Project Euler > Problem 78 > Coin partitions (Java Solution)

Problem:

Let p(n) represent the number of different ways in which n coins can be separated into piles. For example, five coins can separated into piles in exactly seven different ways, so p(5)=7.

OOOOO
OOOO O
OOO OO
OOO O O
OO OO O
OO O O O
O O O O O

Find the least value of n for which p(n) is divisible by one million.


Solution:

669171001

Code:
The solution may include methods that will be found here: Library.java .

public interface EulerSolution{

public String run();

}
/* 
* Solution to Project Euler problem 28
* By Nayuki Minase
*
* http://nayuki.eigenstate.org/page/project-euler-solutions
* https://github.com/nayuki/Project-Euler-solutions
*/


public final class p028 implements EulerSolution {

public static void main(String[] args) {
System.out.println(new p028().run());
}


/*
* From the diagram, let's observe the four corners of an n * n square (where n is odd).
* It's not hard to convince yourself that the top right corner always has the value n^2.
* Working counterclockwise (backwards), the top left corner has the value n^2 - (n - 1),
* the bottom left corner has the value n^2 - 2(n - 1), and the bottom right is n^2 - 3(n - 1).
* Putting it all together, this outermost ring contributes 4n^2 - 6(n - 1) to the final sum.
*
* Incidentally, the closed form of this sum is (4m^3 + 3m^2 + 8m - 9) / 6, where m = size.
*/
private static final int SIZE = 1001; // Must be odd

public String run() {
long sum = 1; // Special case for size 1
for (int n = 3; n <= SIZE; n += 2)
sum += 4 * n * n - 6 * (n - 1);
return Long.toString(sum);
}

}


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