Vanya and Triangles 暴力枚举

枚举法：

枚举法是利用计算机速度快， 精度高的特点， 对要解决的问题所有可能情况进行霸道的， 一个不漏检验， 从中找出符合要求的答案。

特点：

1. 得到的结果一定正确。

2. 可能做了很多无用功，效率低下。

3. 一般会涉及到极值。

4. 数据量大的话可能造成时间崩溃。

结构：

循环结构。

基本思路：

1. 确定枚举对象， 枚举范围， 判定条件。

2. 枚举可能的解， 验证是否是问题的解。

Vanya and Triangles ：

M - Vanya and Triangles

Crawling in process... Crawling failed Time Limit:4000MS     Memory Limit:524288KB     64bit IO Format:%I64d & %I64u

SubmitStatus

Description

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.

Sample Input

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

#include<stdio.h>

struct point{
    int x, y;
}p[3100];

int main()
{
    int n, ans = 0;
    scanf("%d", &n);
    for(int i = 1; i <= n; i++)
        scanf("%d%d", &p[i].x, &p[i].y);
    for(int i = 1; i < n - 1; i++)
        for(int j = i + 1; j < n; j++)
            for(int k = j + 1; k <=n; k++)
            if((p[i].y - p[j].y)*(p[k].x - p[i].x) != (p[k].y - p[i].y)*(p[i].x - p[j].x))
            ans++;
            printf("%d\n", ans);
    return 0;
}