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
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 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.
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;
}