Topic link: https://www.luogu.com.cn/problem/P1652
Code example:
import java.util.Scanner;
public class Main{
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
int circleCount = scanner.nextInt(); // 圆的个数
int circleXCoordinate[] = new int[circleCount]; // 圆的X坐标
int circleYCoordinate[] = new int[circleCount]; // 圆的Y坐标
int circleRCoordinate[] = new int[circleCount]; // 圆的半径
// 输入圆的X坐标
for (int i = 0; i < circleXCoordinate.length; i++) {
circleXCoordinate[i] = scanner.nextInt();
}
// 输入圆的Y坐标
for (int i = 0; i < circleYCoordinate.length; i++) {
circleYCoordinate[i] = scanner.nextInt();
}
// 输入圆的半径
for (int i = 0; i < circleRCoordinate.length; i++) {
circleRCoordinate[i] = scanner.nextInt();
}
int x1 = scanner.nextInt();
int y1 = scanner.nextInt();
int x2 = scanner.nextInt();
int y2 = scanner.nextInt();
int result = 0; // 结果
/**
* 判断两个点到每一个圆的圆心之间的距离 distance
* 如果点到圆心距离小于半径，证明这个点在圆内
* 1.两个点都在圆内，连线经过的边界为 0
* 2.一个圆内一个圆外，连线经过的边界为 1
* 3.都在圆外，连线经过的边界为 0
*/
for (int i = 0; i < circleCount; i++) {
int distanceX1 = (int) Math.sqrt(Math.pow(x1 - circleXCoordinate[i], 2) + Math.pow(y1 - circleYCoordinate[i], 2));
int distanceX2 = (int) Math.sqrt(Math.pow(x2 - circleXCoordinate[i], 2) + Math.pow(y2 - circleYCoordinate[i], 2));
//System.out.println("distanceX1: " + distanceX1 + ", distanceX2 : " + distanceX2 + ",R : " + circleRCoordinate[i]);
// 两圆在圆内
if (distanceX1 < circleRCoordinate[i] && distanceX2 < circleRCoordinate[i]) {
// 都在圆内
}else if (distanceX1 < circleRCoordinate[i] && distanceX2 > circleRCoordinate[i]) {
// 一个圆内一个圆外
result ++;
}else if (distanceX1 > circleRCoordinate[i] && distanceX2 < circleRCoordinate[i]) {
// 一个圆内一个圆外
result ++;
}else if (distanceX1 > circleRCoordinate[i] && distanceX2 > circleRCoordinate[i]) {
// 都在圆外
}
}
System.out.println(result);
}
}
Knowledge point investigation:
Square: Math.pow(a,b) represents the bth power of a Square
root: Math.sqrt()
The distance from the point to the center of the circle: Under the root sign, the square of x1-x2 + the square of y1-y2
Test results:
