最大化给定 0、1 和 2 中的组数，总和可被 3 整除

给定三个整数， C0、C1 和 C2在组S 中的频率为 0、1 和 2。任务是找到总和可被 3 整除的组的最大数量，条件是sum(S)可被 3 整除，并且所有组的并集必须等于S

例子：

Input: C0 = 2, C1 = 4, C2 = 1
Output: 4
Explanation: it can divide the group S = {0, 0, 1, 1, 1, 1, 2} into four groups {0}, {0}, {1, 1, 1}, {1, 2}. It can be proven that 4 is the maximum possible answer.

Input: C0 = 250, C1 = 0, C2 = 0
Output: 250

方法：这个问题可以使用贪心算法来解决。请按照以下步骤解决问题。

• 初始化一个变量maxAns ，比如 0，以存储组的最大数量。
• 将C0添加到 maxAns ，因为每个{0}都可以是一个组，使得总和可以被 3 整除。
• 初始化一个变量k ，比如min(C1, C2) ，并将其添加到 maxAns 中，因为至少可以创建k , {1, 2}组。
• 将abs(C1-C2) /3 添加到 maxAns ，它将贡献剩余的1s 或 2s 。
• 返回maxAns。

下面是上述方法的实现。

C++

// C++ program for above appraoch
#include 
using namespace std;
 
int maxGroup(int c0, int c1, int c2)
{
 
    // Initializing to store maximum number of groups
    int maxAns = 0;
 
    // Adding C0
    maxAns += c0;
 
    // Taking Minimum of c1, c2 as minimum number of
    // pairs must be minimum of c1, c2
    int k = min(c1, c2);
    maxAns += k;
 
    // If there is any remaining element in c1 or c2
    // then it must be the absolute difference of c1 and
    // c2 and dividing it by 3 to make it one pair
    maxAns += abs(c1 - c2) / 3;
 
    return maxAns;
}
 
int main()
{
    int C0 = 2, C1 = 4, C2 = 1;
    cout << maxGroup(C0, C1, C2);
    return 0;
}
 
// This code is contributed by maddler

Java

// Java program for above appraoch
 
import java.io.*;
 
class GFG {
 
    // Function to calculate maximum number of groups
    public static int maxGroup(int c0, int c1, int c2)
    {
 
        // Initializing to store maximum number of groups
        int maxAns = 0;
 
        // Adding C0
        maxAns += c0;
 
        // Taking Minimum of c1, c2 as minimum number of
        // pairs must be minimum of c1, c2
        int k = Math.min(c1, c2);
        maxAns += k;
 
        // If there is any remaining element in c1 or c2
        // then it must be the absolute difference of c1 and
        // c2 and dividing it by 3 to make it one pair
        maxAns += Math.abs(c1 - c2) / 3;
 
        return maxAns;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        // Given Input
        int C0 = 2, C1 = 4, C2 = 1;
 
        // Function Call
        System.out.println(maxGroup(C0, C1, C2));
    }
}

Python3

# python 3 program for above appraoch
def maxGroup(c0, c1, c2):
   
    # Initializing to store maximum number of groups
    maxAns = 0
 
    # Adding C0
    maxAns += c0
 
    # Taking Minimum of c1, c2 as minimum number of
    # pairs must be minimum of c1, c2
    k = min(c1, c2)
    maxAns += k
 
    # If there is any remaining element in c1 or c2
    # then it must be the absolute difference of c1 and
    # c2 and dividing it by 3 to make it one pair
    maxAns += abs(c1 - c2) // 3
 
    return maxAns
 
  # Driver code
if __name__ == '__main__':
    C0 = 2
    C1 = 4
    C2 = 1
    print(maxGroup(C0, C1, C2))
 
    # This code is contributed by ipg2016107.

C#

// C# program for above appraoch
using System;
 
class GFG{
     
// Function to calculate maximum number of groups   
public static int maxGroup(int c0, int c1, int c2)
{
     
    // Initializing to store maximum number
    // of groups
    int maxAns = 0;
 
    // Adding C0
    maxAns += c0;
 
    // Taking Minimum of c1, c2 as minimum number
    // of pairs must be minimum of c1, c2
    int k = Math.Min(c1, c2);
    maxAns += k;
 
    // If there is any remaining element
    // in c1 or c2 then it must be the
    // absolute difference of c1 and c2
    // and dividing it by 3 to make it one pair
    maxAns += Math.Abs(c1 - c2) / 3;
 
    return maxAns;
}
 
// Driver Code
static public void Main()
{
   
    // Given Input
    int C0 = 2, C1 = 4, C2 = 1;
   
    // Function Call
    Console.WriteLine(maxGroup(C0, C1, C2));
}
}
 
// This code is contributed by maddler

Javascript

 
       // JavaScript program for the above approach;
       function maxGroup(c0, c1, c2)
       {
 
           // Initializing to store maximum number of groups
           let maxAns = 0;
 
           // Adding C0
           maxAns += c0;
 
           // Taking Minimum of c1, c2 as minimum number of
           // pairs must be minimum of c1, c2
           let k = Math.min(c1, c2);
           maxAns += k;
 
           // If there is any remaining element in c1 or c2
           // then it must be the absolute difference of c1 and
           // c2 and dividing it by 3 to make it one pair
           maxAns += Math.abs(c1 - c2) / 3;
 
           return maxAns;
       }
 
       let C0 = 2, C1 = 4, C2 = 1;
       document.write(maxGroup(C0, C1, C2));
        
  // This code is contributed by Potta Lokesh
   

输出：

4

时间复杂度： O(1)
辅助空间： O(1)

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