SolveEquation.java

/**
 * Let us look at a boring mathematics problem. :-) We have three different
 * integers, x, y and z, which satisfy the following three relations: • x + y +
 * z = A • xyz = B • x 2 + y 2 + z 2 = C You are asked to write a program that
 * solves for x, y and z for given values of A, B and C. Input The first line of
 * the input file gives the number of test cases N (N < 20). Each of the
 * following N lines gives the values of A, B and C (1 ≤ A, B, C ≤ 10000).
 * Output For each test case, output the corresponding values of x, y and z. If
 * there are many possible answers, choose the one with the least value of x. If
 * there is a tie, output the one with the least value of y. If there is no
 * solution, output the line ‘No solution.’ instead. Sample Input 2 1 2 3 6 6 14
 * Sample Output No solution. 1 2 3
 */

// https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=2612

import java.util.Scanner;

public class SolveEquation {

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

		int numberOfTestCases = input.nextInt();
		while (numberOfTestCases != 0) {

			int A = input.nextInt();
			int B = input.nextInt();
			int C = input.nextInt();
			boolean hasSolution = false;

			for (int x = -22; x <= 22 && !hasSolution; x++) {
				if (x * x <= C) {
					for (int y = -100; y <= 100 && !hasSolution; y++) {
						if (x != y && y * y <= C && (x * x + y * y <= C)) {
							int z = A - x - y;
							if ((z != y && z != x && x * x + y * y + z * z == C)
									&& x * y * z == B) {
								hasSolution = true;
								System.out.println(x + " " + y + " " + z);
							}
						}
					}
				}
			}
			if (!hasSolution) {
				System.out.println("No solution.");
			}

			numberOfTestCases--;
		}

	}
}