CodeForces 18A Triangle好暴力啊→_→

Description

At a geometry lesson Bob learnt that a triangle is called right-angled if it is nondegenerate and one of its angles is right. Bob decided to draw such a triangle immediately: on a sheet of paper he drew three points with integer coordinates, and joined them with segments of straight lines, then he showed the triangle to Peter. Peter said that Bob's triangle is not right-angled, but is almost right-angled: the triangle itself is not right-angled, but it is possible to move one of the points exactly by distance 1 so, that all the coordinates remain integer, and the triangle become right-angled. Bob asks you to help him and find out if Peter tricks him. By the given coordinates of the triangle you should find out if it is right-angled, almost right-angled, or neither of these.

Input

The first input line contains 6 space-separated integers x1, y1, x2, y2, x3, y3 — coordinates of the triangle's vertices. All the coordinates are integer and don't exceed 100 in absolute value. It's guaranteed that the triangle is nondegenerate, i.e. its total area is not zero.

Output

If the given triangle is right-angled, output RIGHT, if it is almost right-angled, output ALMOST, and if it is neither of these, outputNEITHER.

Sample Input

Input
0 0 2 0 0 1
Output
RIGHT
Input
2 3 4 5 6 6
Output
NEITHER
Input
-1 0 2 0 0 1
Output
ALMOST
#include <iostream>
#include<cstdio>

using namespace std;
struct po
{
    int x,y;
}point[4];
bool judge()
{
    if(((point[0].x-point[1].x)*(point[0].x-point[1].x)+(point[0].y-point[1].y)*(point[0].y-point[1].y))&&((point[0].x-point[2].x)*(point[0].x-point[2].x)+(point[0].y-point[2].y)*(point[0].y-point[2].y))&&((point[1].x-point[2].x)*(point[1].x-point[2].x)+(point[1].y-point[2].y)*(point[1].y-point[2].y)))
    return true;
    return false;
}//就是这里没写WA了两次T^T 自己能找到错误也是很开心的
int cos(po a,po b,po c)
{
    return (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y)+(a.x-c.x)*(a.x-c.x)+(a.y-c.y)*(a.y-c.y)-(b.x-c.x)*(b.x-c.x)-(b.y-c.y)*(b.y-c.y);
}
int main()
{
  //  freopen("cin.txt","r",stdin);
    while(~scanf("%d%d%d%d%d%d",&point[0].x,&point[0].y,&point[1].x,&point[1].y,&point[2].x,&point[2].y))
    {
        if(!cos(point[0],point[1],point[2])||!cos(point[1],point[2],point[0])||!cos(point[2],point[0],point[1]))
        {
            if(judge())
            {printf("RIGHT\n");
            continue;}
        }
        bool mark=0;
        for(int i=0;i<3;i++)
        {
            point[i].x++;
            if(!cos(point[0],point[1],point[2])||!cos(point[1],point[2],point[0])||!cos(point[2],point[0],point[1]))
            {
               if(judge()){
                    mark=1;
                    break;
                }
            }
            point[i].x--;point[i].x--;
            if(!cos(point[0],point[1],point[2])||!cos(point[1],point[2],point[0])||!cos(point[2],point[0],point[1]))
            {
                if(judge()){
                    mark=1;
                    break;
                }
            }
            point[i].x++;point[i].y++;
            if(!cos(point[0],point[1],point[2])||!cos(point[1],point[2],point[0])||!cos(point[2],point[0],point[1]))
            {
                if(judge()){
                    mark=1;
                    break;
                }
            }
            point[i].y--;point[i].y--;
            if(!cos(point[0],point[1],point[2])||!cos(point[1],point[2],point[0])||!cos(point[2],point[0],point[1]))
            {
               if(judge()){
                    mark=1;
                    break;
                }
            }
            point[i].y++;
        }
        if(!mark) printf("NEITHER\n");
        if(mark) printf("ALMOST\n");
    }
    return 0;
}

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05-10 16:48
门头沟学院 Java
程序员小白条:主要原因,投递太晚了,快手应该早点溜了,你都从去年9月开始的,半年也差不多3月跑路了,这样的话,至少有5个以上的面试机会
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