OJ Board

Basically I reduce the 4's with 6's following this IDEA:

Let's say we can group the 4's into N groups ( N is even )
where in each group we have three 4's.

Now each group I substitute with 2 elements of value 6.

So for each i = 1 to N, we do n[6] +=2;


Then I reduce the 5's with 6's following this IDEA:

Let's say we can group the 5's into N groups ( N is even )
where in each group we have six 5's.

Now each group I substitute with 5 elements of value 6.

So for each i = 1 to N, we do n[6] +=5

Apparently this is not logically OK. ?!
Or ?

I found the error in my reasoning.

I think I know how to do it now. I will try it.

Yes, right, I have it
Accepted P.E.
now.

I found the problem with my reasoning.

Actually the code I posted here and the reasoning behind it
was completely wrong.

Thanks, little joey!

Yes, it's realy quite simple once you see the light.
This problem took me months of struggling, an account of which can be found elsewhere on the board. But it's a good exercise in logic reasoning and it teaches you not to go after premature conclusions too fast.

Tricks to solve it fast are very cool. but Optimized DP is good enough to get accepted, not very fast around .3 second.

Running time : O(sum)

a[] is the input
t[] is the boolean function.

for (i = 1; i <= 6; i++)
if (a)
for (j = sum; j >= 0; j--)
if (t[j])
for (k = 1; k <= a; k++)
if (j + i * k > sum || t[j + i*k]) break;
else t[j + i * k] = 1;