SZTUOJ 1121

Posted 2021-06-11 是阿夢呀

SZTUOJ 1013 - The Area of a Sector

Description

Given a circle and two points on it, calculate the area of the sector with its central angle no more than 180 degrees.

Input

There are multiple test cases.
Each line contains 6 float numbers denote the center of the circle xc, yc, the two points on the circle x1, y1 and x2, y2.
1 <= xc, yc, x1, y1, x2, y2 <= 10000.

Output

The area of the sector. The result should be accurated to three decimal places.

Sample Input

0 0 1 1 -1 1
2.5 2.5 3.25 3.5 3.5 3.25

Sample Output

1.571
0.222

Hint

With math.h, you could define pi (3.1415926...) as const double pi = acos(-1.0);
Inverse trigonometric function could be obtained by acos, atan, asin, etc..
Be careful that angles like pi in C/C++ are in 3.1415926... type rather than 180.
You’d better use double instead of float. Load double with "%lf" but output it with "%f".

Source

2021 SZTU AI Class Qualifying.

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思路分析

这道题来自2021伦琴AI班选拔机试C题，当时只有包括我在内的两个人做了出来。但题目本身并不难，根据描述按步编写即可。

本题需要求解扇形面积。根据圆的面积公式，我们可以推导出半径为\\(r\\)，圆心角为\\(\\theta\\)的扇形面积公式：

\\[S_\\theta=\\frac{\\theta}{2}r^2\\ (0<\\theta\\leq2\\pi) \\]

根据上述公式，我们了解到，若想求解扇形面积，需要先求出扇形半径\\(r\\)和圆心角\\(\\theta\\).

根据输入数据\\(O(x_0,y_0),P_1(x_1,y_1),P_2(x_2,y_2)\\)，我们可以计算出半径和弦长的欧氏距离：

\\[r=\\sqrt{(x_0-x_1)^2+(y_0-y_1)^2}\\\\ l=\\sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \\]

作\\(\\triangle OP_1P_2\\)的中垂线，根据几何性质，有：

\\[\\sin{\\frac{1}{2}\\angle O}=\\sin{\\frac{\\theta}{2}}=\\frac{l}{2r} \\]

即：

\\[\\theta=2\\arcsin{\\frac{l}{2r}} \\]

至此，求解面积所需的全部参量已经求出，带入扇形面积公式即可得解。

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代码实现

// Language: C
#include<stdio.h>
#include<math.h>

typedef struct Point{
    double x;
    double y;
}point;

int main() {
    point o, p1, p2;
    double ox, oy, x1, y1, x2, y2;
    while (scanf("%lf%lf%lf%lf%lf%lf", &o.x, &o.y, &p1.x, &p1.y, &p2.x, &p2.y) != EOF) {
        double r = hypot(p1.x - o.x, p1.y - o.y);
        double l = hypot(p1.x - p2.x, p1.y - p2.y);
        double theta = 2 * asin(0.5 * l / r);
        printf("%.3f\\n", r * r * theta / 2);
    }
    return 0;
}