how to find 49 combination for 2 array each array size 3...?

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2.33/5 (5 votes)

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Q: Array 1={1,2,3} Array 2={4,5,6}
2n-1=2(pow 3)-1=7
2n-1=2(pow 3)-1=7
7*7 =49

Output Need like that......... Any one can help me thanks
---------------------------------------------------------------------------------
C1=(1,4) C2=(1,5) C3=(1,6) C4=(1,4,5) C5=(1,4,6) C6=(1,5,6) C7=(1,4,5,6)

C8=(2,4) C9=(2,5) C10=(2,6) C11=(2,4,5) C12=(2,4,6) C13=(2,5,6) C14=(2,4,5,6)

C15=(3,4) C16=(3,5) C17=(3,6) C18=(3,4,5) C19=(3,4,6) C20=(3,5,6) C21=(3,4,5,6)

C22=(1,2,4) C23=(1,2,5) C24=(1,2,6) C25=(1,2,4,5) C26=(1,2,4,6) C27=(1,2,5,6) C28=(1,3,4)
C29=(1,3,5) C30=(1,3,6) C31=(1,3,4,5) C32=(1,3,4,6) C33=(1,3,5,6)

C34=(2,3,4) C35=(2,3,5) C36=(2,3,6) C37=(2,3,4,5) C38=(2,3,4,6) C39=(2,3,5,6) C40=(1,2,4,5,6) C41=(1,3,4,5,6) C42=(2,3,4,5,6)

C43=(1,2,3,4) C44=(1,2,3,5) C45=(1,2,3,6) C46=(1,2,3,4,5) C47=(1,2,3,4,6) C48=(1,2,3,5,6) C49=(1,2,3,4,5,6)

i try this code but it wrong....

C++

#include <iostream>
using namespace std;


int main() {
	int i,j;
		int arr1[]={1,2,3};
		int arr2[]={4,5,6};
	for (i=0; i<3; i++){
		for(j=0; j<3; j++)
			cout<<arr1[i]<<", "<<arr2[j]<<endl;
			cout<<arr1[i]<<", "<<arr2[i]<<", "<<arr2[i+1]<<"\n";
			cout<<arr1[i]<<", "<<arr2[i]<<", "<<arr2[i+2]<<"\n";
			cout<<arr1[i]<<", "<<arr2[i+1]<<", "<<arr2[i+2]<<"\n";
			cout<<arr1[i]<<", "<<arr2[i]<<", "<<arr2[i+1]<<", "<<arr2[i+2]<<"\n";

	}
	return 0;
}

Posted 2-Oct-13 2:58am

Member 10310972

Updated 2-Oct-13 3:41am

v5

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Comments

joshrduncan2012 2-Oct-13 9:06am

Why is it wrong?

Member 10310972 2-Oct-13 9:09am

we need output like above...

joshrduncan2012 2-Oct-13 9:15am

That doesn't tell me anything.

What output are you getting versus what you should be expecting?
Have you tried debugging it to see where the problem is? If so, what did you try?
Please "Improve Question" with that information.

Member 10310972 2-Oct-13 9:37am

we need output like that
C1=(1,4) C2=(1,5) C3=(1,6) C4=(1,4,5) C5=(1,4,6) C6=(1,5,6) C7=(1,4,5,6)
C8=(2,4) C9=(2,5) C10=(2,6) C11=(2,4,5) C12=(2,4,6) C13=(2,5,6) C14=(2,4,5,6)
C15=(3,4) C16=(3,5) C17=(3,6) C18=(3,4,5) C19=(3,4,6) C20=(3,5,6) C21=(3,4,5,6)
C22=(1,2,4) C23=(1,2,5) C24=(1,2,6) C25=(1,2,4,5) C26=(1,2,4,6) C27=(1,2,5,6)
C28=(1,3,4) C29=(1,3,5) C30=(1,3,6) C31=(1,3,4,5) C32=(1,3,4,6) C33=(1,3,5,6)
C34=(2,3,4) C35=(2,3,5) C36=(2,3,6) C37=(2,3,4,5) C38=(2,3,4,6) C39=(2,3,5,6) C40=(1,2,4,5,6) C41=(1,3,4,5,6) C42=(2,3,4,5,6)

C43=(1,2,3,4) C44=(1,2,3,5) C45=(1,2,3,6) C46=(1,2,3,4,5) C47=(1,2,3,4,6) C48=(1,2,3,5,6) C49=(1,2,3,4,5,6)

Captain Price 2-Oct-13 9:14am

looks interesting

2 solutions

Solution 1

C++

int main()
{
	int i,j, k = 0;
	int arr1[]={1,2,3};
	int arr2[]={4,5,6};
	for (i=0; i<3; i++)
         {
		for(j=0; j<3; j++)
		{
             	    cout"C"<<k<<" = ("<<arr1[i]<<", "<<arr2[j]<<endl" )";  // you need to format your out put better.
		    cout<<arr1[i]<<", "<<arr2[i]<<", "<<arr2[i+1]<<"\n";
		    cout<<arr1[i]<<", "<<arr2[i]<<", "<<arr2[i+2]<<"\n";
		    cout<<arr1[i]<<", "<<arr2[i+1]<<", "<<arr2[i+2]<<"\n";
		    cout<<arr1[i]<<", "<<arr2[i]<<", "<<arr2[i+1]<<", "<<arr2[i+2]<<"\n";
 
	        }
	return 0;
         }//It appears that you missed a closing brace.
}

Posted 2-Oct-13 22:50pm

The_Inventor

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**Captain Price** · Accepted Answer · 2013-10-06T20:09:00

This is an interesting question. The solution to this problem is much more complex than it sounds. It took me more than an hour to figure out an algorithm.

Here's the code :

C++

#include <iostream>
using namespace std;

//a helper function used by comb().
void print_arr(int *arr, size_t beg, size_t size)
{
	if(size < 1)return;

	for(int i=beg; i < size; i++)
	{
		cout << arr[i];
		cout << ",";
	}

}


void comb(int* a, int *b)
{
	int j=1;

	int al_pos = 0;
	int al_start = 1;

	int k=1;
	int bl_pos = 0;
	int bl_start = 1; 

	while(true)
	{

		if(j > 3)//j==4
		{
			
			al_pos++;
			j = al_pos + 1;

			if(al_pos > 1)//l_pos == 2
			{
				al_start++;
			}
			
		}


		k=1;
		bl_pos = 0;
		bl_start = 1; 

		while(true)
		{
			if(k > 3)//k==4
			{
			
				bl_pos++;
				k = bl_pos + 1;

				if(bl_pos > 1)//bl_pos == 2
				{
					bl_start++;
				}
			
			}

			if(al_start > 2)//l_start ==3
			{
				print_arr(a, 0, 2);
				cout << a[2] << ",";
				
			}
			else
			{
				print_arr(a, al_start-1, al_pos);
				cout << a[j-1] << ",";

			}

			if(bl_start > 2)
			{
				print_arr(b, 0, 2);
				cout << b[2] << endl;
				break;
			}

			print_arr(b, bl_start-1, bl_pos);
			cout << b[k-1] << endl;

			k++;
		}

		if(al_start > 2)
			break;

		j++;
		
	}
}


int main()
{
	int a[3] = {1, 2, 3};
	int b[3] = {4, 5, 6};


	comb(a, b);

        return 0;
}