浙江工商大学23调剂（正方形）

题解：

#include <iostream>
#include <vector>
#include <set>
#include <algorithm>
using namespace std;

int main() {
    vector<pair<int, int>> points(4);
    for (int i = 0; i < 4; ++i) {
        int x, y;
        cin >> x >> y;
        points[i] = {x, y};
    }
    
    // 检查是否有重复点
    set<pair<int, int>> unique_points(points.begin(), points.end());
    if (unique_points.size() != 4) {
        cout << "N" << endl;
        return 0;
    }
    
    // 计算所有六对点的距离平方
    vector<long long> distSq;
    for (int i = 0; i < 4; ++i) {
        for (int j = i + 1; j < 4; ++j) {
            long long dx = points[i].first - points[j].first;
            long long dy = points[i].second - points[j].second;
            distSq.push_back(dx * dx + dy * dy);
        }
    }
    sort(distSq.begin(), distSq.end());
    
    // 判断条件：前四个相等，后两个相等且是前者的两倍
    if (distSq[0] == distSq[1] && distSq[1] == distSq[2] && distSq[2] == distSq[3] &&
        distSq[4] == distSq[5] && distSq[4] == 2 * distSq[0]) {
        cout << "Y" << endl;
    } else {
        cout << "N" << endl;
    }
    
    return 0;
}