原题:
To calculate the circumference of a circle seems to be an easy task - provided you know its diameter. But what if you don't?
You are given the cartesian coordinates of three non-collinear points in the plane.
Your job is to calculate the circumference of the unique circle that intersects all three points.
题意:
给出三角形三个顶点的坐标,求出这个三角形外接圆的周长。
题解:
本来是想先求出圆心,然后再求半径,就是用直线方程联立求圆心,写完之后发现忘了还有斜率不存在的情况,太久不做数学题了......那就直接用三角形三条边长公式来求外接圆半径
代码:AC
#include <iostream>
#include <iomanip>
#include <stdio.h>
#include <math.h>
using namespace std;
const double PI = acos(-1.0);
int main(void)
{
double x1, y1, x2, y2, x3, y3;
while(~scanf("%lf%lf%lf%lf%lf%lf", &x1, &y1, &x2, &y2, &x3, &y3)) {
double a = sqrt((x3 - x2) * (x3 - x2) + (y3 - y2) * (y3 - y2));
double b = sqrt((x1 - x3) * (x1 - x3) + (y1 - y3) * (y1 - y3));
double c = sqrt((x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2));
double p = (a + b + c) * 0.5;
double s = sqrt(p * (p - a) * (p - b) * (p - c));
double d = a * b * c / (2.0 * s);
cout << setiosflags(ios::fixed) << setprecision(2) << d * PI << endl;
}
return 0;
}