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SZTUOJ 1121 – The Area of a Sector

开发技术 开发技术 5天前 7次浏览

SZTUOJ 1013 – The Area of a Sec++tor

Desc++ription

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”.

Sourc++e

2021 SZTU AI Class Qualifying.


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<thetaleq2pi)
]

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}
]

(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=2arcsin{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。


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("%.3fn", r * r * theta / 2);
    }
    return 0;
}

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