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using System;
using System.Linq; // used only for the input
namespace Equasions {
class MainClass {
/* Algorithm that solves a system of equation of the type
* | a1 * x + b1 * y = c1
* | a2 * x + b2 * y = c2
* via substitution.
*
* At the bottom you can find test cases.
*/
public static void Main(string[] args) {
double[] input = Console.ReadLine().Split().Select(double.Parse).ToArray(); // a1 b1 c1 a2 b2 c2
double[] parm = { input[0], input[3], input[1], input[4] }; // a1 a2 b1 b2
double[] eres = { input[2], input[5] }; // c1 c2
double m = 0, n = 0;
string result = null;
for (int i = 0; i < parm.Length; i++) {
for (int j = 0; j < parm.Length; j++) {
if (parm[i] != 0 || parm[j] != 0) {
continue;
}
result = (i < 2) ? "n" : "m";
if (i == j) {
m = eres[i % 2] / parm[(i + 2) % 4];
n = (eres[(i + 1) % 2] - parm[3 - i] * m) / parm[(5 - i) % 4];
}
else if (i + j == 3) {
m = eres[Math.Min(i, j)] / parm[Math.Min(i, j) + 2];
n = eres[Math.Max(i, j) - 2] / parm[Math.Max(i, j) - 2];
}
else {
result = "No solution!";
}
if ((!double.IsNaN(m) && !double.IsInfinity(m)) ||
(!double.IsNaN(n) && !double.IsInfinity(n)) ||
result.Length > 1) // if m or n have normal values or if result is "No solution!"
{
j = i = parm.Length; // "breaking" both cycles
}
}
}
if (result == null) {
m = (parm[0] * eres[1] - parm[1] * eres[0]) / (parm[0] * parm[3] - parm[1] * parm[2]);
n = (eres[0] - parm[2] * m) / parm[0];
result = "n";
}
if (double.IsInfinity(n) || double.IsInfinity(m)) {
result = "No solution!";
}
else if (result == "n") {
result = $"X: {(double.IsNaN(n) ? "Any" : n + "")} Y: {(double.IsNaN(m) ? "Any" : m + "")}";
}
else if (result == "m") {
result = $"X: {(double.IsNaN(m) ? "Any" : m + "")} Y: {(double.IsNaN(n) ? "Any" : n + "")}";
}
Console.WriteLine(result);
}
}
}
/* Example tests (numbers are ordered a1 b1 c1 a2 b2 c2) that:
*
* I. Have a solution (solution is in brackets as (x ; y)):
* 5 -2 1 1 3 7 ( 1; 2) no zeroes
* 0 1 8 1 1 5 (-3; 8) a1 = 0
* 1 0 8 1 1 5 ( 8;-3) a2 = 0
* 1 1 8 0 1 5 ( 3; 5) b1 = 0
* 1 1 8 1 0 5 ( 5; 3) b2 = 0
* 0 1 3 1 0 9 ( 9; 3) a1 = 0 and b2 = 0
* 1 0 3 0 1 9 ( 3; 9) a2 = 0 and b1 = 0
* 0 2 8 0 1 4 (any; 4) a1 = 0 and a2 = 0
* 2 0 8 1 0 4 ( 4;any) b1 = 0 and b2 = 0
* 0 0 0 0 0 0 (any;any) all are 0
*
* II. Don't have a solution:
* 0 0 4 1 1 4 a1 = 0 and b1 = 0
* 1 1 4 0 0 4 a2 = 0 and b2 = 0
* 0 0 1 0 0 0
*
* Problems with this algorithm:
*
* I. Incomplete answers:
* 2 2 2 2 2 2 Answer is (any;any) only when 0 <= x <= 2 and y = 2 - x
*/
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