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p028.java
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35 lines (28 loc) · 1.13 KB
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/*
* Solution to Project Euler problem 28
* Copyright (c) Project Nayuki. All rights reserved.
*
* https://www.nayuki.io/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);
}
}