求满足 2/n = 1/x + 1/y + 1/z 的 x，y，z
原文:https://www . geesforgeks . org/find-x-y-z-submist-2n-1x-1y-1z/
给定 n，求 x，y，z，使 x，y，z 满足等式“2/n = 1/x+1/y+1/z” 有多个 x，y 和 z 满足等式打印其中任何一个，如果不可能，则打印-1。 示例:
Input : 3
Output : 3 4 12
Explanation: here 3 4 and 12 satisfy
the given equation
Input : 7
Output : 7 8 56
注意，对于 n = 1 没有解，对于 n > 1 有解 x = n，y = n+1，z = n (n+1)。 得出这个解，把 2/n 表示为 1/n+1/n，把问题化简为把 1/n 表示为两个分数之和。我们求出 1/n 和 1/(n+1)的差，得到分数 1/(n*(n+1))，这样解就是
2/n = 1/n + 1/(n+1) + 1/(n*(n+1))
C++
// CPP program to find x y z that
// satisfies 2/n = 1/x + 1/y + 1/z...
#include <bits/stdc++.h>
using namespace std;
// function to find x y and z that
// satisfy given equation.
void printXYZ(int n)
{
if (n == 1)
cout << -1;
else
cout << "x is " << n << "\ny is "
<< n + 1 << "\nz is "
<< n * (n + 1);
}
// driver program to test the above function
int main()
{
int n = 7;
printXYZ(n);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to find x y z that
// satisfies 2/n = 1/x + 1/y + 1/z...
import java.io.*;
class Sums {
// function to find x y and z that
// satisfy given equation.
static void printXYZ(int n){
if (n == 1)
System.out.println(-1);
else{
System.out.println("x is "+ n);
System.out.println("y is "+ (n+1));
System.out.println("z is "+ (n * (n + 1)));
}
}
// Driver program to test the above function
public static void main (String[] args) {
int n = 7;
printXYZ(n);
}
}
// This code is contributed by Chinmoy Lenka
Python 3
# Python3 code to find x y z that
# satisfies 2/n = 1/x + 1/y + 1/z...
# function to find x y and z that
# satisfy given equation.
def printXYZ( n ):
if n == 1:
print(-1)
else:
print("x is " , n )
print("y is " ,n + 1)
print("z is " ,n * (n + 1))
# driver code to test the above function
n = 7
printXYZ(n)
# This code is contributed by "Sharad_Bhardwaj".
C
// C# program to find x y z that
// satisfies 2/n = 1/x + 1/y + 1/z...
using System;
class GFG
{
// function to find x y and z that
// satisfy given equation.
static void printXYZ(int n)
{
if (n == 1)
Console.WriteLine(-1);
else
{
Console.WriteLine("x is "+ n);
Console.WriteLine("y is "+ (n+1));
Console.WriteLine("z is "+ (n * (n + 1)));
}
}
// Driver program
public static void Main ()
{
int n = 7;
printXYZ(n);
}
}
// This code is contributed by vt_m
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP program to find x y z that
// satisfies 2/n = 1/x + 1/y + 1/z...
// function to find x y and z that
// satisfy given equation.
function printXYZ($n)
{
if ($n == 1)
echo -1;
else
echo "x is " , $n , "\ny is "
, $n + 1 , "\nz is ",
$n * ($n + 1);
}
// Driver Code
$n = 7;
printXYZ($n);
// This code is contributed by anuj_67.
?>
java 描述语言
<script>
// Javascript program to find x y z that
// satisfies 2/n = 1/x + 1/y + 1/z...
// function to find x y and z that
// satisfy given equation.
function printXYZ(n)
{
if (n == 1)
document.write(-1);
else
document.write("x is " + n + "<br>y is "
+ (n + 1) + "<br>z is "
+ n * (n + 1));
}
// driver program to test the above function
let n = 7;
printXYZ(n);
</script>
输出:
x is 7
y is 8
z is 56
时间复杂度: O(1) 交替解 我们可以写成 2/n = 1/n + 1/n，再进一步写成 2/n = 1/n + 1/2n + 1/2n。
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