elegant dot()

Topic description

Xiaoyi has a circle whose center is at the origin of coordinates, and Xiaoyi knows the square of the radius of the circle. Xiaoyi thinks that the points on the circle and the points whose horizontal and vertical coordinates are integers are elegant. Xiaoyi now wants to find an algorithm to calculate the number of elegant points. Please help him.
For example: if the square of the radius is 25
elegant points have: (+/-3, +/-4), (+/-4, +/-3), (0, +/-5) (+/- 5, 0), a total of 12 points.

Enter description:

The input is an integer, which is the square of the radius of the circle, and the range is in the range of 32-bit int.

Output description:

The output is an integer, which is the number of elegant points

Example 1

enter

25

output

12

1  import java.util.Scanner;
 2  
3  /** 
4  * 
 5  * The elegant point on the circle and the coordinates are integers only need to calculate the first quadrant due to symmetry
 6  * @author Dell
 7  *
 8   */ 
9  public  class Main {
 10  public  static  void main(String[] args) {
 11      Scanner sc = new Scanner(System.in);
 12      // 2 squared 
13      int r_2 = sc.nextInt();
 14      int n = 0 ;
 15      //  On the coordinate axis, either x=0 or y=0 and as long as one of the four numbers matches four, they all meet
 16      // Just need to judge one and finally follow the four quadrants multiplied by 4
 17      // The   judgment range is 0<= x^ 2 < r^2 equal sign takes only one 
18      int x = 0 ;
 19      int y0 = 1 ;
 20      while (x*x< r_2) {
 21          double y_2 = r_2-x* x;
 22          // y1 is of type double y^2 square root 
23          double y1 = Math.sqrt(y_2);
 24          // y0 is y1 forced to get 
25          y0 = ( int )y1;
 26          // By judging whether y0 y1 is equal to determine whether the y1 obtained from the square root is not for integer 
27          if(y0== y1) {
 28              n++ ;
 29          }
 30          x++ ;
 31      }
 32      // Four quadrants multiplied by four after the loop 
33      n*=4 ;
 34      System.out.println(n);
 35  }
 36 }