题意分析

题目给出一个平行四边形两个相邻边的四个坐标，要求求出第四个顶点的坐标。

解题思路

因为是两个相邻边的四个坐标，因此必定有两个坐标是重复的。假设这个坐标是A，其余两个已知坐标是B和C。那么第四个顶点坐标就是$D = B+C-A$（向量表示）

标程

#include <iostream>
#include <sstream>
#include <cmath>
#include <iomanip>
#include <string>
using namespace std;

bool points_equal(double x1, double y1, double x2, double y2) {
    return fabs(x1 - x2) < 1e-5 && fabs(y1 - y2) < 1e-5;
}

double adjust(double val) {
    double r = round(val * 1000.0) / 1000.0;  // 四舍五入到毫米
    if (fabs(r) < 0.0005) return 0.0;         // 消除-0.000
    return r;
}

int main() {
    string line;
    while (getline(cin, line)) {
        if (line.empty()) continue;
        istringstream iss(line);
        double x1, y1, x2, y2, x3, y3, x4, y4;
        iss >> x1 >> y1 >> x2 >> y2 >> x3 >> y3 >> x4 >> y4;
        
        double x, y; // 第四个顶点坐标
        
        if (points_equal(x1, y1, x3, y3)) {
            x = x2 + x4 - x1;
            y = y2 + y4 - y1;
        } else if (points_equal(x1, y1, x4, y4)) {
            x = x2 + x3 - x1;
            y = y2 + y3 - y1;
        } else if (points_equal(x2, y2, x3, y3)) {
            x = x1 + x4 - x2;
            y = y1 + y4 - y2;
        } else { // points_equal(x2, y2, x4, y4)
            x = x1 + x3 - x2;
            y = y1 + y3 - y2;
        }

        x = adjust(x);
        y = adjust(y);
        cout << fixed << setprecision(3) << x << " " << y << endl;
    }
    return 0;
}