从单个整数生成一个毕达哥拉斯三元组
给定一个整数 n 
[1，1000000000]，生成一个毕达哥拉斯三元组，如果可能的话，该三元组的一边包括 n。
例:
Input : 22
Output : Pythagoras Triplets exist i.e. 22 120 122
Input : 4
Output : Pythagoras Triplets exist i.e. 4 3 5
Input : 2
Output : No Pythagoras Triplet exists
解释:
定义:“毕达哥拉斯三元组”是毕达哥拉斯定理的整数解，即它们满足方程
我们的任务是从整数值生成三元组。这可能是一个令人困惑的任务，因为给我们的边可以是斜边或非斜边。
通过把三胞胎放在一个公式中开始计算三胞胎，可以推导出只有对于 1 和 2，才可能没有三胞胎。
再进一步，
如果 n 为偶数，我们的三胞胎用公式
计算如果 n 为奇数，我们的三胞胎用公式
计算证明:
勾股定理也可以写成
即 aa = (c-b)(c+b)
aa x 1 = aa，由此可见
和
一样，如果 n 为奇数，这种解法也有效。
对于偶数解，
，由此，我们得到 n 为偶数时的上式。
代码*
C++
// CPP program to find Pythagoras triplet
// with one side as given number.
#include <bits/stdc++.h>
using namespace std;
// Function, to evaluate the Pythagoras triplet
// with includes 'n' if possible
void evaluate(long long int n)
{
if (n == 1 || n == 2)
printf("No Pythagoras Triplet exists");
else if (n % 2 == 0) {
// Calculating for even case
long long int var = 1LL * n * n / 4;
printf("Pythagoras Triplets exist i.e. ");
printf("%lld %lld %lld", n, var - 1, var + 1);
}
else if (n % 2 != 0) {
// Calculating for odd case
long long int var = 1LL * n * n + 1;
printf("Pythagoras Triplets exist i.e. ");
printf("%lld %lld %lld", n, var / 2 - 1, var / 2);
}
}
// Driver function
int main()
{
long long int n = 22;
evaluate(n);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to find
// Pythagoras triplet
// with one side as
// given number.
import java.io.*;
class GFG
{
// Function, to evaluate
// the Pythagoras triplet
// with includes 'n' if
// possible
static void evaluate( int n)
{
if (n == 1 || n == 2)
System.out.println("No Pythagoras " +
"Triplet exists");
else if (n % 2 == 0)
{
// Calculating for even case
int var = 1 * n * n / 4;
System.out.print("Pythagoras Triplets " +
"exist i.e. ");
System.out.print(n + " ");
System.out.print(var - 1+ " ");
System.out.println(var + 1 +" ");
}
else if (n % 2 != 0)
{
int var = 1 * n * n + 1;
System.out.print("Pythagoras Triplets " +
"exist i.e. ");
System.out.print(n + " ");
System.out.print(var / 2 - 1 + " ");
System.out.println(var / 2 + " ");
}
}
// Driver Code
public static void main(String[] args)
{
int n = 22;
evaluate(n);
}
}
// This code is contributed
// by ajit
Python 3
# Python3 program to find
# Pythagoras triplet with
# one side as given number.
# Function, to evaluate the
# Pythagoras triplet with
# includes 'n' if possible
def evaluate(n):
if (n == 1 or n == 2):
print("No Pythagoras" +
" Triplet exists");
elif (n % 2 == 0):
# Calculating for
# even case
var = n * n / 4;
print("Pythagoras Triplets" +
" exist i.e. ", end = "");
print(int(n), " ", int(var - 1),
" ", int(var + 1));
elif (n % 2 != 0):
# Calculating for odd case
var = n * n + 1;
print("Pythagoras Triplets " +
"exist i.e. ", end = "");
print(int(n), " ", int(var / 2 - 1),
" ", int(var / 2));
# Driver Code
n = 22;
evaluate(n);
# This code is contributed by mits
C
// C# program to find
// Pythagoras triplet
// with one side as
// given number.
using System;
class GFG
{
// Function, to evaluate
// the Pythagoras triplet
// with includes 'n' if
// possible
static void evaluate(int n)
{
if (n == 1 || n == 2)
Console.WriteLine("No Pythagoras " +
"Triplet exists");
else if (n % 2 == 0)
{
// Calculating for even case
int var = 1 * n * n / 4;
Console.Write("Pythagoras Triplets " +
"exist i.e. ");
Console.Write(n + " ");
Console.Write(var - 1+ " ");
Console.WriteLine(var + 1 +" ");
}
else if (n % 2 != 0)
{
int var = 1 * n * n + 1;
Console.Write("Pythagoras Triplets " +
"exist i.e. ");
Console.Write(n + " ");
Console.Write(var / 2 - 1 + " ");
Console.WriteLine(var / 2 + " ");
}
}
// Driver Code
static public void Main ()
{
int n = 22;
evaluate(n);
}
}
// This code is contributed
// by ajit
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP program to find Pythagoras triplet
// with one side as given number.
// Function, to evaluate the
// Pythagoras triplet with
// includes 'n' if possible
function evaluate($n)
{
if ($n == 1 || $n == 2)
echo "No Pythagoras Triplet exists";
else if ($n % 2 == 0) {
// Calculating for even case
$var = $n * $n / 4;
echo "Pythagoras Triplets exist i.e. ";
echo $n, " ", $var - 1, " ", $var + 1;
}
else if ($n % 2 != 0) {
// Calculating for odd case
$var = $n * $n + 1;
echo "Pythagoras Triplets exist i.e. ";
echo $n, " ", $var / 2 - 1, " ", $var / 2;
}
}
// Driver Code
$n = 22;
evaluate($n);
// This code is contributed by ajit
?>
java 描述语言
<script>
// Javascript program to find
// Pythagoras triplet
// with one side as
// given number.
// Function, to evaluate
// the Pythagoras triplet
// with includes 'n' if
// possible
function evaluate(n)
{
if (n == 1 || n == 2)
document.write("No Pythagoras Triplet exists");
else if (n % 2 == 0)
{
// Calculating for even case
let Var = 1 * n * n / 4;
document.write("Pythagoras Triplets " +
"exist i.e. ");
document.write(n + " ");
document.write(Var - 1+ " ");
document.write(Var + 1 +" ");
}
else if (n % 2 != 0)
{
let Var = 1 * n * n + 1;
document.write("Pythagoras Triplets " +
"exist i.e. ");
document.write(n + " ");
document.write(parseInt(Var / 2, 10) - 1 + " ");
document.write(parseInt(Var / 2, 10) + " ");
}
}
let n = 22;
evaluate(n);
</script>
输出:
Pythagoras Triplets exist i.e. 22 120 122
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