给出点集，然后求一个凸包的所有的子凸包的贡献总和，贡献计算是凸包内部含边界上点的数量N，凸包的不包含边界的顶点数S，贡献为$2^{N-S}$
首先很容易想到，凸包上包含内部的所有点构成的子凸包有Sum(i = 3 ->N)C(i,N)种情况，这个式子其实就是二项式的一部分。但是有可能出现多点共线的不合法情况，所以问题转换为求所有点构成的直线中，每条直线上大于2点的点的数目，每条直线都要分别计算，最后减去就行了。求共线可以用叉积可以用斜率，注意判重。
这场比赛迟了10分钟才写，这题开始还在用凸包搞，简直蠢(
/** @Date : 2017-09-02 20:30:47
* @FileName: C.cpp
* @Platform: Windows
* @Author : Lweleth (SoungEarlf@gmail.com)
* @Link : https://github.com/
* @Version : $Id$
*/
#include <bits/stdc++.h>
#define LL long long
#define PII pair<int ,int>
#define MP(x, y) make_pair((x),(y))
#define fi first
#define se second
#define PB(x) push_back((x))
#define MMG(x) memset((x), -1,sizeof(x))
#define MMF(x) memset((x),0,sizeof(x))
#define MMI(x) memset((x), INF, sizeof(x))
using namespace std;
const int INF = 0x3f3f3f3f;
const int N = 210;
const double eps = 1e-6;
const LL mod = 998244353;
LL fa[210], inv[210];
LL fpow(LL a, LL n)
{
LL r = 1LL;
while(n > 0)
{
if(n & 1)
r = r * a % mod;
a = a * a % mod;
n >>= 1;
}
return r;
}
void init()
{
fa[0] = 1;
inv[0] = 1;
for(LL i = 1; i <= 200; i++)
{
fa[i] = fa[i-1] * i % mod;
inv[i] = fpow(fa[i], mod - 2);
}
}
LL C(LL n, LL m)
{
if(n < 0)
return 0;
n >>= 1;
if(n == 0)
return 1LL;
LL ans = 0;
ans = ((fa[n + m] * inv[m] % mod)* inv[n]) % mod;
return ans;
}
struct point
{
double x, y;
point(){}
point(double _x, double _y){x = _x, y = _y;}
point operator -(const point &b) const
{
return point(x - b.x, y - b.y);
}
double operator *(const point &b) const
{
return x * b.x + y * b.y;
}
double operator ^(const point &b) const
{
return x * b.y - y * b.x;
}
bool operator == (const point &b) const
{
return x==b.x && y==b.y;
}
};
double xmult(point p1, point p2, point p0)
{
return (p1 - p0) ^ (p2 - p0);
}
double distc(point a, point b)
{
return sqrt((double)((b - a) * (b - a)));
}
int sign(double x)
{
if(fabs(x) < eps)
return 0;
if(x < 0)
return -1;
else
return 1;
}
struct line
{
point s, t;
line(){}
line(point ss, point tt){
s = ss, t = tt;
}
};
////////
int n;
point stk[N];
point p[N];
int cmpC(point a, point b)//水平序排序
{
return sign(a.x - b.x) < 0 || (sign(a.x - b.x) == 0 && sign(a.y - b.y) < 0);
}
int Graham()//水平序
{
sort(p, p + n, cmpC);
int top = 0;
for(int i = 0; i < n; i++)
{
while(top >= 2 && sign(xmult(stk[top - 2], stk[top - 1], p[i])) < 0)
top--;
stk[top++] = p[i];
}
int tmp = top;
for(int i = n - 2; i >= 0; i--)
{
while(top > tmp && sign(xmult(stk[top - 2],stk[top - 1] ,p[i] )) < 0)
top--;
stk[top++] = p[i];
}
if(n > 1)
top--;
return top;
}
LL check(int m)
{
//cout << m << endl;
LL c = 2;
LL t = 0;
for(int i = 1; i < m; i++)
{
if(sign(xmult(stk[i - 1], stk[(i + 1)%(m)], stk[i])) == 0)
c++;
else t = (t + fpow(2, c) - (1LL + c + c * (c - 1) / 2LL) + mod) % mod, c = 2;
//cout << c << endl;
}
if(c > 2)
t = (t + fpow(2, c) - (1LL + c + c * (c - 1) / 2LL) + mod) % mod;
return t;
}
/////////
int main()
{
while(~scanf("%d", &n))
{
for(int i = 0; i < n; i++)
{
double x, y;
scanf("%lf%lf", &x, &y);
p[i] = point(x, y);
}
LL ans = 0;
LL cnt = Graham();
//cout << cnt;
//ans = (fpow(2, n) - check(cnt) - (1LL + n + (n - 1) * n / 2LL) + mod) % mod;
ans = (fpow(2, n) - (1LL + n) + mod) % mod;
for(int i = 0; i < n; i++)
{
map<LL, int>q;
for(int j = i + 1; j < n; j++)
{
LL t;
if(p[i].x == p[j].x)
t = -1;
else t = ((LL)(p[j].y - p[i].y) * fpow(p[j].x - p[i].x, mod - 2) % mod + mod ) % mod;
q[t]++;
}
for(auto j : q)
{
ans -= fpow(2, j.se) - 1;
ans %= mod;
}
}
while(ans < 0)
ans += mod;
if(cnt > 2)
printf("%lld
", ans);
else printf("0
");
}
return 0;
}