We are provided with a 2D plane and a point (
, 
). We have to find a rectangle (
, 
, 
, 
) such that it
encompasses the given point (, ). The rectangle chosen must satisfy the given condition 
. If
multiple rectangles are possible the we must choose the one with the least Euclid distance between the center of the rectangle and the point (, ).
Image Source – codeforces
Examples :
Input : 70 10 20 5 5 3
Output :12 0 27 9Input :100 100 32 63 2 41
Output :30 18 34 100
The logic behind the problem is as follows. First of all we reduce 
to lowest irreducible form by dividing a and b by 
. We think of the problem in two separate dimensions and independently. We find out the 
after dividing a and b by their gcd to find the maximum distance that we can cover safely by staying in the range of 
. Since we have to find rectangle with minimum distance of its center from the point (, ), therefore we start with the assumption that point (, ) is our center. Then we find and by subtracting and adding half of length to . If either or goes out of range then we shift the coordinates accordingly to bring it inside the range . Similarly we proceed to calculate and .
For the first example according to above logic the answer comes out to be 
.
C++
#include <cmath>#include <iostream>using namespace std;// Function to calculate// GCD of a and bint gcd(int a, int b){ if (a == 0) return b; else return gcd(b % a, a);}// Function to calculate the// coordinates of the rectanglevoid solve(int n, int m, int x, int y, int a, int b){ int k, g; int x1, y1, x2, y2; g = gcd(a, b); // Reducing the ratio to // lowest irreducible form a /= g; b /= g; // Finding the maximum multiple // of length that can be chosen k = min(n / a, m / b); // Assuming the point (X, Y) as the // center of the required rectangle // Finding the lower X coordinate by // subtracting half of total length from X x1 = x - (k * a - k * a / 2); // Finding the upper X coordinate // by adding half of total length to X x2 = x + k * a / 2; // Finding lower Y coordinate by // subtracting half of breadth from Y y1 = y - (k * b - k * b / 2); // Finding upper Y coordinate // by adding half of breadth to Y y2 = y + k * b / 2; // If lower X coordinate // goes below 0 then we shift // the lower coordinate to 0 // and the upper coordinate // accordingly to bring lower // coordinate in range // and to keep center as // close as possible to X, Y if (x1 < 0) { x2 -= x1; x1 = 0; } // If upper X coordinate goes // beyond n, then we shift the // upper X coordinate ton // and we shift the lower coordinate // accordingly to bring the upper // coordinate in range if (x2 > n) { x1 -= x2 - n; x2 = n; } // If lower Y coordinate goes // below 0 then we shift the // lower coordinate to 0 // and the upper coordinate // accordingly to bring lower // coordinate in range // and to keep center as // close as possible to X, Y if (y1 < 0) { y2 -= y1; y1 = 0; } // If upper Y coordinate goes // beyond n, then we shift the // upper X coordinate // to n and we shift the lower // coordinate accordingly to // bring the upper // coordinate in range if (y2 > m) { y1 -= y2 - m; y2 = m; } cout << x1 << " " << y1 << " " << x2 << " " << y2 << endl;}// Driver functionint main(){ int n = 70, m = 10, x = 20, y = 5, a = 5, b = 3; solve(n, m, x, y, a, b); return 0;} |
Java
class GFG { // Function to calculate // GCD of a and b public static int gcd(int a, int b) { if (a == 0) return b; else return gcd(b % a, a); } // Function to calculate the // coordinates of the rectangle public static void solve(int n, int m, int x, int y, int a, int b) { int k, g; int x1, y1, x2, y2; g = gcd(a, b); // Reducing the ratio to // lowest irreducible form a /= g; b /= g; // Finding the maximum multiple // of length that can be chosen k = Math.min(n / a, m / b); // Assuming the point (X, Y) as the // centre of the required rectangle // Finding the lower X coordinate by // subtracting half of total length // from X x1 = x - (k * a - k * a / 2); // Finding the upper X coordinate // by adding half of total length // to X x2 = x + k * a / 2; // Finding lower Y coordinate by // subtracting half of breadth // from Y y1 = y - (k * b - k * b / 2); // Finding upper Y coordinate // by adding half of breadth // to Y y2 = y + k * b / 2; // If lower X coordinate // goes below 0 then we shift // the lower coordinate to 0 // and the upper coordinate // accordingly to bring lower // coordinate in range // and to keep center as // close as possible to X, Y if (x1 < 0) { x2 -= x1; x1 = 0; } // If upper X coordinate goes // beyond n, then we shift the // upper X coordinate ton // and we shift the lower coordinate // accordingly to bring the upper // coordinate in range if (x2 > n) { x1 -= x2 - n; x2 = n; } // If lower Y coordinate goes // below 0 then we shift the // lower coordinate to 0 // and the upper coordinate // accordingly to bring lower // coordinate in range // and to keep center as // close as possible to X, Y if (y1 < 0) { y2 -= y1; y1 = 0; } // If upper Y coordinate goes // beyond n, then we shift the // upper X coordinate // to n and we shift the lower // coordinate accordingly to // bring the upper // coordinate in range if (y2 > m) { y1 -= y2 - m; y2 = m; } System.out.println(x1 + " " + y1 + " " + x2 + " " + y2); } // Driver Code public static void main(String args[]) { int n = 70, m = 10; int x = 20, y = 5; int a = 5, b = 3; solve(n, m, x, y, a, b); }}// This code is contributed by - vkz6198 |
Python 3
# Function to calculate# GCD of a and bdef gcd(a, b): if (a == 0): return b else: return gcd(b % a, a)# Function to calculate the# coordinates of the rectangledef solve(n, m, x, y, a, b): g = int(gcd(a, b)) # Reducing the ratio to # lowest irreducible form a /= g b /= g # Finding the maximum multiple # of length that can be chosen k = int(min(n / a, m / b)) # Assuming the point (X, Y) as the # centre of the required rectangle # Finding the lower X coordinate by # subtracting half of total length # from X x1 = int(x - (k * a - k * a / 2)) # Finding the upper X coordinate # by adding half of total length # to X x2 = int(x + k * a / 2) # Finding lower Y coordinate by # subtracting half of breadth from Y y1 = int(y - (k * b - k * b / 2)) # Finding upper Y coordinate # by adding half of breadth to Y y2 = int(y + k * b / 2) # If lower X coordinate # goes below 0 then we shift # the lower coordinate to 0 # and the upper coordinate # accordingly to bring lower # coordinate in range # and to keep center as # close as possible to X, Y if (int(x1) < 0): x2 -= x1 x1 = 0 # If upper X coordinate goes # beyond n, then we shift the # upper X coordinate ton # and we shift the lower coordinate # accordingly to bring the upper # coordinate in range if (int(x2) > n): x1 -= x2 - n x2 = n # If lower Y coordinate goes # below 0 then we shift the # lower coordinate to 0 # and the upper coordinate # accordingly to bring lower # coordinate in range # and to keep center as # close as possible to X, Y if (int(y1) < 0): y2 -= y1 y1 = 0 # If upper Y coordinate goes # beyond n, then we shift the # upper X coordinate # to n and we shift the lower # coordinate accordingly to # bring the upper # coordinate in range if (int(y2) > m): y1 -= y2 - m y2 = m print(x1, " ", y1, " ", x2, " ", y2, sep="")# Driver functionn = 70m = 10x = 20y = 5a = 5b = 3solve(n, m, x, y, a, b)# This code is contributed by Smitha |
C#
using System;class GFG { // Function to calculate // GCD of a and b public static int gcd(int a, int b) { if (a == 0) return b; else return gcd(b % a, a); } // Function to calculate the // coordinates of the rectangle public static void solve(int n, int m, int x, int y, int a, int b) { int k, g; int x1, y1, x2, y2; g = gcd(a, b); // Reducing the ratio to // lowest irreducible form a /= g; b /= g; // Finding the maximum multiple // of length that can be chosen k = Math.Min(n / a, m / b); // Assuming the point (X, Y) as the // centre of the required rectangle // Finding the lower X coordinate by // subtracting half of total length // from X x1 = x - (k * a - k * a / 2); // Finding the upper X coordinate // by adding half of total length // to X x2 = x + k * a / 2; // Finding lower Y coordinate by // subtracting half of breadth // from Y y1 = y - (k * b - k * b / 2); // Finding upper Y coordinate // by adding half of breadth // to Y y2 = y + k * b / 2; // If lower X coordinate // goes below 0 then we shift // the lower coordinate to 0 // and the upper coordinate // accordingly to bring lower // coordinate in range // and to keep center as // close as possible to X, Y if (x1 < 0) { x2 -= x1; x1 = 0; } // If upper X coordinate goes // beyond n, then we shift the // upper X coordinate ton // and we shift the lower coordinate // accordingly to bring the upper // coordinate in range if (x2 > n) { x1 -= x2 - n; x2 = n; } // If lower Y coordinate goes // below 0 then we shift the // lower coordinate to 0 // and the upper coordinate // accordingly to bring lower // coordinate in range // and to keep center as // close as possible to X, Y if (y1 < 0) { y2 -= y1; y1 = 0; } // If upper Y coordinate goes // beyond n, then we shift the // upper X coordinate // to n and we shift the lower // coordinate accordingly to // bring the upper // coordinate in range if (y2 > m) { y1 -= y2 - m; y2 = m; } Console.Write(x1 + " " + y1 + " " + x2 + " " + y2); } // Driver Code public static void Main() { int n = 70, m = 10; int x = 20, y = 5; int a = 5, b = 3; solve(n, m, x, y, a, b); }}// This code is contributed by Smitha |
