给定一个三角形的三个坐标(x1，y1) ， (x2，y2) ， (x3，y3) 。任务是找出是否三角形可以通过距离1恰好仅移动一个点被转换成直角三角形。
如果可以使三角形成直角，则打印“可能” ，否则打印“不可行” 。
如果三角形已经是直角的，则也应报告该三角形。
例子：
Input:
x1 = -1, y1 = 0
x2 = 2, y2 = 0
x3 = 0, y3 = 1
Output: POSSIBLE
First co-ordinate (-1, 0) can be changed to (0, 0) and make it right-angled.
Input:
x1 = 36, y1 = 1
x2 = -17, y2 = -54
x3 = -19, y3 = 55
Output: POSSIBLE
方法：
众所周知，对于边为a ， b和c的三角形，如果以下等式成立，则该三角形将成直角： a 2 + b 2 = c 2
因此，对于三角形的每一个坐标，找出所有的边，并为他们的3个可能的排列检查它是否已经直角三角形和报告。
如果上述条件不成立，则需要执行以下操作-
我们需要一一更改所有坐标，并检查它是否是直角三角形的有效组合。
看起来可以有4种可能的组合来将每个坐标更改为1。它们是(-1，0)，(0，1)，(1，0)，(0，-1) 。因此，运行一个循环并为每个坐标逐一应用这些更改，并检查公式a 2 + b 2 = c 2是否成立。
如果这是真的，那么它可以将三角形转变为直角三角形，否则不是。
下面是上述代码的实现：
C++
// C++ implementation of
// the above approach
#include
using namespace std;
// Storing all the possible
// changes to make the triangle
// right-angled
int dx[] = { -1, 0, 1, 0 };
int dy[] = { 0, 1, 0, -1 };
// Function to check if the triangle
// is right-angled or not
int ifRight(int x1, int y1,
int x2, int y2,
int x3, int y3)
{
int a = ((x1 - x2) * (x1 - x2))
+ ((y1 - y2) * (y1 - y2));
int b = ((x1 - x3) * (x1 - x3))
+ ((y1 - y3) * (y1 - y3));
int c = ((x2 - x3) * (x2 - x3))
+ ((y2 - y3) * (y2 - y3));
if ((a == (b + c) && a != 0 && b != 0 && c != 0)
|| (b == (a + c) && a != 0 && b != 0 && c != 0)
|| (c == (a + b) && a != 0 && b != 0 && c != 0)) {
return 1;
}
return 0;
}
// Function to check if the triangle
// can be transformed to right-angled
void isValidCombination(int x1, int y1,
int x2, int y2,
int x3, int y3)
{
int x, y;
// Boolean variable to
// return true or false
bool possible = 0;
// If it is already right-angled
if (ifRight(x1, y1,
x2, y2,
x3, y3)) {
cout << "ALREADY RIGHT ANGLED";
return;
}
else {
// Applying the changes on the
// co-ordinates
for (int i = 0; i < 4; i++) {
// Applying on the first
// co-ordinate
x = dx[i] + x1;
y = dy[i] + y1;
if (ifRight(x, y,
x2, y2,
x3, y3)) {
cout << "POSSIBLE";
return;
}
// Applying on the second
// co-ordinate
x = dx[i] + x2;
y = dy[i] + y2;
if (ifRight(x1, y1,
x, y,
x3, y3)) {
cout << "POSSIBLE";
return;
}
// Applying on the third
// co-ordinate
x = dx[i] + x3;
y = dy[i] + y3;
if (ifRight(x1, y1,
x2, y2,
x, y)) {
cout << "POSSIBLE";
return;
}
}
}
// If can't be transformed
if (!possible)
cout << "NOT POSSIBLE" << endl;
}
// Driver Code
int main()
{
int x1 = -49, y1 = 0;
int x2 = 0, y2 = 50;
int x3 = 0, y3 = -50;
isValidCombination(x1, y1,
x2, y2,
x3, y3);
return 0;
} Java
// Java implementation of the approach
import java.util.*;
class GFG
{
// Storing all the possible
// changes to make the triangle
// right-angled
static int dx[] = { -1, 0, 1, 0 };
static int dy[] = { 0, 1, 0, -1 };
// Function to check if the triangle
// is right-angled or not
static boolean ifRight(int x1, int y1,
int x2, int y2,
int x3, int y3)
{
int a = ((x1 - x2) * (x1 - x2)) +
((y1 - y2) * (y1 - y2));
int b = ((x1 - x3) * (x1 - x3)) +
((y1 - y3) * (y1 - y3));
int c = ((x2 - x3) * (x2 - x3)) +
((y2 - y3) * (y2 - y3));
if ((a == (b + c) && a != 0 && b != 0 && c != 0) ||
(b == (a + c) && a != 0 && b != 0 && c != 0) ||
(c == (a + b) && a != 0 && b != 0 && c != 0))
{
return true;
}
return false;
}
// Function to check if the triangle
// can be transformed to right-angled
static void isValidCombination(int x1, int y1,
int x2, int y2,
int x3, int y3)
{
int x, y;
// Boolean variable to
// return true or false
boolean possible = false;
// If it is already right-angled
if (ifRight(x1, y1, x2, y2, x3, y3))
{
System.out.print("ALREADY RIGHT ANGLED");
return;
}
else
{
// Applying the changes on the
// co-ordinates
for (int i = 0; i < 4; i++)
{
// Applying on the first
// co-ordinate
x = dx[i] + x1;
y = dy[i] + y1;
if (ifRight(x, y, x2, y2, x3, y3))
{
System.out.print("POSSIBLE");
return;
}
// Applying on the second
// co-ordinate
x = dx[i] + x2;
y = dy[i] + y2;
if (ifRight(x1, y1, x, y, x3, y3))
{
System.out.print("POSSIBLE");
return;
}
