A. Find K Distinct Points with Fixed Center

time limit per test1 second memory limit per test256 megabytes

I couldn't think of a good title for this problem, so I decided to learn from LeetCode.

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You are given three integers xc, yc, and k (−100≤xc,yc≤100, 1≤k≤1000).

You need to find k distinct points (x1,y1), (x2,y2), …, (xk,yk), having integer coordinates, on the 2D coordinate plane such that:

• their center∗ is (xc,yc)
• −10^9≤xi,yi≤10^9 for all i from 1 to k
  It can be proven that at least one set of kk distinct points always exists that satisfies these conditions.

∗∗The center of kk points (x1,y1x1,y1), (x2,y2x2,y2), ……, (xk,ykxk,yk) is (x1+x2+…+xkk,y1+y2+…+ykk)(x1+x2+…+xkk,y1+y2+…+ykk).

Input

The first line contains tt (1≤t≤1001≤t≤100) — the number of test cases.
Each test case contains three integers xcxc, ycyc, and kk (−100≤xc,yc≤100−100≤xc,yc≤100, 1≤k≤10001≤k≤1000) — the coordinates of the center and the number of distinct points you must output.
It is guaranteed that the sum of kk over all test cases does not exceed 10001000.

Output
For each test case, output kk lines, the ii-th line containing two space separated integers, xixi and yiyi, (−109≤xi,yi≤109−109≤xi,yi≤109) — denoting the position of the ii-th point.
If there are multiple answers, print any of them. It can be shown that a solution always exists under the given constraints.

Example
InputCopy

4
10 10 1
0 0 3
-5 -8 8
4 -5 3

OutputCopy

10 10
-1 -1
5 -1
-4 2
-6 -7
-5 -7
-4 -7
-4 -8
-4 -9
-5 -9
-6 -9
-6 -8
1000 -1000
-996 995
8 -10

Code

import java.util.*;
 
public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        
        while (t-- > 0) {
            int xc = sc.nextInt();
            int yc = sc.nextInt();
            int k = sc.nextInt();
            
            long sumX = 0, sumY = 0;
            
            for (int i = 0; i < k - 1; i++) {
            	// 충분히 작은 값에서 시작
                int x = i - 500000; 
                int y = i - 500000;
                System.out.println(x + " " + y);
                sumX += x;
                sumY += y;
            }
            
            // 마지막 점 계산
            long lastX = (long)k * xc - sumX;
            long lastY = (long)k * yc - sumY;
            
            // 마지막 점의 좌표가 범위를 벗어나지 않도록 보정
            lastX = Math.max(-1000000000, Math.min(1000000000, lastX));
            lastY = Math.max(-1000000000, Math.min(1000000000, lastY));
            
            System.out.println(lastX + " " + lastY);
        }
    }
}