Codeforces Round #308 (Div. 2) D. Vanya and Triangles （判断三角形数量）

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

30 01 12 0

output

1

input

11 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.

题意：

给出n个点，求将n个点相互连线之后，可组成多少个三角形。

分析：

本题时间给的太宽裕，n^3的暴力枚举加快速判断是否能形成三角形就可以解决，不过每个三角形都计算过3次，所以最终结果需要除以3。

更快速的方法是n^2的逐个按点枚举，将斜率存储下来。若该斜率大于1的话，证明该直线上存在2个以上的点，通过计算该斜率出现多少次，即可知道重复计算了哪些三角形。而n个点最多可形成n*(n-1)*(n-2)/3!个三角形，将重复计算的三角形除去即可。为了防止精度问题，将斜率的分子分母约分后，分别存储即可，本题采用了分子+分母*1000，作为该斜率的存储。

#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cstdlib>
#include <string>
#include <vector>
#include <cstdio>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
using namespace std;
#define INF 0x3f3f3f3f
const int N=2015;
const int mod=1e9+7;

struct node {
int x,y;
}p
;

int x[N*100];
long long sum=0;

int gcd(int a,int b) {
return b==0?a:gcd(b, a%b);
}

bool cmp(node a,node b) {
return a.x>b.x;
}

int solve(int i,int j) {
int xx=p[i].x-p[j].x;
int yy=p[i].y-p[j].y;
if (xx<0) {
xx*=-1;
yy*=-1;
}
if (xx==0) {
yy=abs(yy);
}
int t=gcd(xx, abs(yy));
xx/=t;
yy/=t;
yy+=200;

return xx*1000+yy;
}

int main() {
int n;
while (cin>>n) {
for (int i=0; i<n; i++) {
scanf("%d %d",&p[i].x,&p[i].y);
}
sort(p, p+n, cmp);
sum=(long long)n*(n-1)*(n-2);
long long temp=0;
for (int i=0; i<n; i++) {
temp=0;
memset(x, 0, sizeof(x));
for (int j=0; j<n; j++) {
if (i==j) continue;
x[solve(i, j)]++;
}

for (int j=0; j<n; j++) {
if (i==j) continue;
int flag=x[solve(i, j)]+1;
x[solve(i, j)]=0;
if (flag>2) {
temp+=(flag-1)*(flag-2);
}
}
sum-=temp;
}
sum/=6;
cout<<sum<<endl;
}
return 0;
}