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📜  在[L，R]范围内查找数字的XOR

📅  最后修改于: 2021-04-27 05:45:57             🧑  作者: Mango

给定两个整数L和R ，任务是查找范围为[L，R]的元素的XOR。
例子：

天真的方法：将答案归纳为零，从L到R遍历所有数字，并对答案与数字一一进行异或。这将花费O(N)时间。
高效方法：通过遵循此处讨论的方法，我们可以在O(1)时间中从[1，N]范围内找到元素的XOR。
使用这种方法，我们必须找到[1，L – 1]范围和[1，R]范围内元素的异或，然后再次对相应的答案进行异或，以得出[L， R] 。这是因为范围[1，L – 1]中的每个元素都将在结果中进行两次“异或”运算，结果为0 ，当将其与范围[L，R]中的元素进行异或时，将得出结果。
下面是上述方法的实现：

C++
// C++ implementation of the approach
#include 
using namespace std;
 
// Function to return the XOR of elements
// from the range [1, n]
int findXOR(int n)
{
    int mod = n % 4;
 
    // If n is a multiple of 4
    if (mod == 0)
        return n;
 
    // If n % 4 gives remainder 1
    else if (mod == 1)
        return 1;
 
    // If n % 4 gives remainder 2
    else if (mod == 2)
        return n + 1;
 
    // If n % 4 gives remainder 3
    else if (mod == 3)
        return 0;
}
 
// Function to return the XOR of elements
// from the range [l, r]
int findXOR(int l, int r)
{
    return (findXOR(l - 1) ^ findXOR(r));
}
 
// Driver code
int main()
{
    int l = 4, r = 8;
 
    cout << findXOR(l, r);
 
    return 0;
}

Java
// Java implementation of the approach
class GFG
{
    // Function to return the XOR of elements
    // from the range [1, n]
    static int findXOR(int n)
    {
        int mod = n % 4;
 
        // If n is a multiple of 4
        if (mod == 0)
            return n;
 
        // If n % 4 gives remainder 1
        else if (mod == 1)
            return 1;
 
        // If n % 4 gives remainder 2
        else if (mod == 2)
            return n + 1;
 
        // If n % 4 gives remainder 3
        else if (mod == 3)
            return 0;
        return 0;
    }
 
    // Function to return the XOR of elements
    // from the range [l, r]
    static int findXOR(int l, int r)
    {
        return (findXOR(l - 1) ^ findXOR(r));
    }
 
    // Driver code
    public static void main(String[] args)
    {
 
        int l = 4, r = 8;
 
            System.out.println(findXOR(l, r));
    }
}
 
// This code contributed by Rajput-Ji

Python3
# Python3 implementation of the approach
from operator import xor
 
# Function to return the XOR of elements
# from the range [1, n]
def findXOR(n):
    mod = n % 4;
 
    # If n is a multiple of 4
    if (mod == 0):
        return n;
 
    # If n % 4 gives remainder 1
    elif (mod == 1):
        return 1;
 
    # If n % 4 gives remainder 2
    elif (mod == 2):
        return n + 1;
 
    # If n % 4 gives remainder 3
    elif (mod == 3):
        return 0;
 
# Function to return the XOR of elements
# from the range [l, r]
def findXORFun(l, r):
    return (xor(findXOR(l - 1) , findXOR(r)));
 
# Driver code
l = 4; r = 8;
 
print(findXORFun(l, r));
 
# This code is contributed by PrinciRaj1992

C#
// C# implementation of the approach
using System;
 
class GFG
{
    // Function to return the XOR of elements
    // from the range [1, n]
    static int findXOR(int n)
    {
        int mod = n % 4;
 
        // If n is a multiple of 4
        if (mod == 0)
            return n;
 
        // If n % 4 gives remainder 1
        else if (mod == 1)
            return 1;
 
        // If n % 4 gives remainder 2
        else if (mod == 2)
            return n + 1;
 
        // If n % 4 gives remainder 3
        else if (mod == 3)
            return 0;
        return 0;
    }
 
    // Function to return the XOR of elements
    // from the range [l, r]
    static int findXOR(int l, int r)
    {
        return (findXOR(l - 1) ^ findXOR(r));
    }
 
    // Driver code
    public static void Main()
    {
 
        int l = 4, r = 8;
 
            Console.WriteLine(findXOR(l, r));
    }
}
 
// This code is contributed by AnkitRai01

Javascript

输出： 
8