// Applying on the third
// co-ordinate
x = dx[i] + x3;
y = dy[i] + y3;
if (ifRight(x1, y1, x2, y2, x, y))
{
System.out.print("POSSIBLE");
return;
}
}
}
// If can't be transformed
if (!possible)
System.out.println("NOT POSSIBLE");
}
// Driver Code
public static void main(String[] args)
{
int x1 = -49, y1 = 0;
int x2 = 0, y2 = 50;
int x3 = 0, y3 = -50;
isValidCombination(x1, y1, x2, y2, x3, y3);
}
}
// This code is contributed by Rajput-JiPython3
# Python3 implementation of the above approach
# Storing all the possible
# changes to make the triangle
# right-angled
dx = [-1, 0, 1, 0]
dy = [0, 1, 0, -1]
# Function to check if the triangle
# is right-angled or not
def ifRight(x1, y1, x2, y2, x3, y3):
a = ((x1 - x2) * (x1 - x2)) + \
((y1 - y2) * (y1 - y2))
b = ((x1 - x3) * (x1 - x3)) + \
((y1 - y3) * (y1 - y3))
c = ((x2 - x3) * (x2 - x3)) + \
((y2 - y3) * (y2 - y3))
if ((a == (b + c) and a != 0 and b != 0 and c != 0) or
(b == (a + c) and a != 0 and b != 0 and c != 0) or
(c == (a + b) and a != 0 and b != 0 and c != 0)):
return 1
# Function to check if the triangle
# can be transformed to right-angled
def isValidCombination(x1, y1, x2, y2, x3, y3):
x, y = 0, 0
# Boolean variable to
# return true or false
possible = 0
# If it is already right-angled
if (ifRight(x1, y1, x2, y2, x3, y3)):
print("ALREADY RIGHT ANGLED")
return
else:
# Applying the changes on the
# co-ordinates
for i in range(4):
# Applying on the first
# co-ordinate
x = dx[i] + x1
y = dy[i] + y1
if (ifRight(x, y, x2, y2, x3, y3)):
print("POSSIBLE")
return
# Applying on the second
# co-ordinate
x = dx[i] + x2
y = dy[i] + y2
if (ifRight(x1, y1, x, y, x3, y3)):
print("POSSIBLE")
return
# Applying on the third
# co-ordinate
x = dx[i] + x3
y = dy[i] + y3
if (ifRight(x1, y1, x2, y2, x, y)):
print("POSSIBLE")
return
# If can't be transformed
if (possible == 0):
print("NOT POSSIBLE")
# Driver Code
x1 = -49
y1 = 0
x2 = 0
y2 = 50
x3 = 0
y3 = -50
isValidCombination(x1, y1, x2, y2, x3, y3)
# This code is contributed by Mohit KumarC#
// C# implementation of the approach
using System;
class GFG
{
// Storing all the possible
// changes to make the triangle
// right-angled
static int []dx = { -1, 0, 1, 0 };
static int []dy = { 0, 1, 0, -1 };
// Function to check if the triangle
// is right-angled or not
static bool ifRight(int x1, int y1,
int x2, int y2,
int x3, int y3)
{
int a = ((x1 - x2) * (x1 - x2)) +
((y1 - y2) * (y1 - y2));
int b = ((x1 - x3) * (x1 - x3)) +
((y1 - y3) * (y1 - y3));
int c = ((x2 - x3) * (x2 - x3)) +
((y2 - y3) * (y2 - y3));
if ((a == (b + c) && a != 0 && b != 0 && c != 0) ||
(b == (a + c) && a != 0 && b != 0 && c != 0) ||
(c == (a + b) && a != 0 && b != 0 && c != 0))
{
return true;
}
return false;
}
// Function to check if the triangle
// can be transformed to right-angled
static void isValidCombination(int x1, int y1,
int x2, int y2,
int x3, int y3)
{
int x, y;
// Boolean variable to
// return true or false
bool possible = false;
// If it is already right-angled
if (ifRight(x1, y1, x2, y2, x3, y3))
{
Console.WriteLine("ALREADY RIGHT ANGLED");
return;
}
else
{
// Applying the changes on the
// co-ordinates
for (int i = 0; i < 4; i++)
{
// Applying on the first
// co-ordinate
x = dx[i] + x1;
y = dy[i] + y1;
if (ifRight(x, y, x2, y2, x3, y3))
{
Console.WriteLine("POSSIBLE");
return;
}
// Applying on the second
// co-ordinate
x = dx[i] + x2;
y = dy[i] + y2;
if (ifRight(x1, y1, x, y, x3, y3))
{
Console.WriteLine("POSSIBLE");
return;
}
// Applying on the third
// co-ordinate
x = dx[i] + x3;
y = dy[i] + y3;
if (ifRight(x1, y1, x2, y2, x, y))
{
Console.Write("POSSIBLE");
return;
}
}
}
// If can't be transformed
if (!possible)
Console.WriteLine("NOT POSSIBLE");
}
// Driver Code
static public void Main ()
{
int x1 = -49, y1 = 0;
int x2 = 0, y2 = 50;
int x3 = 0, y3 = -50;
isValidCombination(x1, y1, x2, y2, x3, y3);
}
}
// This code is contributed by AnkitRai01输出：
POSSIBLE