题解 | #直角三角形#

#include <iostream>
#include <cmath>
#include <algorithm>
#include <numeric>
#include <iomanip>
using namespace std;
float distance(int x1, int y1, int x2, int y2) {
    return sqrt(pow((x2 - x1), 2) + pow((y2 - y1), 2));
}
int main() {
    int m;
    cin >> m;
    while (m--) {
        int x1, y1, x2, y2, x3, y3;
        cin >> x1 >> y1 >> x2 >> y2 >> x3 >> y3;
        float d1 = distance(x1, y1, x2, y2);
        float d2 = distance(x2, y2, x3, y3);
        float d3 = distance(x3, y3, x1, y1);
        float d[] = {d1, d2, d3};
        sort(d, d + 3);
        if (abs(pow(d[0], 2) + pow(d[1], 2) - pow(d[2], 2)) < 1e-4) {
            cout << "Yes" << endl;
        } else {
            cout << "No" << endl;
        }
        cout << fixed << setprecision(2) << accumulate(d, d + 3, 0.0) << endl;
    }
    return 0;
}