Problem

Given a perimeter p, find the number of triangles that have sides a, b, and c that add up to p. Make sure to not repeat triangles, e.g 0, 0, 9 = 9, 0, 0 (same triangle but flipped).

Solution

For a faster run time, and to avoid duplicates, sort the sides from smallest to greatest (9, 5, 6 will become 5, 6, 9). We don't need to check all possibilities since a + b + c = p, and an equilateral triangle means a = b = c, so therefore, we only need to check from 0 to p/3 for one side. A 2D array is used to make search easier and faster. Variable a will represent the rows, b the columns, and c the contents inside [a][b].

Code (java)

import java.util.Scanner;

public class Main {

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);

        System.out.print("Enter perimeter: ");
        int p = scanner.nextInt();
        int[][] mat = new int[p / 3 + 1][p / 2 + 1];

        for (int i = 0; i < mat.length; i++) {
            for (int i1 = 0; i1 < mat[i].length; i1++) {
                mat[i][i1] = -1;
            }
        }

        //printMat(mat);

        scanner.close();

        int num = 0;

        for (int a = 0; a < p/3 + 1; a++) {

            for (int b = 0; b + a < p + 1; b++) {
                int c = p - (a + b);

                // sort (to hash)
                int a1 = smallest(a, b, c);
                int b1;
                int c1 = biggest(a, b, c);

                if (a1 == a) {
                    b1 = min(b, c);
                } else if (a1 == b) {
                    b1 = min(a, c);
                } else {
                    b1 = min(a, b);
                }

                if (mat[a1][b1] == -1) {
                    num++;
                    mat[a1][b1] = c1;
                    System.out.println("a: " + a1 + ", b: " + b1 + ", c: " + c1);
                }
            }

        }

        System.out.println("TOTAL: " + num);
        printMat(mat);
    }

    private static void printMat(int[][] mat) {
        System.out.println("Printing matrix...");
        for (int[] ints : mat) {
            for (int anInt : ints) {
                System.out.printf("%-5s", (anInt == -1) ? "" : anInt);
            }

            System.out.println();
        }
    }

    private static int smallest(int a1, int b1, int c1) {
        return min(a1, min(b1, c1));
    }

    private static int biggest(int a1, int b1, int c1) {
        return max(a1, max(b1, c1));
    }

    private static int min(int a1, int b1) {
        if (a1 < b1)
            return a1;
        return b1;
    }

    private static int max(int a1, int b1) {
        if (a1 > b1)
            return a1;
        return b1;
    }
}

Output

Enter perimeter: 9
a: 0, b: 0, c: 9
a: 0, b: 1, c: 8
a: 0, b: 2, c: 7
a: 0, b: 3, c: 6
a: 0, b: 4, c: 5
a: 1, b: 1, c: 7
a: 1, b: 2, c: 6
a: 1, b: 3, c: 5
a: 1, b: 4, c: 4
a: 2, b: 2, c: 5
a: 2, b: 3, c: 4
a: 3, b: 3, c: 3
TOTAL: 12
Printing matrix...
9  8  7  6  5   
   7  6  5  4   
     5  4     
        3