UVa11565 - Simple Equations

In this very first task, 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
• x² + y² + z² = C

You are asked to write a program that solves for x, y and z for given values of A, B and C.

For the Extreme version of this problem, please click here.

Input

The first line of the input file gives the number of test cases 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

---

#include <cstdio>

using namespace std;

int a, b, c;

bool input();
void solve();


int main()
{
#ifndef ONLINE_JUDGE
	freopen("d:\\OJ\\uva_in.txt", "r", stdin);
#endif

	int n;
	scanf("%d", &n);
	
	while (n--) {
		input();
		solve();
	}

	return 0;
}

bool input()
{
	scanf("%d%d%d", &a, &b, &c);
	return true;
}


void solve()
{
	bool found = false;
	
	int x, y, z;
	
	for (x = -22; x <= 22 && !found; x++) {
		if (x * x <= c) {
			for (y = -100; y <= 100 && !found; y++) {
				if (y != x && x * x + y * y <= c) {
					for (z = -100; z <= 100 && !found; z++)  {
						if (z != x && z != y && x + y + z == a
						    && x * y * z == b
						&& x * x + y * y + z * z == c) {
								printf("%d %d %d\n", x, y, z);
								found = true;
							}
					}
				}
			}
		}
	}
	
	if (!found) {
		printf("No solution.\n");
	}
}