标题:三角形面积
已知三角形三个顶点在直角坐标系下的坐标分别为:
(2.3, 2.5)
(6.4, 3.1)
(5.1, 7.2)
求该三角形的面积。
注意,要提交的是一个小数形式表示的浮点数。
要求精确到小数后3位,如不足3位,需要补零。
利用叉积计算三角形面积即可。
// #pragma GCC optimize(2)
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <cmath>
#include <string>
#include <vector>
#include <stack>
#include <map>
#include <sstream>
#include <cstring>
#include <set>
#include <cctype>
#include <bitset>
#define IO \
ios::sync_with_stdio(false); \
// cout.tie(0);
using namespace std;
// int dis[8][2] = {0, 1, 1, 0, 0, -1, -1, 0, 1, -1, 1, 1, -1, 1, -1, -1};
typedef unsigned long long ULL;
typedef long long LL;
typedef pair<int, int> P;
const int maxn = 1e5 + 10;
const int maxm = 2e5 + 10;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int inf = 0x3f3f3f3f;
const LL mod = 1e9 + 7;
const double eps = 1e-8;
const double pi = acos(-1);
//int dis[4][2] = {1, 0, 0, -1, 0, 1, -1, 0};
int dis[2][2] = {1, 0, 0, 1};
//int m[13] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
int cmp(double x)
{
if (fabs(x) < eps)
return 0;
if (x > 0)
return 1;
return -1;
}
inline double sqr(double x)
{
return x * x;
}
struct Point
{
double x, y;
Point() {}
Point(double a, double b) : x(a), y(b) {}
friend Point operator+(const Point &a, const Point &b)
{
return Point(a.x + b.x, a.y + b.y);
}
friend Point operator-(const Point &a, const Point &b)
{
return Point(a.x - b.x, a.y - b.y);
}
friend bool operator==(const Point &a, const Point &b)
{
return cmp(a.x - b.x) == 0 && cmp(a.y - b.y) == 0;
}
friend Point operator*(const double &a, const Point &b)
{
return Point(a * b.x, a * b.y);
}
friend Point operator*(const Point &a, const Point &b)
{
return Point(a.x * b.x, a.y * b.y);
}
friend Point operator/(const Point &a, const double &b)
{
return Point(a.x / b, a.y / b);
}
double norm()
{
return sqrt(sqr(x) + sqr(y));
}
} a, b, c;
double dist(const Point &a, const Point &b) // ?????
{
return (a - b).norm();
}
double dot(const Point &a, const Point &b) // ??
{
return a.x * b.x + a.y * b.y;
}
double det(const Point &a, const Point &b) // ??
{
return a.x * b.y - a.y * b.x;
}
int main()
{
#ifdef WXY
freopen("in.txt", "r", stdin);
// freopen("out.txt", "w", stdout);
#endif
IO;
double x, y;
cin >> x >> y;
a = Point(x, y);
cin >> x >> y;
b = Point(x, y);
cin >> x >> y;
c = Point(x, y);
double s = det(c - a, c - b) / 2.0;
printf("%.3lf", s);
return 0;
}