Java 解法, 执行用时: 43ms, 内存消耗: 10720K, 提交时间: 2023-08-08 11:32:34

import java.util.Scanner;
public class Main {
	public static void main(String []args)
	{
		Scanner sc=new Scanner(System.in);
		long a=sc.nextInt();
		long b=sc.nextInt();
		long c=sc.nextInt();
		long d=sc.nextInt();
		long t=b*d/gcd(b,d);
		long x=a*t/b;
		long y=c*t/d;
		long s=x+y;
		System.out.println(s/gcd(s,t)+" "+t/gcd(s,t));
		if(x-y==0)
			System.out.println(0+" "+0);
		else
		{
			if(x-y>0)
			{
				s=x-y;
				System.out.println(s/gcd(s,t)+" "+t/gcd(s,t));
			}
			else
			{
				s=y-x;
				System.out.println(-s/gcd(s,t)+" "+t/gcd(s,t));
			}
		}
		System.out.println(a*c/gcd(a*c,b*d)+" "+b*d/gcd(a*c,b*d));
		System.out.println(a*d/gcd(a*d,b*c)+" "+b*c/gcd(a*d,b*c));
	}
	public static long gcd(long a,long b)
	{
		return (b==0)?a:gcd(b,a%b);
	}
}

Python3 解法, 执行用时: 43ms, 内存消耗: 4592K, 提交时间: 2023-08-08 11:31:09

x1, y1, x2, y2 = map(int, input().strip().split())

def gcd(x, y):
    a, b = x, y
    while a and b:
        a, b = abs(a-b), a
    return x//(a+b), y//(a+b)

for x ,y in [[x1*y2+x2*y1, y1*y2], [x1*y2-x2*y1, y1*y2], [x1*x2, y1*y2], [x1*y2, x2*y1]]:
    if x == 0:
        print(0, 0)
    elif x*y < 0:
        r1, r2 = gcd(abs(x), abs(y))
        print(-1*r1, r2)
    else:
        r1, r2 = gcd(abs(x), abs(y))
        print(r1, r2)

C++ 解法, 执行用时: 3ms, 内存消耗: 416K, 提交时间: 2023-08-08 11:54:14

#include <bits/stdc++.h>
using namespace std;
void f(int x,int y)
{
    int t;
    t=__gcd(x,y);
    printf("%d %d\n",x/t,y/t);
}
int main()
{
    int x1,y1,x2,y2;
    cin>>x1>>y1>>x2>>y2;
    f(x1*y2+x2*y1,y1*y2);
    if(x1*y2-x2*y1==0)
        cout<<"0 0"<<endl;
    else
        f(x1*y2-x2*y1,y1*y2);
    f(x1*x2,y1*y2);
    f(x1*y2,x2*y1);
}