ConvexScore
时间限制: 1 Sec 内存限制: 128 MB题目描述
You are given N points (xi,yi) located on a two-dimensional plane. Consider a subset S of the N points that forms a convex polygon. Here, we say a set of points S forms a convex polygon when there exists a convex polygon with a positive area that has the same set of vertices as S. All the interior angles of the polygon must be strictly less than 180°. 
For example, in the figure above, {A,C,E} and {B,D,E} form convex polygons; {A,C,D,E}, {A,B,C,E}, {A,B,C}, {D,E} and {} do not. For a given set S, let n be the number of the points among the N points that are inside the convex hull of S (including the boundary and vertices). Then, we will define the score of S as 2n−|S|. Compute the scores of all possible sets S that form convex polygons, and find the sum of all those scores. However, since the sum can be extremely large, print the sum modulo 998244353. Constraints 1≤N≤200 0≤xi,yi<104(1≤i≤N) If i≠j, xi≠xj or yi≠yj. xi and yi are integers.
输入
The input is given from Standard Input in the following format: N x1 y1 x2 y2 : xN yN
输出
Print the sum of all the scores modulo 998244353.
样例输入
40 00 11 01 1
样例输出
5
提示
We have five possible sets as S, four sets that form triangles and one set that forms a square. Each of them has a score of 20=1, so the answer is 5.
题意：给定 N 个点，对于一个凸 n 边形，称其的 n 个顶点构成一个集合 S，并且这个多边形内及其边上有 k 个顶点，定义这个 S 的 score为 2^(k-n)
对所有的 score 求和，输出 mod 998244353 的值。
题解：可以转化为固定某个凸包，在里面或边上添加或不添加点的方案数
即所有的可能 - 只有一个点 - 空集 - 共线的情况
共线的情况 可以枚举 确定某两点，枚举第三个点（0(n**3)），共线用向量的点乘判断
c++ code:
#includeusing namespace std;typedef long long ll;const int N = 2000+10;const int mod = 998244353;int x[N],y[N];ll f[N];int cal(int i ,int j,int k){ return (x[j] - x[i])*(y[k] - y[j]) == (y[j] - y[i])*(x[k] - x[j]);}int main(){ for(int i = 0;i < N;i++) f[i] = i == 0?1:(f[i-1]<<1)%mod; int n; scanf("%d",&n); for(int i = 1;i <= n ;i++) scanf("%d%d",&x[i],&y[i]); ll ans = f[n] - n - 1; for(int i = 1;i <= n;i++) for(int j = i+1;j <= n;j++) { int tot = 0; for(int k = j+1;k <= n;k++) if(cal(i,j,k)) tot++; // 共线 ans = (ans - f[tot] + mod )%mod; } printf("%lld\n",ans); return 0;}