枚举直线，对于直线的某个点在直线的左端还是右端，能够状压出一个数。用bitset记录。

然后三角形就是3个bitset&一下

#include <cstdio>
#include <cstring>
#include <bitset>
#include <algorithm>
using namespace std;
typedef long long ll;
const int N = 101;
const int M = 1005;
bitset<M> b1[N*N], b2[N*N];
int x[N], y[N], px[M], py[M];
bool check(int i, int j, int k) {if(x[i] != x[j]) {double yy = (double)(y[i] - y[j]) / (x[i] - x[j]) * (x[k] - x[i]) + y[i];if(y[k] >= yy) return true;else return false;} else {if(x[k] >= x[i])return true;else return false;}
}
void put(bitset<M> x) {for(int i = 0; i < M; i ++) {if(x[i]) printf("%d ", i);}puts("*");
}
int main() {int n, m, cas = 0;while(~scanf("%d%d", &n, &m)) {for(int i = 0; i < n; i ++) {scanf("%d%d", &x[i], &y[i]);}for(int i = 0; i < m; i ++) {scanf("%d%d", &px[i], &py[i]);}for(int i = 0; i < n; i ++) {for(int j = i+1; j < n; j ++) {for(int k = 0; k < m; k ++) {//        printf("%d %d  ", i, j);if(x[i] != x[j]) {double yy = (double)(y[i] - y[j]) / (x[i] - x[j]) * (px[k] - x[i]) + y[i];if(py[k] == yy) {b1[i*n+j].set(k);b2[i*n+j].set(k);}else if(py[k] > yy) {b1[i*n+j].set(k);//               printf("u1-%d", k);} else {b2[i*n+j].set(k);//               printf("d1-%d", k);}} else {if(px[k] == x[i]) {b1[i*n+j].set(k);b2[i*n+j].set(k);}else if(px[k] > x[i])  {b1[i*n+j].set(k);//             printf("u2-%d", k);} else {b2[i*n+j].set(k);//               printf("d2-%d", k);}}//       puts("");}//          printf("  %d %d %d %d\n", i, j, b1[i*n+j].count(), b2[i*n+j].count());}}bitset<M> tmp(0);int ans = 0;for(int i = 0; i < n; i ++) {for(int j = i+1; j < n; j ++) {for(int k = j+1; k < n; k ++) {if(check(i, j, k)) {tmp = b1[i*n+j];//                 printf("UU1 ");//                 put(b1[i*n+j]);}else {tmp = b2[i*n+j];//             put(b2[i*n+j]);}if(check(i, k, j)) {tmp &= b1[i*n+k];//                  printf("UU2 ");//             put(b1[i*n+k]);}else {tmp &= b2[i*n+k];//                    put(b2[i*n+k]);}if(check(j, k, i)) {tmp &= b1[j*n+k];//              printf("UU3 ");//         put(b1[j*n+k]);}else {tmp &= b2[j*n+k];//                    put(b2[j*n+k]);}//         printf("***%d %d %d %d\n", i, j, k, tmp.count());if(tmp.count() & 1) ans ++;}}}printf("Case %d: %d\n", ++cas, ans);for(int i = 0; i < n*n; i ++) {b1[i].reset();b2[i].reset();}}return 0;
}/*
3 1
0 0
0 100
100 0
0 03 1
0 0
0 100
100 0
50 503 1
0 0
0 100
100 0
100 03 1
0 0
0 100
100 0
0 -14 4
0 0
0 100
100 0
-2 500 0
0 100
100 0
-1 50*/