求给定方程中 X 和 Y 的值
给定两个数字和。求方程中 X 和 Y 的值。
- A = X + Y
- B = X 异或 Y
任务是让 X 尽可能最小。如果无法找到 X 和 Y 的任何有效值,则打印-1。 例:
Input : A = 12, B = 8
Output : X = 2, Y = 10
Input : A = 12, B = 9
Output : -1
让我们看看 X 中的某个位,它等于 1。如果 Y 中的相应位等于 0,则可以交换这两个位,从而减少 X 和增加 Y,而不改变它们的总和和异或。我们可以得出结论,如果 X 中的某个位等于 1,那么 Y 中的相应位也等于 1。这样,Y = X + B .考虑到 X + Y = X + X + B = A,我们可以得到以下求 X 和 Y 的公式:
- x =(A–B)/2
- Y = X + B = (A + B) / 2
人们还应该注意到,如果 A < B 或者 A 和 B 具有不同的奇偶性,那么答案不存在,输出为-1。如果 X 和(A–X)不等于 X,那么答案也是-1。 以下是上述办法的实施情况:
C++
// CPP program to find the values of
// X and Y using the given equations
#include <bits/stdc++.h>
using namespace std;
// Function to find the
// values of X and Y
void findValues(int a, int b)
{
// base condition
if ((a - b) % 2 == 1) {
cout << "-1";
return;
}
// required answer
cout << (a - b) / 2 << " " << (a + b) / 2;
}
// Driver Code
int main()
{
int a = 12, b = 8;
findValues(a, b);
return 0;
}
Java 语言(一种计算机语言,尤用于创建网站)
// Java program to find the values of
// X and Y using the given equations
import java.io.*;
class GFG
{
// Function to find the
// values of X and Y
static void findValues(int a, int b)
{
// base condition
if ((a - b) % 2 == 1)
{
System.out.println ("-1");
return;
}
// required answer
System.out.println (((a - b) / 2)+ " " +
((a + b) / 2));
}
// Driver Code
public static void main (String[] args)
{
int a = 12, b = 8;
findValues(a, b);
}
}
// This code is contributed by ajit...
Python 3
# Python3 program to find the values of
# X and Y using the given equations
# Function to find the values of X and Y
def findValues(a, b):
# base condition
if ((a - b) % 2 == 1):
print("-1");
return;
# required answer
print((a - b) // 2, (a + b) // 2);
# Driver Code
a = 12; b = 8;
findValues(a, b);
# This code is contributed
# by Akanksha Rai
C
// C# program to find the values of
// X and Y using the given equations
using System;
class GFG
{
// Function to find the
// values of X and Y
static void findValues(int a, int b)
{
// base condition
if ((a - b) % 2 == 1)
{
Console.Write ("-1");
return;
}
// required answer
Console.WriteLine(((a - b) / 2)+ " " +
((a + b) / 2));
}
// Driver Code
static public void Main ()
{
int a = 12, b = 8;
findValues(a, b);
}
}
// This code is contributed by @Tushil..
服务器端编程语言(Professional Hypertext Preprocessor 的缩写)
<?php
// PHP program to find the values of
// X and Y using the given equations
// Function to find the values
// of X and Y
function findValues($a, $b)
{
// base condition
if (($a - $b) % 2 == 1)
{
echo "-1";
return;
}
// required answer
echo ($a - $b) / 2, " " ,
($a + $b) / 2;
}
// Driver Code
$a = 12;
$b = 8;
findValues($a, $b);
// This code is contributed by jit_t
?>
java 描述语言
<script>
// Javascript program to find the values of
// X and Y using the given equations
// Function to find the values
// of X and Y
function findValues(a, b)
{
// base condition
if ((a - b) % 2 == 1)
{
document.write( "-1");
return;
}
// required answer
document.write( (a - b) / 2+ " " +
(a + b) / 2);
}
// Driver Code
let a = 12;
let b = 8;
findValues(a, b);
// This code is contributed
// by bobby
</script>
Output:
2 10
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