Codeforces Round #308 (Div. 2) D. Vanya and Triangles 水题

D. Vanya and Triangles

Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://codeforces.com/contest/552/problem/D

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

4
0 0
1 1
2 0
2 2

Sample Output

3

HINT

 

题意

平面上给你n个点，然后问你能构成多少个三角形

题解：

n^3暴力跑一法就过了……

是数据太水，还是我太屌？

代码：

 

#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define test freopen("test.txt","r",stdin)  
#define maxn 2000001
#define mod 10007
#define eps 1e-5
const int inf=0x3f3f3f3f;
const ll infll = 0x3f3f3f3f3f3f3f3fLL;
inline ll read()
{
    ll x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
//**************************************************************************************
#include<cstdio>
#include<cmath>
using namespace std;
struct point
{
    int x,y;
    friend int operator *(point A,point B){return A.x*B.y-A.y*B.x;}
    friend point operator -(point A,point B){point C;C.x=A.x-B.x;C.y=A.y-B.y;return C;}
}p[3100];
int ans,n;
int main()
{
    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-2;i++)
        for(int j=i+1;j<n;j++)
            for(int k=j+1;k<=n;k++)
                if((p[i]-p[j])*(p[j]-p[k]))ans++;
                    printf("%d",ans);
    return 0;
}