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diff --git a/C++/FindSpiralNumber.cpp b/C++/FindSpiralNumber.cpp
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+#include <iostream>
+
+using namespace std;
+
+int main()
+{
+ int x, y;
+ cout << "x = ";
+ cin >> x;
+ cout << "y = ";
+ cin >> y;
+
+ int n, v = max(abs(x), abs(y)), g = 8 * (v * (v + 1) / 2);
+
+ if (y == v || x == v) { // Top row or right column
+ n = g - x + y - 6 * v;
+ }
+ else { // Left column or bottom row
+ n = g + x - y - 2 * v;
+ }
+
+ printf("n = %d", n);
+
+ return 0;
+}
+
+/* --------------
+ * Task condition
+ * --------------
+ * You have a spiral of whole numbers, starting from 0, like this:
+ *
+ * 36 < 35 < 34 < 33 < 32 < 31 < 30
+ * V ^
+ * 37 16 < 15 < 14 < 13 < 12 29
+ * V V ^ ^
+ * 38 17 4 < 3 < 2 11 28
+ * V V V ^ ^ ^
+ * 39 18 5 0 > 1 10 27 ...
+ * V V V ^ ^ ^
+ * 40 19 6 > 7 > 8 > 9 26 51
+ * V V ^ ^
+ * 41 20 > 21 > 22 > 23 > 24 > 25 50
+ * V ^
+ * 42 > 43 > 44 > 45 > 46 > 47 > 48 > 49
+ *
+ * The distance between two neighbouring numbers is 1, meaning 0 is at (0;0), 1 is at (1;0), 38 is at (-3;1).
+ * You are given the coordinates of a number and you have to find the actual number.
+ *
+ * --------------------------
+ * Explanation of my solution
+ * --------------------------
+ *
+ * My solution is a very mathematical one. First, I divide the spiral into squares,
+ * depending on the offset of 0;0.
+ * Example:
+ * - Square 1 (because x and y are between -1 and 1 (one of them is always 1 or -1)):
+ * 4 < 3 < 2
+ * V ^
+ * 5 1
+ * V
+ * 6 > 7 > 8
+ *
+ * - Square 2 (because x and y are between -2 and 2 (one of them is always 2 or -2)):
+ * 16 < 15 < 14 < 13 < 12
+ * V ^
+ * 17 11
+ * V ^
+ * 18 10
+ * V ^
+ * 19 9
+ * V
+ * 20 > 21 > 22 > 23 > 24
+ *
+ * - Square 3 (because x and y are between -3 and 3 (one of them is always 3 or -3)):
+ * 36 < 35 < 34 < 33 < 32 < 31 < 30
+ * V ^
+ * 37 29
+ * V ^
+ * 38 28
+ * V ^
+ * 39 27
+ * V ^
+ * 40 26
+ * V ^
+ * 41 25
+ * V
+ * 42 > 43 > 44 > 45 > 46 > 47 > 48
+ *
+ * And so on. I call the number of the square "v".
+ * We can find it by getting the absolute value of both coordinates and get the biggest number.
+ *
+ * Then there is the biggest number in the square, the one in the bottom right corner, I call that "g".
+ * g for square 1 (v = 1) is 8, for v = 2: g = 24, for v = 3: g = 48 and so on.
+ * We can see that each g is divisble by 8, but the multiplier is slightly tricky to find.
+ * The multiplier is the sum of all whole number from 1 up until v.
+ *
+ * Example:
+ * v = 2
+ * g = 8 * (Sum of numbers from 1 to v) = 8 * (1 + 2) = 8 * 3 = 24
+ *
+ * v = 3
+ * g = 8 * (Sum of numbers from 1 to v) = 8 * (1 + 2 + 3) = 8 * 6 = 48
+ *
+ * Then, we divide the whole square into two corners. We take the diagonal from top left to bottom right, and check in which part the coordinates are.
+ * We do that because we have two formulas. In brief, what they do is tranform the coordinates into the number from which we'll subtract from g.
+ *
+ * Example:
+ * v = 3, g = 48
+ *
+ * Lets take a look at the formula for the bottom of the square. We aren't interested in g, so the important part is "x - y - 2*v".
+ * What it does is make the coordinates for x and y correspond to a number from -12 to 0.
+ * So this:
+ * 36
+ * V
+ * 37
+ * V
+ * 38
+ * V
+ * 39
+ * V
+ * 40
+ * V
+ * 41
+ * V
+ * 42 > 43 > 44 > 45 > 46 > 47 > 48
+ * Becomes:
+ * -12
+ * V
+ * -11
+ * V
+ * -10
+ * V
+ * -9
+ * V
+ * -8
+ * V
+ * -7
+ * V
+ * -6 > -5 > -4 > -3 > -2 > -1 > 0
+ *
+ * "- 2 * v" is -6, the number in the bottom left corner. Then, if we increase Y (meaning we want to go up the left column),
+ * we want the number to also decrease (make it further away from 0). We also have to account for X, because there it's always -3.
+ * So, we need them to sum up to 0 when we're on the bottom left corner, so we can just subtract Y from X. As we said,
+ * X is always -3 on the left column, so by increasing Y, we also increase the negative number.
+ *
+ * Ok, here are some example numbers:
+ * X | Y | X - Y
+ * ---|----|------
+ * -3 | -3 | 0
+ * -3 | -2 | -1
+ * -3 | 0 | -3
+ *
+ * This actually works out for us just fine when we're traversing the bottom row. There, Y is always -3, and
+ * when increasing X we want to increase our number (make it close to 0). So, X - Y become a positive number,
+ * which summed up with the negative number ("-2*v") makes the result number closer to 0, which is what we want.
+ *
+ * Here are more example numbers:
+ * X | Y | X - Y
+ * ---|----|------
+ * -3 | -3 | 0
+ * -2 | -3 | 1
+ * 0 | -3 | 3
+ *
+ * The formula for the top of the square (top row and right column) works the same , but there our corner number isn't "-2*v" but "-6*v",
+ * and we want the reverse effect: increase number (make smaller) on increasing X and decerease number on incrasing Y,
+ * so we make "X - Y" to be "Y - X"
+ *
+ * -13 < -14 < -15 < -16 < -17 < -18 < -19
+ * ^
+ * -20
+ * ^
+ * -21
+ * ^
+ * -22
+ * ^
+ * -23
+ * ^
+ * -24
+ */