给定一个半径圆 R 在以原点或（0，0）为中心的二维中。任务是找到圆周上的总格点。格点是二维空间中坐标为整数的点。 例子：
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Input : r = 5.Output : 12Below are lattice points on a circle withradius 5 and origin as (0, 0).(0,5), (0,-5), (5,0), (-5,0),(3,4), (-3,4), (-3,-4), (3,-4),(4,3), (-4,3), (-4,-3), (4,-3).are 12 lattice point.
为了找到格点，我们基本上需要找到满足方程x的（x，y）值 2. +y 2. =r 2. . 对于满足上述方程的（x，y）的任何值，我们实际上总共有4个满足方程的不同组合。例如，如果r=5和（3，4）是满足方程的一对，那么实际上有4个组合（3，4），（-3，4），（-3，-4），（3，-4）。但是有一个例外，在（0，r）或（r，0）的情况下，实际上有2个点，因为没有负0。
// Initialize result as 4 for (r, 0), (-r. 0),// (0, r) and (0, -r)result = 4Loop for x = 1 to r-1 and do following for every x. If r*r - x*x is a perfect square, then add 4 tor result.
下面是上述想法的实现。
CPP
// C++ program to find countLattice points on a circle #include<bits/stdc++.h> using namespace std; // Function to count Lattice points on a circle int countLattice( int r) { if (r <= 0) return 0; // Initialize result as 4 for (r, 0), (-r. 0), // (0, r) and (0, -r) int result = 4; // Check every value that can be potential x for ( int x=1; x<r; x++) { // Find a potential y int ySquare = r*r - x*x; int y = sqrt (ySquare); // checking whether square root is an integer // or not. Count increments by 4 for four // different quadrant values if (y*y == ySquare) result += 4; } return result; } // Driver program int main() { int r = 5; cout << countLattice(r); return 0; } |
JAVA
// Java program to find // countLattice points on a circle class GFG { // Function to count // Lattice points on a circle static int countLattice( int r) { if (r <= 0 ) return 0 ; // Initialize result as 4 for (r, 0), (-r. 0), // (0, r) and (0, -r) int result = 4 ; // Check every value that can be potential x for ( int x= 1 ; x<r; x++) { // Find a potential y int ySquare = r*r - x*x; int y = ( int )Math.sqrt(ySquare); // checking whether square root is an integer // or not. Count increments by 4 for four // different quadrant values if (y*y == ySquare) result += 4 ; } return result; } // Driver code public static void main(String arg[]) { int r = 5 ; System.out.println(countLattice(r)); } } // This code is contributed by Anant Agarwal. |
Python3
# Python3 program to find # countLattice podefs on a circle import math # Function to count Lattice # podefs on a circle def countLattice(r): if (r < = 0 ): return 0 # Initialize result as 4 for (r, 0), (-r. 0), # (0, r) and (0, -r) result = 4 # Check every value that can be potential x for x in range ( 1 , r): # Find a potential y ySquare = r * r - x * x y = int (math.sqrt(ySquare)) # checking whether square root is an defeger # or not. Count increments by 4 for four # different quadrant values if (y * y = = ySquare): result + = 4 return result # Driver program r = 5 print (countLattice(r)) # This code is contributed by # Smitha Dinesh Semwal |
C#
// C# program to find countLattice // points on a circle using System; class GFG { // Function to count Lattice // points on a circle static int countLattice( int r) { if (r <= 0) return 0; // Initialize result as 4 // for (r, 0), (-r. 0), // (0, r) and (0, -r) int result = 4; // Check every value that // can be potential x for ( int x = 1; x < r; x++) { // Find a potential y int ySquare = r*r - x*x; int y = ( int )Math.Sqrt(ySquare); // checking whether square root // is an integer or not. Count // increments by 4 for four // different quadrant values if (y*y == ySquare) result += 4; } return result; } // Driver code public static void Main() { int r = 5; Console.Write(countLattice(r)); } } // This code is contributed by nitin mittal. |
PHP
<?php // PHP program to find countLattice // points on a circle // Function to count Lattice // points on a circle function countLattice( $r ) { if ( $r <= 0) return 0; // Initialize result as 4 // for (r, 0), (-r. 0), // (0, r) and (0, -r) $result = 4; // Check every value that // can be potential x for ( $x = 1; $x < $r ; $x ++) { // Find a potential y $ySquare = $r * $r - $x * $x ; $y = ceil (sqrt( $ySquare )); // checking whether square // root is an integer // or not. Count increments // by 4 for four different // quadrant values if ( $y * $y == $ySquare ) $result += 4; } return $result ; } // Driver Code $r = 5; echo countLattice( $r ); // This code is contributed by nitin mittal ?> |
Javascript
<script> // javascript program to find // countLattice points on a circle // Function to count // Lattice points on a circle function countLattice(r) { if (r <= 0) return 0; // Initialize result as 4 for (r, 0), (-r. 0), // (0, r) and (0, -r) var result = 4; // Check every value that can be potential x for (x = 1; x < r; x++) { // Find a potential y var ySquare = r * r - x * x; var y = parseInt( Math.sqrt(ySquare)); // checking whether square root is an integer // or not. Count increments by 4 for four // different quadrant values if (y * y == ySquare) result += 4; } return result; } // Driver code var r = 5; document.write(countLattice(r)); // This code is contributed by umadevi9616 </script> |
输出：
12
参考： http://mathworld.wolfram.com/CircleLatticePoints.html 本文由 希瓦姆·普拉丹（anuj_charm） .如果你喜欢GeekSforgek，并想贡献自己的力量，你也可以使用 写极客。组织 或者把你的文章寄去评论-team@geeksforgeeks.org.看到你的文章出现在Geeksforgeks主页上，并帮助其他极客。 如果您发现任何不正确的地方，或者您想分享有关上述主题的更多信息，请写下评论。
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