Knapsack problem

/* A Naive recursive implementation  
of 0-1 Knapsack problem */
#include <stdio.h> 
  
// 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 program to test above function 
int main() 
{ 
    int val[] = { 60, 100, 120 }; 
    int wt[] = { 10, 20, 30 }; 
    int W = 50; 
    int n = sizeof(val) / sizeof(val[0]); 
    printf("%d", knapSack(W, wt, val, n)); 
    return 0; 
}