无界小背包

给定n个项目的权重和值，任务是将这些项目放入容量为W的背包中，以在背包中获得最大的总价值，我们可以重复放置相同的项目，也可以放置项目的分数。

例子：

Input: val[] = {14, 27, 44, 19}, wt[] = {6, 7, 9, 8}, W = 50
Output: 244.444

Input: val[] = {100, 60, 120}, wt[] = {20, 10, 30}, W = 50
Output: 300

做法：这里的想法是只要找到它具有最大的价值与重量比的项目。然后仅用此物品填充整个背包，以使背包的最终价值最大化。

以下是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
  
// Function to return the maximum required value
float knapSack(int W, float wt[], float val[], int n)
{
  
    // maxratio will store the maximum value to weight
    // ratio we can have for any item and maxindex
    // will store the index of that element
    float maxratio = INT_MIN;
    int maxindex = 0;
  
    // Find the maximum ratio
    for (int i = 0; i < n; i++) {
        if ((val[i] / wt[i]) > maxratio) {
            maxratio = (val[i] / wt[i]);
            maxindex = i;
        }
    }
  
    // The item with the maximum value to
    // weight ratio will be put into
    // the knapsack repeatedly until full
    return (W * maxratio);
}
  
// Driver code
int main()
{
    float val[] = { 14, 27, 44, 19 };
    float wt[] = { 6, 7, 9, 8 };
    int n = sizeof(val) / sizeof(val[0]);
    int W = 50;
  
    cout << knapSack(W, wt, val, n);
  
    return 0;
}

Java

// Java implementation of the approach
class GFG 
{
  
    // Function to return the maximum required value
    static float knapSack(int W, float wt[], 
                            float val[], int n) 
    {
  
        // maxratio will store the maximum value to weight
        // ratio we can have for any item and maxindex
        // will store the index of that element
        float maxratio = Integer.MIN_VALUE;
        int maxindex = 0;
  
        // Find the maximum ratio
        for (int i = 0; i < n; i++) 
        {
            if ((val[i] / wt[i]) > maxratio)
            {
                maxratio = (val[i] / wt[i]);
                maxindex = i;
            }
        }
  
        // The item with the maximum value to
        // weight ratio will be put into
        // the knapsack repeatedly until full
        return (W * maxratio);
    }
  
    // Driver code
    public static void main(String[] args) 
    {
        float val[] = {14, 27, 44, 19};
        float wt[] = {6, 7, 9, 8};
        int n = val.length;
        int W = 50;
  
        System.out.println(knapSack(W, wt, val, n));
    }
}
  
// This code is contributed by 29AjayKumar

Python3

# Python implementation of the approach
  
import sys 
  
# Function to return the maximum required value
def knapSack(W, wt, val, n):
  
    # maxratio will store the maximum value to weight
    # ratio we can have for any item and maxindex
    # will store the index of that element
    maxratio = -sys.maxsize-1;
    maxindex = 0;
  
    # Find the maximum ratio
    for i in range(n):
        if ((val[i] / wt[i]) > maxratio):
            maxratio = (val[i] / wt[i]);
            maxindex = i;
  
  
    # The item with the maximum value to
    # weight ratio will be put into
    # the knapsack repeatedly until full
    return (W * maxratio);
  
  
# Driver code
  
val = [ 14, 27, 44, 19 ];
wt = [ 6, 7, 9, 8 ];
n = len(val);
W = 50;
  
print(knapSack(W, wt, val, n));
  
# This code is contributed by Rajput-Ji

C#

// C# implementation of the approach
using System;
  
class GFG 
{
  
    // Function to return the maximum required value
    static float knapSack(int W, float []wt, 
                            float []val, int n) 
    {
  
        // maxratio will store the maximum value to weight
        // ratio we can have for any item and maxindex
        // will store the index of that element
        float maxratio = int.MinValue;
        int maxindex = 0;
  
        // Find the maximum ratio
        for (int i = 0; i < n; i++) 
        {
            if ((val[i] / wt[i]) > maxratio)
            {
                maxratio = (val[i] / wt[i]);
                maxindex = i;
            }
        }
  
        // The item with the maximum value to
        // weight ratio will be put into
        // the knapsack repeatedly until full
        return (W * maxratio);
    }
  
    // Driver code
    public static void Main() 
    {
        float []val = {14, 27, 44, 19};
        float []wt = {6, 7, 9, 8};
        int n = val.Length;
        int W = 50;
  
        Console.WriteLine(knapSack(W, wt, val, n));
    }
}
  
// This code is contributed by AnkitRai01

输出：

244.444