PHP
<?php// Function to calculate// GCD of a and bfunction gcd($a, $b){ if ($a == 0) return $b; else return gcd($b % $a, $a);}// Function to calculate the// coordinates of the rectanglefunction solve($n, $m, $x, $y, $a, $b){ $k; $g; $x1; $y1; $x2; $y2; $g = gcd($a, $b); // Reducing the ratio to // lowest irreducible form $a /= $g; $b /= $g; // Finding the maximum multiple // of length that can be chosen $k = min($n / $a, $m / $b); // Assuming the point (X, Y) // as the centre of the required // rectangle Finding the lower // X coordinate by subtracting // half of total length from X $x1 = $x - ($k * $a - $k * $a / 2); // Finding the upper X // coordinate by adding // half of total length to X $x2 = $x + $k * $a / 2; // Finding lower Y coordinate by // subtracting half of breadth from Y $y1 = $y - ($k * $b - $k * $b / 2); // Finding upper Y coordinate // by adding half of breadth to Y $y2 = $y + $k * $b / 2; // If lower X coordinate // goes below 0 then we shift // the lower coordinate to 0 // and the upper coordinate // accordingly to bring lower // coordinate in range // and to keep center as // close as possible to X, Y if ($x1 < 0) { $x2 -= $x1; $x1 = 0; } // If upper X coordinate goes // beyond n, then we shift the // upper X coordinate ton // and we shift the lower coordinate // accordingly to bring the upper // coordinate in range if ($x2 > $n) { $x1 -= $x2 - $n; $x2 = $n; } // If lower Y coordinate goes // below 0 then we shift the // lower coordinate to 0 // and the upper coordinate // accordingly to bring lower // coordinate in range // and to keep center as // close as possible to X, Y if ($y1 < 0) { $y2 -= $y1; $y1 = 0; } // If upper Y coordinate goes // beyond n, then we shift the // upper X coordinate // to n and we shift the lower // coordinate accordingly to // bring the upper // coordinate in range if ($y2 > $m) { $y1 -= $y2 - $m; $y2 = $m; } echo $x1, " ", $y1, " ", $x2, " ", $y2, "\n";}// Driver Code$n = 70; $m = 10; $x = 20;$y = 5; $a = 5; $b = 3;solve($n, $m, $x, $y, $a, $b);// This code is contributed by aj_36?> |
Javascript
<script> // Function to calculate // GCD of a and b function gcd(a, b) { if (a == 0) return b; else return gcd(b % a, a); } // Function to calculate the // coordinates of the rectangle function solve(n, m, x, y, a, b) { let k, g; let x1, y1, x2, y2; g = gcd(a, b); // Reducing the ratio to // lowest irreducible form a = a / g; b = b / g; // Finding the maximum multiple // of length that can be chosen k = Math.min(parseInt(n / a, 10), parseInt(m / b, 10)); // Assuming the point (X, Y) as the // centre of the required rectangle // Finding the lower X coordinate by // subtracting half of total length // from X x1 = x - (k * a - k * parseInt(a / 2, 10)); // Finding the upper X coordinate // by adding half of total length // to X x2 = x + k * parseInt(a / 2, 10); // Finding lower Y coordinate by // subtracting half of breadth // from Y y1 = y - (k * b - k * parseInt(b / 2, 10)); // Finding upper Y coordinate // by adding half of breadth // to Y y2 = y + k * parseInt(b / 2, 10); // If lower X coordinate // goes below 0 then we shift // the lower coordinate to 0 // and the upper coordinate // accordingly to bring lower // coordinate in range // and to keep center as // close as possible to X, Y if (x1 < 0) { x2 -= x1; x1 = 0; } // If upper X coordinate goes // beyond n, then we shift the // upper X coordinate ton // and we shift the lower coordinate // accordingly to bring the upper // coordinate in range if (x2 > n) { x1 -= x2 - n; x2 = n; } x1 += 1; x2 += 1; // If lower Y coordinate goes // below 0 then we shift the // lower coordinate to 0 // and the upper coordinate // accordingly to bring lower // coordinate in range // and to keep center as // close as possible to X, Y if (y1 < 0) { y2 -= y1; y1 = 0; } // If upper Y coordinate goes // beyond n, then we shift the // upper X coordinate // to n and we shift the lower // coordinate accordingly to // bring the upper // coordinate in range if (y2 > m) { y1 -= y2 - m; y2 = m; } document.write(x1 + " " + y1 + " " + x2 + " " + y2); } let n = 70, m = 10; let x = 20, y = 5; let a = 5, b = 3; solve(n, m, x, y, a, b); </script> |
Output
12 0 27 9
Time complexity: O(log(min(a,b)))
Auxiliary space: O(log(min(a,b))
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