Sword finger office secant

subject ： I'll give you a length of n Rope , Please cut the rope into m paragraph （m、n Are integers. ,n>1 also m>1）. The length of each piece of rope is marked as k[0]、k[1]、……、k[m].k[0] * k[1]*…*k[m] What's the maximum possible product ？ For example, when the length of the rope is 8 when , We cut it into lengths of 2、3、3 Three paragraphs of , At this point we get the maximum product 18.

/************************* Dynamic programming ******************************/
int maxProductAfterCutting_solution1(int length) 
{
	if (length < 2)
		return 0;
	if (length == 2)
		return 1;
	if (length == 3)
		return 2;

	int* products = (int*)malloc((length + 1)*sizeof(int));
	assert(products != NULL);
	memset(products, 0, (length + 1)*sizeof(int));
	
	/**** The length is 0、1、2、3 Length of rope ********/
	products[0] = 0;
	products[1] = 1;
	products[2] = 2;
	products[3] = 3;

	int max = 0;
	int i = 0;
	int j = 0;
	
	for (i = 4; i <= length; i++)
	{
		max = 0;
		for (j = 1; j <= i / 2; j++)
		{
			int product = products[j] * products[i - j];
			if (max < product)
				max = product;

			products[i] = max;
		}
	}
	max = products[length];

	free(products);
	products = NULL;

	return max;
}

products[0] = 0;
products[1] = 1;
products[2] = 2;
products[3] = 3;
 Why? 	products[2] = 2;products[3] = 3; Well ？ because n=2,3 No more points .