Simple problem

[andreistefan]

Hi, I was thinking for a few days at a C++ problem and I couldn't solve it, I tried but I'm a begginer.
So n and k are two natural numbers with max. five numbers in them. (Example : 11111).
It will need to output the sum of N in K even numbers.
Examples:
For n = 18 and k = 3 the output will be 2+6+10=18
For n = 10 and k = 4 it will say "it's impossible)
English isn't my native language so please be kind.
Thank you!

[LB]

What part did you get stuck on? This is basically just one for loop adding to a variable. For i from n to k, if i is even add i to sum.

[keskiverto]

If n is odd, then it is impossible. That you can test easily.

Why is n=10, k=4 impossible? No repeats and no 0 allowed?

18/2 = 9
1+2+3 = 6
9-6 = 3

2*1 = 2
2*2 = 4
2*(3+3) = 12
2+4+12 = 18

[Aceix]

I don't get it. Maybe you could write in your native language, and use google translate?

Aceix.

[LB]

I misread.

"It will need to output the sum of N in K even numbers."

n is the number, which has to be even for this to work, and k is how many even numbers you have to use as its sum.

That's why n=10, k=4 is impossible - There is no way with four natural even numbers you can add to 10 (2+4+6+8 is already too high)

I'll do some thinking about how to solve it.

[Aceix]

Ok got it! So do you need tips on how to write a code for that?

Aceix.

[andreistefan]

Done it!Thanks guys.

#include <iostream.h>
int main()
{
	int n,k,suma,i;
	cin >> n;
	cin >> k;
	if(n<(k*(k+1)) || n%2!=0)
	{
		cout << "No";
	}
	
	suma=0;
	for(i=1;i<=k;i++)
	{
		if(i<k)
		{
			cout << i*2 << "+";
			suma+=i*2;
		}
		else
		{
			cout << n-suma;
		}
	}
}

[LB]

Ah, so there are multiple solutions to most problems - that simplifies the program a lot.

Fixed compile errors:
http://ideone.com/g6SKvb

You should note that <iostream.h> is deprecated, you are supposed to use <iostream> without the .h

[andreistefan]

Thank you L B, keskiverto and Aceix!