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

àì夳堔傛蜴生んèń 2022-08-21 06:49 9阅读 0赞

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 *x**i*, *y**i* ( - 100 ≤ *x**i*, *y**i* ≤ 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

input

3
    0 0
    1 1
    2 0

output

input

1
    1 1

output

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;
    }