codechef Chef and The Right Triangles 题解

Chef and The Right Triangles

The Chef is given a list of N triangles. Each triangle is identfied by the coordinates of its three corners in the 2-D cartesian plane. His job is to figure out how many

of the given triangles are right triangles. A right triangle is a triangle in which one angle is a 90 degree angle. The vertices

of the triangles have integer coordinates and all the triangles given are valid( three points aren't colinear ).

Input

The first line of the input contains an integer N denoting the number of triangles. Each of the following N

lines contain six space separated integers x1 y1 x2 y2 x3 y3 where (x1, y1),

(x2, y2) and (x3, y3) are the vertices of a triangle.

Output

Output one integer, the number of right triangles among the given triangles.

Constraints

1 ≤ N ≤ 100000 (105)
0 ≤ x1, y1, x2, y2, x3, y3 ≤ 20

Example

Input:
5
0 5 19 5 0 0
17 19 12 16 19 0
5 14 6 13 8 7
0 4 0 14 3 14
0 2 0 14 9 2

Output:
3

推断是否是直角三角形，两种方法：

1 a*a + b*b = c*c

2 A dot B == 0 //dot是向量的点乘

重载操作符：

1 定义一个Point

2 定义Point的操作： 1） 减法 2） *乘号代表dot运算

#pragma once
#include <stdio.h>

class ChefandTheRightTriangles
{
struct Point
{
int x, y;
explicit Point(int a = 0, int b = 0): x(a), y(b) {}
Point operator-(const Point &p) const
{
return Point(x - p.x, y - p.y);
}
int operator*(const Point &p) const
{
return x * p.x + y * p.y;
}
};

int getInt()
{
char c = getchar();
while (c < '0' || '9' < c)
{
c = getchar();
}
int num = 0;
while ('0' <= c && c <= '9')
{
num = (num<<3) + (num<<1) + (c - '0');
c = getchar();
}
return num;
}
public:
ChefandTheRightTriangles()
{
int N = 0, C = 0;
N = getInt();
Point p1, p2, p3, v1, v2, v3;
while (N--)
{
p1.x = getInt(), p1.y = getInt();
p2.x = getInt(), p2.y = getInt();
p3.x = getInt(), p3.y = getInt();
v1 = p1 - p2, v2 = p2 - p3, v3 = p3 - p1;
if (v1 * v2 == 0 || v2 * v3 == 0 || v3 * v1 == 0)
C++;
}
printf("%d", C);
}
};

int chefandTheRightTriangles()
{
ChefandTheRightTriangles();
return 0;
}