Subset sum, C++ (gcc)

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// A Dynamic Programming solution
// for subset sum problem
#include <stdio.h>

// Returns true if there is a subset of set[]
// with sun equal to given sum
bool isSubsetSum(int set[], int n, int sum)
{
	// The value of subset[i][j] will be true if
	// there is a subset of set[0..j-1] with sum
	// equal to i
	bool subset[n + 1][sum + 1];

// If sum is 0, then answer is true
	for (int i = 0; i <= n; i++)
		subset[i][0] = true;

// If sum is not 0 and set is empty,
	// then answer is false
	for (int i = 1; i <= sum; i++)
		subset[0][i] = false;

// Fill the subset table in botton up manner
	for (int i = 1; i <= n; i++) {
		for (int j = 1; j <= sum; j++) {
			if (j < set[i - 1])
				subset[i][j] = subset[i - 1][j];
			if (j >= set[i - 1])
				subset[i][j] = subset[i - 1][j]
							|| subset[i - 1][j - set[i - 1]];
		}
	}

// uncomment this code to print table
	for (int i = 1; i <= n; i++)
	{
	for (int j = 1; j <= sum; j++)
		printf ("%4d", subset[i][j]);
	printf("\n");
	}
	return subset[n][sum];
}

// Driver code
int main()
{
	int set[] = { 2, 3, 7, 8, 10 };
	int sum = 11;
	int n = sizeof(set) / sizeof(set[0]);
	if (isSubsetSum(set, n, sum) == true)
		printf("Found a subset with given sum");
	else
		printf("No subset with given sum");
	return 0;
}
// This code is contributed by Arjun Tyagi.

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Compilation time: 0.12 sec, absolute running time: 0.14 sec, cpu time: 0.01 sec, memory peak: 5 Mb, absolute service time: 0,34 sec