/* A Naive recursive implementation of
0-1 Knapsack problem */
#include <bits/stdc++.h>
using namespace std;
// A utility function that returns
// maximum of two integers
int max(int a, int b) { return (a > b) ? a : b; }
// Returns the maximum value that
// can be put in a knapsack of capacity W
int knapSack(int W, int wt[], int val[], int n)
{
// Base Case
if (n == 0 || W == 0)
return 0;
// If weight of the nth item is more
// than Knapsack capacity W, then
// this item cannot be included
// in the optimal solution
if (wt[n - 1] > W)
return knapSack(W, wt, val, n - 1);
// Return the maximum of two cases:
// (1) nth item included
// (2) not included
else
return max(
val[n - 1]
+ knapSack(W - wt[n - 1],
wt, val, n - 1),
knapSack(W, wt, val, n - 1));
}
// Driver code
int main()
{
int val[] = { 60, 100, 120 };
int wt[] = { 10, 20, 30 };
int W = 50;
int n = sizeof(val) / sizeof(val[0]);
cout << knapSack(W, wt, val, n);
return 0;
}
// This code is contributed by rathbhupendra