我们得到了 长度、宽度 和 身高 比如说S，指的是长方体。这项任务是找到可以达到的最大体积，这样边的和就是S。 长方体的体积=长度*宽度*高度 例如：
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Input : s = 4Output : 2Only possible dimensions are some combination of 1, 1, 2.Input : s = 8Output : 18All possible edge dimensions:[1, 1, 6], volume = 6[1, 2, 5], volume = 10[1, 3, 4], volume = 12[2, 2, 4], volume = 16[2, 3, 3], volume = 18
方法1：（暴力） 我们的想法是三个嵌套，一个是长度，一个是宽度，一个是高度。对于每次迭代，计算体积并与最大体积进行比较。 以下是该方法的实施情况：
C++
#include <bits/stdc++.h> using namespace std; // Return the maximum volume. int maxvolume( int s) { int maxvalue = 0; // for length for ( int i = 1; i <= s - 2; i++) { // for breadth for ( int j = 1; j <= s - 1; j++) { // for height int k = s - i - j; // calculating maximum volume. maxvalue = max(maxvalue, i * j * k); } } return maxvalue; } // Driven Program int main() { int s = 8; cout << maxvolume(s) << endl; return 0; } |
JAVA
// Java code to Maximize volume of // cuboid with given sum of sides class GFG { // Return the maximum volume. static int maxvolume( int s) { int maxvalue = 0 ; // for length for ( int i = 1 ; i <= s - 2 ; i++) { // for breadth for ( int j = 1 ; j <= s - 1 ; j++) { // for height int k = s - i - j; // calculating maximum volume. maxvalue = Math.max(maxvalue, i * j * k); } } return maxvalue; } // Driver function public static void main (String[] args) { int s = 8 ; System.out.println(maxvolume(s)); } } // This code is contributed by Anant Agarwal. |
Python3
# Python3 code to Maximize volume of # cuboid with given sum of sides # Return the maximum volume. def maxvolume (s): maxvalue = 0 # for length i = 1 for i in range (s - 1 ): j = 1 # for breadth for j in range (s): # for height k = s - i - j # calculating maximum volume. maxvalue = max (maxvalue, i * j * k) return maxvalue # Driven Program s = 8 print (maxvolume(s)) # This code is contributed by "Sharad_Bhardwaj". |
C#
// C# code to Maximize volume of // cuboid with given sum of sides using System; class GFG { // Return the maximum volume. static int maxvolume( int s) { int maxvalue = 0; // for length for ( int i = 1; i <= s - 2; i++) { // for breadth for ( int j = 1; j <= s - 1; j++) { // for height int k = s - i - j; // calculating maximum volume. maxvalue = Math.Max(maxvalue, i * j * k); } } return maxvalue; } // Driver function public static void Main () { int s = 8; Console.WriteLine(maxvolume(s)); } } // This code is contributed by vt_m. |
PHP
<?php // PHP code to Maximize volume of // cuboid with given sum of sides // Return the maximum volume. function maxvolume( $s ) { $maxvalue = 0; // for length for ( $i = 1; $i <= $s - 2; $i ++) { // for breadth for ( $j = 1; $j <= $s - 1; $j ++) { // for height $k = $s - $i - $j ; // calculating maximum volume. $maxvalue = max( $maxvalue , $i * $j * $k ); } } return $maxvalue ; } // Driver Code $s = 8; echo (maxvolume( $s )); // This code is contributed by vt_m. ?> |
Javascript
<script> // javascript code to Maximize volume of // cuboid with given sum of sides // Return the maximum volume. function maxvolume( s) { let maxvalue = 0; // for length for (let i = 1; i <= s - 2; i++) { // for breadth for (let j = 1; j <= s - 1; j++) { // for height let k = s - i - j; // calculating maximum volume. maxvalue = Math.max(maxvalue, i * j * k); } } return maxvalue; } // Driver code let s = 8; document.write(maxvolume(s)); // This code is contributed by gauravrajput1 </script> |
输出：
18
时间复杂性： O（n） 2. ) 方法2：（有效方法） 这样做的目的是尽可能平均地分割边缘。 所以 长度=楼层（s/3） 宽度=楼层（（s-长度）/2）=楼层（（s-楼层（s/3）/2） 高度=s–长度–宽度=s–楼层（s/3）-楼层（s–楼层（s/3））/2） 以下是该方法的实施情况：
C++
#include <bits/stdc++.h> using namespace std; // Return the maximum volume. int maxvolume( int s) { // finding length int length = s / 3; s -= length; // finding breadth int breadth = s / 2; // finding height int height = s - breadth; return length * breadth * height; } // Driven Program int main() { int s = 8; cout << maxvolume(s) << endl; return 0; } |
JAVA
// Java code to Maximize volume of // cuboid with given sum of sides import java.io.*; class GFG { // Return the maximum volume. static int maxvolume( int s) { // finding length int length = s / 3 ; s -= length; // finding breadth int breadth = s / 2 ; // finding height int height = s - breadth; return length * breadth * height; } // Driven Program public static void main (String[] args) { int s = 8 ; System.out.println ( maxvolume(s)); } } // This code is contributed by vt_m. |
Python3
# Python3 code to Maximize volume of # cuboid with given sum of sides # Return the maximum volume. def maxvolume( s ): # finding length length = int (s / 3 ) s - = length # finding breadth breadth = s / 2 # finding height height = s - breadth return int (length * breadth * height) # Driven Program s = 8 print ( maxvolume(s) ) # This code is contributed by "Sharad_Bhardwaj". |
C#
// C# code to Maximize volume of // cuboid with given sum of sides using System; class GFG { // Return the maximum volume. static int maxvolume( int s) { // finding length int length = s / 3; s -= length; // finding breadth int breadth = s / 2; // finding height int height = s - breadth; return length * breadth * height; } // Driven Program public static void Main () { int s = 8; Console.WriteLine( maxvolume(s)); } } // This code is contributed by vt_m. |
PHP
<?php // Return the maximum volume. function maxvolume( $s ) { // finding length $length = (int)( $s / 3); $s -= $length ; // finding breadth $breadth = (int)( $s / 2); // finding height $height = $s - $breadth ; return $length * $breadth * $height ; } // Driven Code $s = 8; echo (maxvolume( $s )); // This code is contributed by Ajit. ?> |
Javascript
<script> // Return the maximum volume. function maxvolume( s) { // finding length let length = parseInt(s / 3); s -= length; // finding breadth let breadth = parseInt(s / 2); // finding height let height = s - breadth; return length * breadth * height; } // Driven Program let s = 8; document.write(maxvolume(s)); // This code is contributed by aashish1995 </script> |
输出：
18
时间复杂性： O（1） 这是怎么回事？
我们基本上需要最大化 三个数字，x，y和z，其和是给定的。 给定s=x+y+z 最大化P=x*y*z =x*y*（s-x-y） =x*y*s–x*x*s–x*y*y 我们得到dp/dx=sy-2xy-y*y dp/dy=sx-2xy-x*x 我们得到dp/dx=0和dp/dy=0 x=s/3，y=s/3 所以z=s-x-y=s/3
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