         {                       
                                   
                                                 // Sum of divisors                       
                                                 $result                                     = 0;                       
                                   
                                                 // find all divisors                       
                                                 // which divides 'num'                       
                                                 for                                     (                                     $i                                     = 2;                                     $i                                     <= sqrt(                                     $n                                     );                                     $i                                     ++)                       
                                                 {                       
                                   
                                                 // if 'i' is divisor of 'n'                       
                                                 if                                     (                                     $n                                     %                                     $i                                     == 0)                       
                                                 {                       
                                   
                                                 // if both divisors are same                       
                                                 // then add it once else add                       
                                                 // both                       
                                                 if                                     (                                     $i                                     == (                                     $n                                     /                                     $i                                     ))                       
                                                 $result                                     +=                                     $i                                     ;                       
                                                 else                       
                                                 $result                                     += (                                     $i                                     +                                     $n                                     /                                     $i                                     );                       
                                                 }                       
                                                 }                       
                                   
                                                 // Add 1 and n to result                       
                                                 // as above loop considers                       
                                                 // proper divisors greater                       
                                                 // than 1.                       
                                                 return                                     (                                     $result                                     + 1);                       
                         }                       
                                   
                         // Returns true if x and y                       
                         // are Amicable else false.                       
                         function                                     areAmicable(                                     $x                                     ,                                     $y                                     )                       
                         {                       
                                                 if                                     (divSum(                                     $x                                     ) !=                                     $y                                     )                       
                                                 return                                     false;                       
                                   
                                                 return                                     (divSum(                                     $y                                     ) ==                                     $x                                     );                       
                         }                       
                                   
                                                 // Driver Code                       
                                                 $x                                     = 220;                       
                                                 $y                                     = 284;                       
                                                 if                                     (areAmicable(                                     $x                                     ,                                     $y                                     ))                       
                                                 echo                                     "Yes"                                     ;                       
                                                 else                       
                                                 echo                                     "No"                                     ;                       
                                   
                         // This code is contributed by anuj_67.                       
                         ?>                       

Javascript

                         <script>                       
                                                 // Javascript program to check if two                       
                         // numbers are Amicable or not.                       
                                   
                         // Function to calculate sum of all                       
                         // proper divisors of a given number                       
                         function                                     divSum(n)                       
                         {                       
                                   
                                                 // Sum of divisors                       
                                                 let result = 0;                       
                                   
                                                 // find all divisors                       
                                                 // which divides 'num'                       
                                                 for                                     (let i = 2; i <= Math.sqrt(n); i++)                       
                                                 {                       
                                   
                                                 // if 'i' is divisor of 'n'                       
                                                 if                                     (n % i == 0)                       
                                                 {                       
                                   
                                                 // if both divisors are same                       
                                                 // then add it once else add                       
                                                 // both                       
                                                 if                                     (i == (n / i))                       
                                                 result += i;                       
                                                 else                       
                                                 result += (i + n / i);                       
                                                 }                       
                                                 }                       
                                   
                                                 // Add 1 and n to result                       
                                                 // as above loop considers                       
                                                 // proper divisors greater                       
                                                 // than 1.                       
                                                 return                                     (result + 1);                       
                         }                       
                                   
                         // Returns true if x and y                       
                         // are Amicable else false.                       
                         function                                     areAmicable(x, y)                       
                         {                       
                                                 if                                     (divSum(x) != y)                       
                                                 return                                     false                                     ;                       
                                   
                                                 return                                     (divSum(y) == x);                       
                         }                       
                                   
                                                 // Driver Code                       
                                                 let x = 220;                       
                                                 let y = 284;                       
                                                 if                                     (areAmicable(x, y))                       
                                                 document.write(                                     "Yes"                                     );                       
                                                 else                       
                                                 document.write(                                     "No"                                     );                       
                                   
                         // This code is contributed by _saurabh_jaiswal.                       
                                   
                         </script>                       

输出：

Yes

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