牛客多校(2020第三场)C Operation Love

题目链接： https://ac.nowcoder.com/acm/contest/5668/C

题意：给了右手手掌的形状，然后给你20个连续的点（可顺可逆），问是左手还是右手

题解：

　　寻找最长边的那俩个点，和紧挨它的短的那条边的点，共三个点，求叉积判断左右手

#include<iostream>
using namespace std;

const int N = 30;
double x[N], y[N];

double cal(double x1, double y1, double x2, double y2) { //计算俩点距离的平方
    return (x2 - x1) * (x2 - x1) + (y2 - y1) * (y2 - y1);
}

int main() {
    int t;
    scanf("%d", & t);
    while (t--) {
        for (int i = 1; i <= 20; i++) {
            scanf("%lf %lf", &x[i], &y[i]);
        }  
        x[0] = x[20], y[0] = y[20];
        x[21] = x[1], y[21] = y[1];
        x[22] = x[2], y[22] = y[2]; //若不懂这三行，且先别急，看完33-39就明白了

        double max_len = -1;

        int pos1 = 0, pos2 = 0; //存最长边10的俩个端点
        for (int i = 1; i <= 20; i++) {
            double temp = cal(x[i], y[i], x[i+1], y[i+1]);
            if (max_len < temp) {
                max_len = temp;
                pos1 = i, pos2 = i+1; //记录下最长边的俩端
            }
        }

        int l, mid, r;
        if (cal(x[pos1], y[pos1], x[pos1 - 1], y[pos1 -1]) < cal(x[pos2], y[pos2], x[pos2+1], y[pos2+1])) {
            l = pos1 - 1, mid = pos1, r = pos2;
        }
        else {
            l = pos2 + 1, mid = pos2, r = pos1;
        }

        if (((x[r] - x[mid]) * (y[l] -y[mid]) - (x[l] - x[mid]) * (y[r] - y[mid])) < 0) {
            printf("left
");
        }
        else {
            printf("right
");
        }
    }
    
    return 0;
}