CodeForces - 1612A Distance

Posted 2022-03-08 海岛Blog

A. Distance
time limit per test3 seconds
memory limit per test512 megabytes

Let’s denote the Manhattan distance between two points p1 (with coordinates (x1,y1)) and p2 (with coordinates (x2,y2)) as d(p1,p2)=|x1−x2|+|y1−y2|. For example, the distance between two points with coordinates (1,3) and (4,2) is |1−4|+|3−2|=4.

You are given two points, A and B. The point A has coordinates (0,0), the point B has coordinates (x,y).

Your goal is to find a point C such that:

both coordinates of C are non-negative integers;
d(A,C)=d(A,B)/2 (without any rounding);
d(B,C)=d(A,B)/2 (without any rounding).
Find any point C that meets these constraints, or report that no such point exists.

Input
The first line contains one integer t (1≤t≤3000) — the number of test cases.

Each test case consists of one line containing two integers x and y (0≤x,y≤50) — the coordinates of the point B.

Output
For each test case, print the answer on a separate line as follows:

if it is impossible to find a point C meeting the constraints, print “-1 -1” (without quotes);
otherwise, print two non-negative integers not exceeding 106 — the coordinates of point C meeting the constraints. If there are multiple answers, print any of them. It can be shown that if any such point exists, it’s possible to find a point with coordinates not exceeding 106 that meets the constraints.
Example
input
10
49 3
2 50
13 0
0 41
42 0
0 36
13 37
42 16
42 13
0 0
output
23 3
1 25
-1 -1
-1 -1
21 0
0 18
13 12
25 4
-1 -1
0 0
Note
Explanations for some of the test cases from the example:

·In the first test case, the point B has coordinates (49,3). If the point C has coordinates (23,3), then the distance from A to B is |49−0|+|3−0|=52, the distance from A to C is |23−0|+|3−0|=26, and the distance from B to C is |23−49|+|3−3|=26.
·In the second test case, the point B has coordinates (2,50). If the point C has coordinates (1,25), then the distance from A to B is |2−0|+|50−0|=52, the distance from A to C is |1−0|+|25−0|=26, and the distance from B to C is |1−2|+|25−50|=26.
·In the third and the fourth test cases, it can be shown that no point with integer coordinates meets the constraints.
·In the fifth test case, the point B has coordinates (42,0). If the point C has coordinates (21,0), then the distance from A to B is |42−0|+|0−0|=42, the distance from A to C is |21−0|+|0−0|=21, and the distance from B to C is |21−42|+|0−0|=21.

问题链接：CodeForces - 1612A Distance
问题简述：（略）
问题分析：（略）
AC的C语言程序如下：

/* CodeForces - 1612A Distance */

#include <stdio.h>

int main()

    int t, x, y;
    scanf("%d", &t);
    while (t--) 
        scanf("%d%d", &x, &y);
        if ((x ^ y) & 1)
            puts("-1 -1");
        else if ((x & 1) && (y & 1))
            printf("%d %d\\n", x / 2, y / 2 + 1);
        else
            printf("%d %d\\n", x / 2, y / 2);
    

    return 0;