0/1Knapsack, C++ (gcc)

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// A Dynamic Programming based 
// solution for 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) 
{ 
	int i, w; 
	int K[n + 1][W + 1];

// Build table K[][] in bottom up manner 
	for (i = 0; i <= n; i++) 
	{ 
		for (w = 0; w <= W; w++) 
		{ 
			if (i == 0 || w == 0) 
				K[i][w] = 0; 
			else if (wt[i - 1] <= w) 
				K[i][w] = max(val[i - 1] 
						+ K[i - 1][w - wt[i - 1]], 
						K[i - 1][w]); 
			else
				K[i][w] = K[i - 1][w]; 
		} 
	} 
	for(int i=0; i<=n; i++)
    {
      printf("\n");
      for(int j=0; j<=W; j++)
      {
        printf("%4d\t", K[i][j]);
      }
      printf("\n");
    }
	return K[n][W]; 
}

// Driver Code 
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; 
}

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Compilation time: 0.12 sec, absolute running time: 0.16 sec, cpu time: 0.01 sec, memory peak: 6 Mb, absolute service time: 0,39 sec