给定两个整数L和R ，任务是从L开头的范围[L，R]中查找连续数字的偶数和奇数按位XOR值的计数。
例子：
Input: L = 2, R = 7
Output: Even = 3, Odd = 3
Explanation: Taking bitwise XOR of continuous numbers:
2
2 ^ 3 = 1
2 ^ 3 ^ 4 = 5
2 ^ 3 ^ 4 ^ 5 = 0
2 ^ 3 ^ 4 ^ 5 ^ 6 = 6
2 ^ 3 ^ 4 ^ 5 ^ 6 ^ 7 = 1
Therefore, Bitwise XOR values obtained are {2, 1, 5, 0, 6, 1}.
Therefore, count of even XOR values is 3 and odd XOR values is 3.
Input: L = 1, R = 7
Output: Even = 3, Odd = 4
天真的方法：最简单的方法是遍历[L，R]范围内的所有数字 并执行从L开始的连续数字的按位XOR 。最后，计算获得的偶数和奇数值的按位XOR值的数量。
时间复杂度： O(R – L)
辅助空间： O(1)
高效方法：为了优化上述方法，该思想基于以下观察结果：可以在恒定时间内计算从1到N的按位XOR：
- 除以4时，求出N的余数。
- 如果获得的余数为0 ，则XOR等于N。
- 如果获得的余数为1 ，则XOR等于1 。
- 如果获得的余数为2 ，则XOR等于N +1 。
- 如果获得的余数为3 ，则获得的XOR将为0 。
从以上观察结果可以得出结论，偶数XOR值为0或4的倍数。请按照以下步骤解决问题：
- 将[L，R]范围内的元素数存储在变量X中。
- 通过将X除以4并乘以2来存储偶数XOR值的计数，该变量为Even 。
- 如果L为奇数且X％4等于3 ，则将偶数增加1 。
- 否则，如果L为偶数且X％4大于0 ，则将偶数增加1 。
- 将奇数XOR值的计数存储在变量Odd = X – Even中。
- 完成上述步骤后，将结果打印为偶数和奇数。
下面是上述方法的实现。
C++
// C++ program for the above approach
#include
using namespace std;
// Print count of even and odd numbers
// of XOR value from L to R
void countEvenOdd(int L, int R)
{
// Store the number of elements
// between L and R
int range = R - L + 1;
// Count of even XOR values
int even = (range / 4) * 2;
// If L is odd and range % 4 = 3
if ((L & 1) && (range % 4 == 3)) {
// Increment even by 1
even++;
}
// If L is even and range % 4 !=0
else if (!(L & 1) && (range % 4)) {
// Increment even by 1
even++;
}
// Print the answer
cout << "Even = " << even
<< ", Odd = " << range - even;
}
// Driver Code
int main()
{
int L = 2, R = 7;
countEvenOdd(L, R);
return 0;
} Java
// Java program for the above approach
import java.io.*;
import java.util.*;
class GFG
{
// Print count of even and odd numbers
// of XOR value from L to R
static void countEvenOdd(int L, int R)
{
// Store the number of elements
// between L and R
int range = R - L + 1;
// Count of even XOR values
int even = (range / 4) * 2;
// If L is odd and range % 4 = 3
if ((L & 1) != 0 && (range % 4 == 3))
{
// Increment even by 1
even++;
}
// If L is even and range % 4 !=0
else if ((L & 1) == 0 && (range % 4 != 0))
{
// Increment even by 1
even++;
}
// Print the answer
System.out.print("Even = " + even +
", Odd = " + (range - even));
}
// Driver Code
public static void main(String[] args)
{
int L = 2, R = 7;
countEvenOdd(L, R);
}
}
// This code is contributed by sanjoy_62.Python3
# Python program for the above approach
# Prcount of even and odd numbers
# of XOR value from L to R
def countEvenOdd(L, R):
# Store the number of elements
# between L and R
range = R - L + 1;
# Count of even XOR values
even = (range // 4) * 2;
# If L is odd and range % 4 = 3
if ((L & 1) != 0 and (range % 4 == 3)):
# Increment even by 1
even += 1;
# If L is even and range % 4 !=0
elif ((L & 1) == 0 and (range % 4 != 0)):
# Increment even by 1
even += 1;
# Prthe answer
print("Even = " , even ,\
", Odd = " , (range - even));
# Driver Code
if __name__ == '__main__':
L = 2; R = 7;
countEvenOdd(L, R);
# This code is contributed by shikhasingrajputC#
// C# program for the above approach
using System;
class GFG
{
// Print count of even and odd numbers
// of XOR value from L to R
static void countEvenOdd(int L, int R)
{
// Store the number of elements
// between L and R
int range = R - L + 1;
// Count of even XOR values
int even = (range / 4) * 2;
// If L is odd and range % 4 = 3
if ((L & 1) != 0 && (range % 4 == 3))
{
// Increment even by 1
even++;
}
// If L is even and range % 4 !=0
else if ((L & 1) == 0 && (range % 4 != 0))
{
// Increment even by 1
even++;
}
// Print the answer
Console.Write("Even = " + even +
", Odd = " + (range - even));
}
// Driver code
static void Main()
{
int L = 2, R = 7;
countEvenOdd(L, R);
}
}
// This code is contributed by divyeshrabadiya07.Javascript
输出：
Even = 3, Odd = 3时间复杂度： O(1)
辅助空间： O(1)