Integer Partitions from 1 to N source code

This snippet submitted by Christopher Brown on 2012-12-28. It has been viewed 90724 times.
Rating of 7.4 with 497 votes

 #include <stdlib.h>
#include <stdio.h>

int NUM_OBJECTS = 20;

typedef struct{
	int* val;
	void* next;
} n_obj;

#define Q_INT malloc(sizeof(int))
#define Q_OBJ malloc(sizeof(n_obj))
#define Q_SEQ(x) calloc((x), sizeof(int))

int main()
{
	int clear_i;
	int i, left_i, cur_sz, cur_num;
	n_obj **cl;
	n_obj *temp;
	n_obj *cur_list;
	n_obj *cur_seq;
	
	cl = malloc(sizeof(n_obj*)*NUM_OBJECTS);
	
	for(i=0;i<NUM_OBJECTS;i++)
	{
		cl[i] = Q_OBJ;
		cl[i]->val = NULL;
		
		temp = Q_OBJ;
		temp->val = Q_SEQ(1);
		
		temp->val[0] = i + 1;
		
		temp->next = cl[i]; 
		cl[i] = temp;
		
		cur_list = cl[i];
		for(left_i=0; left_i<i; left_i++)
		{
			for(cur_seq=cl[left_i];cur_seq->val!=NULL;cur_seq=cur_seq->next)
			{				
				if(cur_seq->val[0]<=(i - left_i))
				{
					temp = Q_OBJ;
					temp->val = Q_SEQ(left_i + 3);
					
					temp->val[0] =  i - left_i;
					
					temp->next = cur_list->next;
					cur_list->next = temp; 
					cur_list = temp;
					
					for(cur_num=0;cur_num<=left_i;cur_num++)
						cur_list->val[cur_num+1] = cur_seq->val[cur_num];
				}
			}
		}
	}
	
	for(i = 0; i < NUM_OBJECTS; i++)
	{
		cur_sz = 0;
		printf(\"---------------------------\n\");
		for(cur_list = cl[i]; cur_list->val != NULL; cur_list = cur_list->next)
		{
			for(clear_i = 0; cur_list->val[clear_i] != 0; clear_i++)
			{
				printf(\"%d \", cur_list->val[clear_i]);
			}
			printf(\"\n\");
			cur_sz++;
		}
		printf(\"Number of partitions = %d\n\", cur_sz);
	
	}
	return 0;
}

 




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