wiskey的眼神

WisKey的眼镜有500多度,所以眼神不大好,而且他有个习惯,就是走路喜欢看着地(不是为了拣钱哦^_^),

 

所以大家下次碰见他的时候最好主动打下招呼,呵呵.

 

但是Rabbit总是喜欢扮神秘,一天WisKey去食堂排队等着买饭,突然收到一道短消息,是Rabbit发的,”呵呵,又看见你了,你没看到我吧”.

 

WisKey马上拉长脖子扫描食堂,可是就是看不到,再发短信问Rabbit在哪,Rabbit回信曰”我已经在寝室了”.WisKey无语....

 

假设食堂是个正方形,食堂中心坐标为(0,0),长度为2*L, WisKey保证在食堂内.

 

因为是吃饭高峰期,所以每个点上都站着人,当某些人处在同一直线上时就有可能被前面的人挡住.

 

聪明的ACMer请你帮帮WisKey,告诉他能不能看见Rabbit.

 

 Input

输入L,sx,sy,px,py; L<=1000,sx,sy是WisKey的坐标,px,py是Rabbit的坐标.

以L=0为结束.

Output

 对于每组输入数据，能看见输出”Yes”,看不见输出”No”.

Rabbit不在食堂输出”Out Of Range”. Problem Description

 

 

Sample Input

5 0 0 1 1

5 0 0 2 0

5 0 0 6 6

5 0 0 -1 -1

0

 

 

Sample Output

Yes

No

Out Of Range

Yes

 

#include<iostream>

#include<stdio.h>

using namespace std;

int main()

{

    int x1,y1;

    int x2,y2;

    int n;

    while(~scanf("%d",&n))

    {

        if(n==0)

            break;

        scanf("%d %d %d %d",&x1,&y1,&x2,&y2);

        int t=0;

        if(x2>n||x2<-n||y2>n||y2<-n)

        {

            printf("Out Of Range\n");

        }

        else

        {

            if(x1==x2)

            {

                if(y1-y2==1||y2-y1==1)

                {

                    printf("Yes\n");

                }

                else

                {

                    printf("No\n");

                }

            }

             if(y1==y2)

            {

                if(x1-x2==1||x2-x1==1)

                {

                    printf("Yes\n");

                }

                else

                {

                    printf("No\n");

                }

            }

            if(x1!=x2&&y1!=y2)

            {

               int a,b,c;//a是横边长，b是纵边长

               if(x1>x2)

                a=x1-x2;

               else

                a=x2-x1;

                if(y1>y2)

                b=y1-y2;

               else

                b=y2-y1;

                c=a%b;

                while(c!=0)

                {

                    a=b;

                    b=c;

                    c=a%b;

                }

                if(b==1)//判断最大公约数是否是1

                    printf("Yes\n");

                else

                    printf("No\n");

 

            }

        }

    }

}