可以使用数组中的所有数字将一个数整除为3-yiteyi-C++库

给定一个整数数组，任务是找出是否有可能使用这些数字的所有数字构造一个整数，使其可以被3整除。如果可能，则打印“是”，如果不可能，则打印“否”。例如：

null

Input : arr[] = {40, 50, 90}Output : YesWe can construct a number which isdivisible by 3, for example 945000. So the answer is Yes. Input : arr[] = {1, 4}Output : NoThe only possible numbers are 14 and 41,but both of them are not divisible by 3, so the answer is No.

建议：请在“上解决” 实践 “首先，在讨论解决方案之前。

这个想法基于这样一个事实：一个数可以被3整除，如果它的数字之和可以被3整除 .所以我们只需要求数组元素的和。如果总和可以被3整除，我们的答案是肯定的，否则是否定的。

CPP

                         // C++ program to find if it is possible                       
                         // to make a number divisible by 3 using                       
                         // all digits of given array                       
                         #include <bits/stdc++.h>                       
                         using                                     namespace                                     std;                       
           
                         bool                                     isPossibleToMakeDivisible(                                     int                                     arr[],                                     int                                     n)                       
                         {                       
                                                 // Find remainder of sum when divided by 3                       
                                                 int                                     remainder = 0;                       
                                                 for                                     (                                     int                                     i=0; i<n; i++)                       
                                                 remainder = (remainder + arr[i]) % 3;                       
           
                                                 // Return true if remainder is 0.                       
                                                 return                                     (remainder == 0);                       
                         }                       
           
                         // Driver code                       
                         int                                     main()                       
                         {                       
                                                 int                                     arr[] = { 40, 50, 90 };                       
                                                 int                                     n =                                     sizeof                                     (arr) /                                     sizeof                                     (arr[0]);                       
                                                 if                                     (isPossibleToMakeDivisible(arr, n))                       
                                                 printf                                     (                                     "Yes"                                     );                       
                                                 else                       
                                                 printf                                     (                                     "No"                                     );                       
           
                                                 return                                     0;                       
                         }                       

JAVA

                         // Java program to find if it is possible                       
                         // to make a number divisible by 3 using                       
                         // all digits of given array                       
           
                         import                                     java.io.*;                       
                         import                                     java.util.*;                       
           
                         class                                     GFG                       
                         {                       
                                                 public                                     static                                     boolean                                     isPossibleToMakeDivisible(                                     int                                     arr[],                                     int                                     n)                       
                                                 {                       
                                                 // Find remainder of sum when divided by 3                       
                                                 int                                     remainder =                                     0                                     ;                       
                                                 for                                     (                                     int                                     i=                                     0                                     ; i<n; i++)                       
                                                 remainder = (remainder + arr[i]) %                                     3                                     ;                       
           
                                                 // Return true if remainder is 0.                       
                                                 return                                     (remainder ==                                     0                                     );                       
                                                 }                       
           
                                                 public                                     static                                     void                                     main (String[] args)                       
                                                 {                       
                                                 int                                     arr[] = {                                     40                                     ,                                     50                                     ,                                     90                                     };                       
                                                 int                                     n =                                     3                                     ;                       
                                                 if                                     (isPossibleToMakeDivisible(arr, n))                       
                                                 System.out.print(                                     "Yes"                                     );                       
                                                 else                       
                                                 System.out.print(                                     "No"                                     );                       
                                                 }                       
                         }                       
           
                         // Code Contributed by Mohit Gupta_OMG <(0_o)>                       

Python3

                         # Python program to find if it is possible                       
                         # to make a number divisible by 3 using                       
                         # all digits of given array                       
           
                         def                                     isPossibleToMakeDivisible(arr, n):                       
                                                 # Find remainder of sum when divided by 3                       
                                                 remainder                                     =                                     0                       
                                                 for                                     i                                     in                                     range                                     (                                     0                                     , n):                       
                                                 remainder                                     =                                     (remainder                                     +                                     arr[i])                                     %                                     3                       
           
                                                 # Return true if remainder is 0.                       
                                                 return                                     (remainder                                     =                                     =                                     0                                     )                       
           
