Knapsack problem greedy algorithm

[problem description] n items and a backpack are given. The weight of item i is Wi, its value is Vi, and the capacity of backpack is c. How to choose the items in the backpack to maximize the total value of the items in the backpack

[problem analysis]

·General knapsack problem!

Optimal idea: considering the comprehensive effect of value growth and capacity consumption, i.e. each time I choose the item with the largest value to weight ratio vi/wi, i.e. first load it into the backpack, which is the final greedy strategy.

Manually input knapsack capacity c and item number n, the value of weight array w[i] of random biological product I and the value array v[i] of item I.

Program source code:

package SF;
import java.util.Scanner;
public class Greedy knapsack problem {
	public static void main(String[] args) {
		Scanner reader = new Scanner(System.in);
		System.out.println("Please enter the number of items n: ");
		int n = reader.nextInt();
		System.out.println("Please input the backpack capacity c: ");
		float c = reader.nextFloat();
		float[] w = new float[n+1];		//Weight of articles
		float[] v = new float[n+1];		//Value of goods
		for(int i=0;i<n;i++){//Random number less than 20 for weight of goods
			w[i] = (float) Math.floor(Math.random()*20);
		}
		System.out.println("The weight of the item is:");
		for(int j=0;j<n;j++){
		System.out.print(w[j]+" ");
		}
		System.out.println("");
		for(int i=0;i<n;i++){//Value of goods is within random number (20)
			v[i] = (float) Math.floor(Math.random()*20);
		}
		System.out.println("The value of the item is:");
		for(int j=0;j<n;j++){
		System.out.print(v[j]+" ");
		}
		System.out.println("");
		float[] x = new float[n+1];		//Solution to the problem
		//Greedy knapsack problem p = new greedy knapsack problem ();
		p.Sort(n, v, w);
		float s = p.Knapsack(w, v, x, c, n);
		System.out.println("The final value of the optimal solution is:"+s);
	}
	float Knapsack(float w[],float v[],float x[],float c,int n){
		float total = 0;
		int i;
		Sort(n,v,w);
		for(i=0;i<n;i++)
			x[i] = 0;
		i = 1;
		while(w[i]<c){
			x[i] = 1;
			total = total + v[i];
			c = c - w[i];
			i++;
		}
		x[i] = c/w[i];
		total = total + x[i]*v[i];
		System.out.println("The solution of the problem is as follows:");
		for(int j=0;j<n;j++){
		System.out.print(x[j]+" ");
		}
		System.out.println("");
		return total;
	}
	private void Sort(int n,float v[],float w[]) {//Sort from large to small
		int i;
		float j;
		float k;
		for(i =0;i<n;i++){
			if( v[i+1]/w[i+1] > v[i]/w[i] ){
				j = v[i];
				k = w[i];
				v[i] = v[i+1];
				w[i] = w[i+1];
				v[i+1] = j;
				w[i+1] = k;
			}
		}
	}
}

Running screenshot: