Question

In a programming contest, a problem was:

Count all solutions to the equation: x + 4y + 4z = n. You will be
given n and you will determine the count of solutions. Assume x, y and z are positive integers.

I have considered using triple for loops (brute force), but it was unefficient, causing TIME LIMIT EXCEED. (since the n may be = 1000,000):

int sol = 0;
for (int i = 1; i <= n; i++)
{
 for (int j = 1; j <= n / 4; j++)
 {
  for (int k = 1; k <= n / 4; k++)
   {
      if (i + 4 * j + 4 * k == n)
         sol++;
   }
 }
}

My friend could solve the problem. When I asked him, he said that he didn’t use brute force at all. Instead, he converted the equation to a ‘series’ (i.e. summition). I asked him to tell how me but he refused 🙂

Can I know how?

Answer 1

This is particular case of coin change problem, which is solved in general by dynamic programming.

But here we can elaborate simple solution. I consider x,y,z > 0

x + 4*(y+z)=n
Let y + z = q = p + 1 (q > 1, p > 0)

x+4*q=n

x+4*p=n-4

There are M = Floor((n-5)/4) variants for x and p, hence there are M possible values of
q = 2..M+1
For every q>1 there are (q-1) variants of y and z: q = 1 + (q-1) = 2 + (q-2) +..+(q-1)+1

So we have N=1 + 2 + 3 + … + M = M * (M + 1)/2 solutions

Example:

n = 15;

M = (15 – 5) div 4 = 2

N = 3

(3,1,2),(3,2,1),(7,1,1)