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Diffstat (limited to 'C_C++/FindSpiralNumberCoord.cpp')
| -rw-r--r-- | C_C++/FindSpiralNumberCoord.cpp | 147 |
1 files changed, 147 insertions, 0 deletions
diff --git a/C_C++/FindSpiralNumberCoord.cpp b/C_C++/FindSpiralNumberCoord.cpp new file mode 100644 index 0000000..07b2071 --- /dev/null +++ b/C_C++/FindSpiralNumberCoord.cpp @@ -0,0 +1,147 @@ +#include <iostream> + +using namespace std; + +int main() +{ + int n, v, g, x, y; + cout << "n = "; + cin >> n; + + while (g < n) { + v++; + g = 8 * (v * (v + 1) / 2); + } + + if (n >= g - 2 * v) { // Bottom row + y = -v; + x = v - (g - n); + } + else if (n >= g - 4 * v) { // Left column + x = -v; + y = ((g - 2 * v) - n) - v; + } + else if (n >= g - 6 * v) { // Top row + y = v; + x = ((g - 4 * v) - n) - v; + } + else { // Right column + x = v; + y = v - ((g - 6 * v) - n); + } + + printf("(%d;%d)\n", x, y); + + 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 + * + * You are given a number n and your goal is to find the coordinates of the number. + * The distance between two neighbouring numbers is 1, meaning 0 is at (0;0), 1 is at (1;0), 38 is at (-3;1). + * + * -------------------------- + * 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". + * + * 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 + * + * After we've found g and v, we start looking at the numbers in the corners. + * The bottom right corner is g by definition, the bottom left corner is g - 2 * v, + * the top left is g - 4 * v and the top right is g - 6 * v. + * By comparing the values in the corners we can determine where the number is: + * top row, bottom row, left column or right column. + * + * Example: + * v = 2, g = 24, n = 11 + * + * if (n > bottom left corner) number is on bottom row (between 20 and 24) + * else if (n > top left corner) number is on left column (between 16 and 20) + * else if (n > top right corner) number is on top row (between 16 and 12) + * else number is on right column (between 12 and 9) + * + * After we find on which side the number is, we can immediately determine the x or y coordinate, + * as it will be +v or -v. + * The final step is to determine the other coordinate. The formula differs slightly depending on a side, + * but the gist of it is that we look at the difference between a corner number and our number, which gives + * us distance from the corner number and we offset the searched coordinate. + * + * Example: + * v = 2, g = 24, n = 11 + * We have determined that our number is in the right column. + * This means we know for sure that the X coordinate of n is 2. + * + * offset = corner number - n = (g - 6 * v) - 11 = 12 - 11 = 1 + * + * The Y coordinate of the corner number is 2 (12 is located at (2;2)). + * In our case, we want to subtract our offset from that coordinate, so + * + * Y coordinate of n = Y coordinate of corner number - offset = 2 - 1 = 1. + * + * And so we found that n is located at (2;1) + */ |