                         # main()                       
           
                         arr                                     =                                     [                                     40                                     ,                                     50                                     ,                                     90                                     ];                       
                         n                                     =                                     3                       
                         if                                     (isPossibleToMakeDivisible(arr, n)):                       
                                                 print                                     (                                     "Yes"                                     )                       
                         else                                     :                       
                                                 print                                     (                                     "No"                                     )                       
           
                         # Code Contributed by Mohit Gupta_OMG <(0_o)>                       

C#

                         // C# program to find if it is possible                       
                         // to make a number divisible by 3 using                       
                         // all digits of given array                       
                         using                                     System;                       
           
                         class                                     GFG                       
                         {                       
                                                 public                                     static                                     bool                                     isPossibleToMakeDivisible(                                     int                                     []arr,                                     int                                     n)                       
                                                 {                       
                                                 // Find remainder of sum when divided by 3                       
                                                 int                                     remainder = 0;                       
                                   
                                                 for                                     (                                     int                                     i = 0; i < n; i++)                       
                                                 remainder = (remainder + arr[i]) % 3;                       
           
                                                 // Return true if remainder is 0.                       
                                                 return                                     (remainder == 0);                       
                                                 }                       
           
                                                 public                                     static                                     void                                     Main ()                       
                                                 {                       
                                                 int                                     []arr = { 40, 50, 90 };                       
                                                 int                                     n = 3;                       
                                   
                                                 if                                     (isPossibleToMakeDivisible(arr, n))                       
                                                 Console.WriteLine(                                     "Yes"                                     );                       
                                                 else                       
                                                 Console.WriteLine(                                     "No"                                     );                       
                                                 }                       
                         }                       
           
                         // This code is contributed by vt_m                       

PHP

                         <?php                       
                         // PHP program to find if it is possible                       
                         // to make a number divisible by 3 using                       
                         // all digits of given array                       
           
                         function                                     isPossibleToMakeDivisible(                                     $arr                                     ,                                     $n                                     )                       
                         {                       
                                   
                                                 // Find remainder of sum                       
                                                 // when divided by 3                       
                                                 $remainder                                     = 0;                       
                                                 for                                     (                                     $i                                     = 0;                                     $i                                     <                                     $n                                     ;                                     $i                                     ++)                       
                                                 $remainder                                     = (                                     $remainder                                     +                                     $arr                                     [                                     $i                                     ]) % 3;                       
           
                                                 // Return true if remainder is 0.                       
                                                 return                                     (                                     $remainder                                     == 0);                       
                         }                       
           
                         // Driver code                       
                         $arr                                     =                                     array                                     ( 40, 50, 90 );                       
                         $n                                     = sizeof(                                     $arr                                     );                       
                         if                                     (isPossibleToMakeDivisible(                                     $arr                                     ,                                     $n                                     ))                       
                                                 echo                                     (                                     "Yes"                                     );                       
                         else                       
                                                 echo                                     (                                     "No"                                     );                       
           
                         // This code is contributed by Ajit.                       
                         ?>                       

Javascript

                         <script>                       
           
                         // javascript program to find if it is possible                       
                         // to make a number divisible by 3 using                       
                         // all digits of given array                       
           
                         function                                     isPossibleToMakeDivisible(arr , n)                       
                         {                       
                                                 // Find remainder of sum when divided by 3                       
                                                 var                                     remainder = 0;                       
                                                 for                                     (i=0; i<n; i++)                       
                                                 remainder = (remainder + arr[i]) % 3;                       
           
                                                 // Return true if remainder is 0.                       
                                                 return                                     (remainder == 0);                       
                         }                       
           
                         var                                     arr = [ 40, 50, 90 ];                       
                         var                                     n = 3;                       
                         if                                     (isPossibleToMakeDivisible(arr, n))                       
                                                 document.write(                                     "Yes"                                     );                       
                         else                       
                                                 document.write(                                     "No"                                     );                       
           
                         // This code contributed by Princi Singh                       
           
                         </script>                       

输出：

Yes

时间复杂度：O（n）

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