Shift Dot

    科技2022-07-13  106

    Description 给出平面直角坐标系中的一点,并顺序给出n个向量,求该点根据给定的n个向量位移后的位置。

    Input 多组输入,第一行是三个整数x,y,n,表示点的坐标(x,y),和向量的个数n。接下来n行,每行两个数xi,yi,表示第i个向量。题目中所有数据不会超出整形范围。

    Output 每组输入输出一行,"(x,y)"表示点的最终位置。

    Sample Input 0 0 1 2 3 0 0 2 1 2 2 3 Output (2,3) (3,5)

    import java.util.Scanner; class Point { int x; int y; public Point(int x, int y) { this.x = x; this.y = y; } public Point sum(Point po) { x = x + po.x; y = y + po.y; return new Point(x, y); } } public class Main { public static void main(String[] args) { Scanner reader = new Scanner(System.in); while(reader.hasNext()) { int x = reader.nextInt(); int y = reader.nextInt(); Point point = new Point(x, y); int n = reader.nextInt(); for(int i = 0; i< n; i++) { int x1 = reader.nextInt(); int y1 = reader.nextInt(); Point point1 = new Point(x1, y1); point = point.sum(point1); } System.out.println("("+point.x+","+point.y+")"); } reader.close(); } }
    Processed: 0.011, SQL: 8