Clockwise or Counterclockwise

思路：
AB(向量）X BC(向量），根据右手定则，可判断BC向量是在AB右侧，还是左侧

参考代码：

#include <cstdio>
#include <algorithm>
#include <iostream>
#include <vector>
#include <map>
#include <queue>
#include <set>
#include <ctime>
#include <cstring>
#include <cstdlib>
#include <math.h>
using namespace std;
typedef long long ll;
//#define ll long long
const ll N = 2009;
const ll maxn = 1e5 + 20;
const ll mod = 1000000007;
ll inv[maxn], vis[maxn], dis[maxn], head[maxn];
ll fac[maxn], a[maxn], q[maxn], b[N], c[N], pre[maxn];
vector<ll> vec;
//typedef pair<ll, ll> p;
//priority_queue<p, vector<p>, greater<p> > m;
// ll sum[maxn], a[maxn];
ll max(ll a, ll b) { return a > b ? a : b; }
ll min(ll a, ll b) { return a < b ? a : b; }
ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }
ll lcm(ll a, ll b) { return a * b / gcd(a, b); }
map<ll, ll> mp;
ll ksm(ll a, ll b)
{a %= mod;ll ans = 1ll;while (b){if (b & 1)ans = (ans * a) % mod;a = (a * a) % mod;b >>= 1ll;}return ans;
}
ll dp[maxn][2][2], cnt, num[2], ans;
struct node
{ll x, y;
} p[maxn];
int main()
{ios::sync_with_stdio(false);cin.tie(0);ll t, n, m, k;cin >> t;while (t--){ll xa, ya, xb, yb, xc, yc;cin >> xa >> ya >> xb >> yb >> xc >> yc;ll x1 = xb - xa, y1 = yb - ya;ll x2 = xc - xb, y2 = yc - yb;ll ans = x1 * y2 - x2 * y1;if (ans < 0)cout << "Clockwise" << endl;elsecout << "Counterclockwise" << endl;}
}

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