查找面积=(S / 2)的三角形的坐标

给定整数S ，任务是查找面积为(S / 2)的三角形的坐标。
例子：

Input: S = 4 
Output: 
(0, 0) 
(1000000000, 1) 
(999999996, 1)
Input: S = 15 
Output: 
(0, 0) 
(1000000000, 1) 
(999999985, 1) 
 

方法：

• 知道坐标为(X1，Y1) ， (X2，Y2)和(X3，Y3)的三角形的面积由A =((X1 * Y2)+(X2 * Y3)+(X3 * Y1)–(X1 * Y3)–(X2 * Y1)–(X3 * Y2))/ 2 。
• 现在将(X1，Y1)固定为(0，0)得出A =((X2 * Y3)–(X3 * Y2))/ 2 。
• 假设A = S / 2 ，这意味着S =(X2 * Y3)–(X3 * Y2) 。
• 现在修复(X2，Y2)至(10 ^9，1)和等式变成S = 10 ⁹ * Y3 – X3这可以通过取给定的其他变量的整数值的变量的整数值来解决。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
const long MAX = 1000000000;
 
// Function to find the traingle
// with area = (S / 2)
void findTriangle(long S)
{
 
    // Fix the two pairs of coordinates
    long X1 = 0, Y1 = 0;
    long X2 = MAX, Y2 = 1;
 
    // Find (X3, Y3) with integer coordinates
    long X3 = (MAX - S % MAX) % MAX;
    long Y3 = (S + X3) / MAX;
 
    cout << "(" << X1 << ", " << Y1 << ")\n";
    cout << "(" << X2 << ", " << Y2 << ")\n";
    cout << "(" << X3 << ", " << Y3 << ")";
}
 
// Driver code
int main()
{
 
    long S = 4;
 
    findTriangle(S);
 
    return 0;
}

Java

// Java implementation of the approach
class GFG
{
    static final long MAX = 1000000000;
     
    // Function to find the traingle
    // with area = (S / 2)
    static void findTriangle(long S)
    {
     
        // Fix the two pairs of coordinates
        long X1 = 0, Y1 = 0;
        long X2 = MAX, Y2 = 1;
     
        // Find (X3, Y3) with integer coordinates
        long X3 = (MAX - S % MAX) % MAX;
        long Y3 = (S + X3) / MAX;
     
        System.out.println("(" + X1 +
                           ", " + Y1 + ")");
        System.out.println("(" + X2 +
                           ", " + Y2 + ")");
        System.out.println("(" + X3 +
                           ", " + Y3 + ")");
    }
     
    // Driver code
    public static void main (String[] args)
    {
        long S = 4;
     
        findTriangle(S);
    }
}
 
// This code is contributed by AnkitRai01

Python3

# Python3 implementation of the approach
MAX = 1000000000;
 
# Function to find the traingle
# with area = (S / 2)
def findTriangle(S) :
 
    # Fix the two pairs of coordinates
    X1 = 0; Y1 = 0;
    X2 = MAX; Y2 = 1;
 
    # Find (X3, Y3) with integer coordinates
    X3 = (MAX - S % MAX) % MAX;
    Y3 = (S + X3) / MAX;
 
    print("(", X1, ",", Y1, ")");
    print("(", X2, ",", Y2, ")");
    print("(", X3, ",", Y3, ")");
 
# Driver code
if __name__ == "__main__" :
 
    S = 4;
 
    findTriangle(S);
 
# This code is contributed by kanugargng

C#

// C# implementation of the above approach
using System;
     
class GFG
{
    static readonly long MAX = 1000000000;
     
    // Function to find the traingle
    // with area = (S / 2)
    static void findTriangle(long S)
    {
     
        // Fix the two pairs of coordinates
        long X1 = 0, Y1 = 0;
        long X2 = MAX, Y2 = 1;
     
        // Find (X3, Y3) with integer coordinates
        long X3 = (MAX - S % MAX) % MAX;
        long Y3 = (S + X3) / MAX;
     
        Console.WriteLine("(" + X1 +
                         ", " + Y1 + ")");
        Console.WriteLine("(" + X2 +
                         ", " + Y2 + ")");
        Console.WriteLine("(" + X3 +
                         ", " + Y3 + ")");
    }
     
    // Driver code
    public static void Main (String[] args)
    {
        long S = 4;
     
        findTriangle(S);
    }
}
 
// This code is contributed by PrinciRaj1992

Javascript

 
// Javascript implementation of the approach
 
let MAX = 1000000000;
 
// Function to find the traingle
// with area = (S / 2)
function findTriangle( S)
{
 
    // Fix the two pairs of coordinates
    let X1 = 0, Y1 = 0;
    let X2 = MAX, Y2 = 1;
 
    // Find (X3, Y3) with integer coordinates
    let X3 = (MAX - S % MAX) % MAX;
    let Y3 = (S + X3) / MAX;
 
    document.write( "(" + X1 + ", " + Y1 + ")<br/>");
    document.write( "(" + X2 + ", " + Y2 + ")<br/>");
    document.write( "(" + X3 + ", " + Y3 + ")<br/>")
    }
 
// Driver code
 
    let S = 4;
 
    findTriangle(S);
 
// This code contributed by aashish1995
 

输出：

(0, 0)
(1000000000, 1)
(999999996, 1)

时间复杂度： O(1)