Note
Note to the first sample test. There are3triangles 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 is1triangle formed:(0,0)-(1,1)-(2,0).
Note to the third sample test. A single point doesn’t form a single triangle.
题目要求是求出n个点所能组成的三角形的个数，我们知道，若都能组成三角形，则总个数为c(n,3)个，现在目标就是求出不能组成三角形的个数，则
可以这样求，以一个点为起点，其它点与这个点的斜率保存在map中，则对于每个则有c(m,2)个不能组成三角形，因为重复算了三次，最后要除以3，，
还要用long long 保存结果，保存斜率可以用分数a/b,这里a,b互质，分数可以用a*N+b一个整数保存起来，就没有误差了！
#define INF9000000000000000000#define EPS(double)1e-9#define mod1000000007#define PI3.14159265358979//*******************************************************************************/#endif#define N 2005#define MOD 1000000007int n;struct node{int x,y;node(int xx,int yy){x = xx,y = yy;}node(){x = 0;y=0;}};node nodes[N];map<int,int> mymaps;map<int,int>::iterator it;int get(int a,int b){//a/bif(a == 0) return 0;if(b == 0) return N * N + N;int sa = a>=0?1:-1,sb= b>=0?1:-1;sa = sa * sb;sb = 1;a = abs(a);b=abs(b);int g = gcd(a,b);a/=g;b/=g;return sa * a * N + sa * b;}int main(){//printf("%d",get(-1,2));while (S(n) != EOF){FI(n){S2(nodes[i].x,nodes[i].y);}long long ans = (long long)n*((long long)n-1)*((long long)n-2)/6,anst = 0;FI(n){mymaps.clear();FJ(n){if(i != j){mymaps[get(nodes[j].y – nodes[i].y,nodes[j].x – nodes[i].x)]++;}}for(it = mymaps.begin();it!=mymaps.end();it++){int num = it->second;anst += num>=2? num*(num-1)/2:0;}}cout<<ans – anst/3<<endl;}return 0;}
穷则思变，差则思勤！没有比人更高的山没有比脚更长的路。
