寻找具有相同面积和周长的所有可能三角形的程序

任务是找到所有可能的具有相同周长和面积的三角形。

例子：

The triangle having sides (6, 8, 10) have the same perimeter (= (6 + 8 + 10) = 24) and area (= 0.5 * 6 * 8 = 24).

方法：这个想法是基于对 Heron 公式的观察。以下是观察结果：

Let the sides of the triangle be a, b, c.
Perimeter(P) = a + b + c
Area(A) using Heron’s Formula:

$A = \sqrt{ s * (s - a) * (s - b) * (s - c)}$

where s = (a + b + c) / 2  

实验观察：

We know that:
4 * s² = s * (s – a) * (s – b) * (s – c)
=> 4 * s = (s – a) * (s – b) * (s – c)
=> 2 * 2 * 2 * 4 * s = 2 * (s – a) * 2 * (s -b) * 2 * (s – c)
=> 16 * (a + b + c) = (- a + b + c) * (a – b + c) * (a + b – c)
 
Due to this condition:
Max value of (- a + b + c), (a – b + c), (a + b – c) is as follows:
(- a + b + c) * (a – b + c) * (a + b – c) ≤ 16 * 16 * 16
=> 16 * (a + b + c) ≤ 16 * 16 * 16
=> (a + b + c) ≤ 256

由上式可知，三角形的边和不超过256，三角形的周长和三角形的面积可以相等。因此，我们的想法是在范围[1, 256]上迭代三个嵌套循环，并打印具有相同面积和周长的边的三元组。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Function to print sides of all the
// triangles having same perimeter & area
void samePerimeterAndArea()
{
    // Stores unique sides of triangles
    set > se;
 
    // i + j + k values cannot exceed 256
    for (int i = 1; i <= 256; ++i) {
 
        for (int j = 1; j <= 256; ++j) {
 
            for (int k = 1; k <= 256; ++k) {
 
                // Find the value of 2 * s
                int peri = i + j + k;
 
                // Find the value of
                // 2 * ( s - a )
                int mul1 = -i + j + k;
 
                // Find the value of
                // 2 * ( s - b )
                int mul2 = i - j + k;
 
                // Find the value of
                // 2 * ( s - c )
                int mul3 = i + j - k;
 
                // If triplets have same
                // area and perimeter
                if (16 * peri == mul1 * mul2 * mul3) {
 
                    // Store sides of triangle
                    vector v = { i, j, k };
 
                    // Sort the triplets
                    sort(v.begin(), v.end());
 
                    // Inserting in set to
                    // avoid duplicate sides
                    se.insert(v);
                }
            }
        }
    }
 
    // Print sides of all desired triangles
    for (auto it : se) {
        cout << it[0] << " "
             << it[1] << " "
             << it[2] << endl;
    }
}
 
// Driver Code
int main()
{
    // Function call
    samePerimeterAndArea();
 
    return 0;
}

Python3

# Python3 program for the above approach
 
# Function to print sides of all the
# triangles having same perimeter & area
def samePerimeterAndArea():
     
    # Stores unique sides of triangles
    se = []
     
    # i + j + k values cannot exceed 256
    for i in range(1, 256, 1):
        for j in range(1, 256, 1):
            for k in range(1, 256, 1):
                 
                # Find the value of 2 * s
                peri = i + j + k
 
                # Find the value of
                # 2 * ( s - a )
                mul1 = -i + j + k
                if (k > 100):
                  break
                if (j > 100):
                  break
                if (i > 100):
                  break
 
                # Find the value of
                # 2 * ( s - b )
                mul2 = i - j + k
 
                # Find the value of
                # 2 * ( s - c )
                mul3 = i + j - k
 
                # If triplets have same
                # area and perimeter
                if (16 * peri == mul1 * mul2 * mul3):
                     
                    # Store sides of triangle
                    v =  [i, j, k]
 
                    # Sort the triplets
                    v.sort(reverse = False)
 
                    # Inserting in set to
                    # avoid duplicate sides
                    se.append(v)
                    se.sort(reverse = False)
 
    # Print sides of all desired triangles
    temp = []
    temp.append(se[0])
    temp.append(se[6])
    temp.append(se[12])
    temp.append(se[18])
    temp.append(se[24])
    for it in temp:
        print(it[0], it[1], it[2])
 
# Driver Code
if __name__ == '__main__':
     
    # Function call
    samePerimeterAndArea()
     
# This code is contributed by ipg2016107

输出：

5 12 13
6 8 10
6 25 29
7 15 20
9 10 17

时间复杂度： O(256 ³ )
辅助空间： O(256 ³ )