检查给定范围内所有数字的异或是偶数还是奇数
原文:https://www . geesforgeks . org/check-给定范围内所有数字的异或是偶数还是奇数/
给定一个范围[ L，R ]，任务是找出范围 L 到 R(包括两者)内所有自然数的异或值是偶数还是奇数。如果范围内所有数字的异或为偶数，则打印“偶数”，否则打印奇数。 例:
Input: L = 1, R= 10
Output: Odd
Input: L= 5, R=15
Output: Even
一个简单的解决方案是计算范围【L，R】内所有数字的异或，然后检查结果异或值是偶数还是奇数。 这种方法的时间复杂度为 O(n)。 一种高效解决方案基于以下事实:
odd ^ odd = even
odd ^ even = odd
even ^ odd = odd
even ^ even = even
所有偶数的异或将是偶数(与范围大小无关)，如果奇数的计数是奇数，那么最终的异或将是奇数，如果是偶数，那么最终的异或将是偶数。 现在可以得出结论，
- If the count of odd numbers is even, XOR of all odd numbers = even numbers XOR of all even numbers = even numbers Final XOR = even numbers even numbers = even numbers
- If the count of odd numbers is odd, XOR of all odd numbers = odd numbers XOR of all even numbers = even numbers Final xor = odd numbers even numbers = odd numbers.
因此，我们所要做的就是计算 L 到 r 范围内的奇数。 方法:
- 计算范围[ L，R ]内的奇数。
- 检查奇数的计数是偶数还是奇数。
- 如果计数为偶数，则打印“偶数”，否则打印“奇数”。
以下是上述方法的实现:
C++
// C++ program to check if XOR of
// all numbers in range [L, R]
// is Even or odd
#include <bits/stdc++.h>
using namespace std;
// Function to check if XOR of all numbers
// in range [L, R] is Even or Odd
string isEvenOrOdd(int L, int R)
{
// Count odd Numbers in range [L, R]
int oddCount = (R - L) / 2;
if (R % 2 == 1 || L % 2 == 1)
oddCount++;
// Check if count of odd Numbers
// is even or odd
if (oddCount % 2 == 0)
return "Even";
else
return "Odd";
}
// Driver Code
int main()
{
int L = 5, R = 15;
cout << isEvenOrOdd(L, R);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to check if XOR of
// all numbers in range [L, R]
// is Even or odd
class GFG {
// Function to check if XOR of all numbers
// in range [L, R] is Even or Odd
static String isEvenOrOdd(int L, int R)
{
// Count odd Numbers in range [L, R]
int oddCount = (R - L) / 2;
if (R % 2 == 1 || L % 2 == 1)
oddCount++;
// Check if count of odd Numbers
// is even or odd
if (oddCount % 2 == 0)
return "Even";
else
return "Odd";
}
// Driver Code
public static void main(String[] args)
{
int L = 5, R = 15;
System.out.println(isEvenOrOdd(L, R));
}
}
C
// C# program to check if XOR of
// all numbers in range [L, R]
// is Even or odd
using System;
class GFG {
// Function to check if XOR of all numbers
// in range [L, R] is Even or Odd
static string isEvenOrOdd(int L, int R)
{
// Count odd Numbers in range [L, R]
int oddCount = (R - L) / 2;
if (R % 2 == 1 || L % 2 == 1)
oddCount++;
// Check if count of odd Numbers
// is even or odd
if (oddCount % 2 == 0)
return "Even";
else
return "Odd";
}
// Driver Code
public static void Main()
{
int L = 5, R = 15;
Console.WriteLine(isEvenOrOdd(L, R));
}
}
Python 3
# Python3 program to check if XOR of
# all numbers in range [L, R]
# is Even or odd
# Function to check if XOR of all numbers
# in range [L, R] is Even or Odd
def isEvenOrOdd( L, R ):
# Count odd Numbers in range [L, R]
oddCount = (R - L )/2
if( R % 2 == 1 or L % 2 == 1):
oddCount = oddCount + 1
# Check if count of odd Numbers
# is even or odd
if(oddCount % 2 == 0 ):
return "Even"
else :
return "Odd"
# Driver Code
L = 5
R = 15
print(isEvenOrOdd(L, R));
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP program to check if XOR of all
// numbers in range [L, R] is Even or odd
// Function to check if XOR of all numbers
// in range [L, R] is Even or Odd
function isEvenOrOdd($L, $R)
{
// Count odd Numbers in range [L, R]
$oddCount = floor(($R - $L) / 2);
if ($R % 2 == 1 || $L % 2 == 1)
$oddCount++;
// Check if count of odd Numbers
// is even or odd
if ($oddCount % 2 == 0)
return "Even";
else
return "Odd";
}
// Driver Code
$L = 5;
$R = 15;
echo isEvenOrOdd($L, $R);
// This code is contributed by Ryuga
?>
java 描述语言
<script>
// JavaScript program to check if XOR of
// all numbers in range [L, R]
// is Even or odd
// Function to check if XOR of all numbers
// in range [L, R] is Even or Odd
function isEvenOrOdd(L, R)
{
// Count odd Numbers in range [L, R]
let oddCount = Math.floor((R - L) / 2);
if (R % 2 == 1 || L % 2 == 1)
oddCount++;
// Check if count of odd Numbers
// is even or odd
if (oddCount % 2 == 0)
return "Even";
else
return "Odd";
}
// Driver Code
let L = 5, R = 15;
document.write(isEvenOrOdd(L, R));
// This code is contributed by Surbhi Tyagi.
</script>
Output:
Even
时间复杂度:O(1) 辅助空间:O(1)
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