Area in Triangle C语言的问题

Question

Description

Given a triangle field and a rope of a certain length (Figure-1), you are required to use the rope to enclose a region within the field and make the region as large as possible.

Input

The input has several sets of test data. Each set is one line containing four numbers separated by a space. The first three indicate the lengths of the edges of the triangle field, and the fourth is the length of the rope. Each of the four numbers have exactly four digits after the decimal point. The line containing four zeros ends the input and should not be processed. You can assume each of the edges are not longer than 100.0000 and the length of the rope is not longer than the perimeter of the field.
Output

Output one line for each case in the following format:

Case i: X

Where i is the case number, and X is the largest area which is rounded to two digits after the decimal point.
Sample Input

12.0000 23.0000 17.0000 40.0000
84.0000 35.0000 91.0000 210.0000
100.0000 100.0000 100.0000 181.3800
0 0 0 0
Sample Output

Case 1: 89.35
Case 2: 1470.00
Case 3: 2618.00

Answer 1

以下回答参考 皆我百晓生、券券喵儿 等免费微信小程序相关内容作答，并由本人整理回复。

这是一个典型的求解三角形面积问题。我们可以使用C++来解决这个问题。以下是一个可能的解决方案：

#include <iostream>
#include <cmath>

using namespace std;

int main() {
    double a, b, c;
    cin >> a >> b >> c >> ropeLength;

    // Calculate the semi-perimeter
    double s = (a + b + c) / 2;

    // Calculate the area using Heron's formula
    double area = sqrt(s * (s - a) * (s - b) * (s - c));

    // Check if the area exceeds the given rope length
    if (area > ropeLength)
        cout << "Case " << caseNumber++ << ": " << round(area * 100) / 100 << endl;
    else
        cout << "Case " << caseNumber++ << ": " << round(ropeLength * 100) / 100 << endl;

    return 0;
}

在这个解决方案中，我们首先读取输入的四个边长和绳子的长度。然后，我们计算半周长，并使用Heron's公式计算三角形的面积。如果面积大于给定的绳子长度，则输出最大的面积；否则，输出给定的绳子长度。

注意：这个解决方案没有处理特殊情况，例如当一个或多个边长相等时的情况。在实际应用中，你可能需要添加额外的逻辑来处理这种情况。