定义一个小车类Car,其初始位置为(0,0),属性包括:当前位置,行驶速度,行驶朝向(东1、南2、西3、北4);方向变化:直行(1)、左转(2)、右转(3);速度变化(正数为加速,负数为减速)。
约束条件:(1)采用直角坐标系,上北下南左西右东;(2)速度为0时,车辆停止,减速无效;(3)速度达到20时,加速无效。输入
小车的初始位置(x,y)、初始速度(v)和初始朝向(f),时刻数量(<=1000),每个时刻内的方向变化(c)、速度变化(s)
输出
当前小车的位置
样例输入 Copy
0 0 0 1
10
1 1 1 1 1 1 1 0 1 0 1 0 2 0 1 0 1 2 1 -1
样例输出 Copy
15,15
该回答引用ChatGPT
以下是Java代码实现,可以通过LeetCode的Java环境进行测试:
public class Car {
private int x;
private int y;
private int velocity;
private int direction;
public Car(int x, int y, int velocity, int direction) {
this.x = x;
this.y = y;
this.velocity = velocity;
this.direction = direction;
}
public void move(int changeDirection, int changeVelocity) {
if (velocity == 0) {
return;
}
// 更新速度
velocity += changeVelocity;
if (velocity < 0) {
velocity = 0;
} else if (velocity > 20) {
velocity = 20;
}
// 更新方向
direction = (direction + changeDirection) % 4;
if (direction == 0) {
direction = 4;
}
// 根据方向和速度更新位置
switch (direction) {
case 1:
x += velocity;
break;
case 2:
y -= velocity;
break;
case 3:
x -= velocity;
break;
case 4:
y += velocity;
break;
}
}
public String getPosition() {
return x + "," + y;
}
}
public class Main {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
int x = scanner.nextInt();
int y = scanner.nextInt();
int velocity = scanner.nextInt();
int direction = scanner.nextInt();
int n = scanner.nextInt();
Car car = new Car(x, y, velocity, direction);
for (int i = 0; i < n; i++) {
int changeDirection = scanner.nextInt();
int changeVelocity = scanner.nextInt();
car.move(changeDirection, changeVelocity);
}
System.out.println(car.getPosition());
}
}