最大化x + y + z的值，使得ax + by + cz = n

给定整数n ， a ， b和c ，任务是找到x + y + z的最大值，使得ax + by + cz = n 。

例子：

Input: 
n = 10
a = 5
b = 3
c = 4 
Output: 
3 
Explanation:
x = 0, y = 2 and z = 1
Input: 
n = 50
a = 8
b = 10
c = 2 
Output: 
25 
Explanation:
x = 0, y = 0 and z = 25 

方法：固定x和y的值，则z的值可以计算为z =(n –(ax + by))/ c 。如果z的当前值为整数，则更新到目前为止找到的x + y + z的最大值。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
// Function to return the maximum value of (x + y + z)
// such that (ax + by + cz = n)
int maxResult(int n, int a, int b, int c)
{
    int maxVal = 0;
 
    // i represents possible values of a * x
    for (int i = 0; i <= n; i += a)
    {
        // j represents possible values of b * y
        for (int j = 0; j <= n - i; j += b)
        {
            float z = (float)(n - (i + j)) / (float)(c);
 
            // If z is an integer
            if (floor(z) == ceil(z))
            {
                int x = i / a;
                int y = j / b;
                maxVal = max(maxVal, x + y + (int)z);
            }
        }
    }
 
    return maxVal;
}
 
// Driver code
int main()
{
    int n = 10, a = 5, b = 3, c = 4;
   
      // Function Call
    cout << maxResult(n, a, b, c);
 
    return 0;
}

Java

// Java implementation of the approach
import java.util.*;
 
class GFG {
 
    // Function to return the maximum value of (x + y + z)
    // such that (ax + by + cz = n)
    static int maxResult(int n, int a, int b, int c)
    {
        int maxVal = 0;
 
        // i represents possible values of a * x
        for (int i = 0; i <= n; i += a)
 
            // j represents possible values of b * y
            for (int j = 0; j <= n - i; j += b) {
                float z = (float)(n - (i + j)) / (float)c;
 
                // If z is an integer
                if (Math.floor(z) == Math.ceil(z)) {
                    int x = i / a;
                    int y = j / b;
                    maxVal
                        = Math.max(maxVal, x + y + (int)z);
                }
            }
 
        return maxVal;
    }
 
    // Driver code
    public static void main(String args[])
    {
        int n = 10, a = 5, b = 3, c = 4;
       
          // Function Call
        System.out.println(maxResult(n, a, b, c));
    }
}
 
// This code is contributed by
// Surendra_Gangwar

Python 3

# Python3 implementation of the approach
from math import *
 
# Function to return the maximum value
# of (x + y + z) such that (ax + by + cz = n)
 
 
def maxResult(n, a, b, c):
    maxVal = 0
 
    # i represents possible values of a * x
    for i in range(0, n + 1, a):
 
        # j represents possible values of b * y
        for j in range(0, n - i + 1, b):
            z = (n - (i + j)) / c
 
            # If z is an integer
            if (floor(z) == ceil(z)):
                x = i // a
                y = j // b
                maxVal = max(maxVal, x + y + int(z))
 
    return maxVal
 
 
# Driver code
if __name__ == "__main__":
 
    n = 10
    a = 5
    b = 3
    c = 4
 
    # Function Call
    print(maxResult(n, a, b, c))
 
# This code is contributed by Ryuga

C#

// C# implementation of the approach
using System;
 
class GFG {
 
    // Function to return the maximum value of (x + y + z)
    // such that (ax + by + cz = n)
    static int maxResult(int n, int a, int b, int c)
    {
        int maxVal = 0;
 
        // i represents possible values of a * x
        for (int i = 0; i <= n; i += a)
 
            // j represents possible values of b * y
            for (int j = 0; j <= n - i; j += b) {
                float z = (float)(n - (i + j)) / (float)c;
 
                // If z is an integer
                if (Math.Floor(z) == Math.Ceiling(z)) {
                    int x = i / a;
                    int y = j / b;
                    maxVal
                        = Math.Max(maxVal, x + y + (int)z);
                }
            }
        return maxVal;
    }
 
    // Driver code
    public static void Main(String[] args)
    {
        int n = 10, a = 5, b = 3, c = 4;
       
          // Function Call
        Console.WriteLine(maxResult(n, a, b, c));
    }
}
 
// This code has been contributed by 29AjayKumar

输出

3

时间复杂度： O(N ² )