CodeForces 552D-Vanya and Triangles【计算整数三点能否组成三角形】

D. Vanya and Triangles

time limit per test

4 seconds

memory limit per test

512 megabytes

input

standard input

output

standard output

Vanya got bored and he painted n distinct points on the plane. After that he connected all the points pairwise and saw that as a result many triangles were formed with vertices in the painted points. He asks you to count the number of the formed triangles with the non-zero area.

Input

The first line contains integer n (1 ≤ n ≤ 2000) — the number of the points painted on the plane.

Next n lines contain two integers each xi, yi ( - 100 ≤ xi, yi ≤ 100) — the coordinates of the i-th point. It is guaranteed that no two given points coincide.

Output

In the first line print an integer — the number of triangles with the non-zero area among the painted points.

Examples

input

4
0 0
1 1
2 0
2 2

output

3

input

3
0 0
1 1
2 0

output

1

input

1
1 1

output

0

Note

Note to the first sample test. There are 3 triangles formed: (0, 0) - (1, 1) - (2, 0); (0, 0) - (2, 2) - (2, 0); (1, 1) - (2, 2) - (2, 0).

Note to the second sample test. There is 1 triangle formed: (0, 0) - (1, 1) - (2, 0).

Note to the third sample test. A single point doesn't form a single triangle.

解题思路：

就是给出一些点看能找出多少三角形。因为是整数点，所以很好判。

暴力可以过。

#include<stdio.h>
#include<string.h>
#include<cmath>
#include<algorithm>
using namespace std;
const double eps=1e-8;
struct node
{
	int x,y;
}num[1000005];
/*bool cc(node p1,node p2,node p3)
{
	  if(p2.x>=min(p1.x,p3.x)&&p2.x<=max(p1.x,p3.x)
	  &&p2.y>=min(p1.y,p3.y)&&p2.y<=max(p1.y,p3.y))
	  {
		return fabs((p2.x-p1.x)*1.0*(p3.y-p1.y)-(p3.x-p1.x)*1.0*(p2.y-p1.y))<=eps;
	  }
}*/
bool cc(int i,int j,int k)
{
	if((num[i].y-num[j].y)*(num[k].x-num[i].x)!=(num[k].y-num[i].y)*(num[i].x-num[j].x))
	return 1;
	return 0;
}
int main()
{
	int n;
	while(scanf("%d",&n)!=EOF)
	{
		int i,j,k;
		for(i=1;i<=n;i++)
		{
			scanf("%d%d",&num[i].x,&num[i].y);
		}
		if(n<=2)
		{
			printf("0\n");
			continue;
		}
		int ans=0;
		for(i=1;i<n-1;i++)
		{
			for(j=i+1;j<n;j++)
			{
				for(k=j+1;k<=n;k++)
				{
					if(cc(i,j,k)==1)
					ans++;
				}
			}
		}
		printf("%d\n",ans);
	}
	return 0;
}