python/TangDou/AoSaiYiBenTong/YBT1034.cpp

#include <bits/stdc++.h>

using namespace std;

int main() {
    //cin读入优化
    std::ios::sync_with_stdio(false);

    float x1, y1, x2, y2, x3, y3;
    cin >> x1 >> y1 >> x2 >> y2 >> x3 >> y3;

    //求三条边的长度，然后再用海伦公式进行求解
    double a, b, c;
    a = sqrt((x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2));
    b = sqrt((x1 - x3) * (x1 - x3) + (y1 - y3) * (y1 - y3));
    c = sqrt((x3 - x2) * (x3 - x2) + (y3 - y2) * (y3 - y2));

    double p = (a + b + c) / 2;
    //假设在平面内,有一个三角形,边长分别为a、b、c
    //三角形的面积S可由以下公式求得：
    //S=√[p(p-a)(p-b)(p-c)]
    //而公式里的p为半周长：p=(a+b+c)/2
    cout << fixed << setprecision(2) << sqrt(p * (p - a) * (p - b) * (p - c)) << endl;
    return 0;
}