计算与x的和等于与x的XOR的数字

给定整数“ x”，找到满足以下条件的“ a”值的数量：

1. 0 <=一个<= x
2. 一个XOR x = a + x

例子 ：

Input : 5
Output : 2
Explanation: 
For x = 5, following 2 values
of 'a' satisfy the conditions:
5 XOR 0 = 5+0
5 XOR 2 = 5+2 

Input : 10
Output : 4
Explanation: 
For x = 10, following 4 values
of 'a' satisfy the conditions:
10 XOR 0 = 10+0
10 XOR 1 = 10+1
10 XOR 4 = 10+4
10 XOR 5 = 10+5

天真的方法：
一种简单的方法是检查“ a”在0到“ x”(包括两个端点)之间的所有值，并计算与x的XOR，并检查条件2是否满足。

下面是上述想法的实现：

C++

// C++ program to find count of values whose XOR
// with x is equal to the sum of value and x
// and values are smaller than equal to x
#include 
using namespace std;
 
int FindValues(int x)
{
    // Initialize result
    int count = 0;
 
    // Traversing through all values between
    // 0 and x both inclusive and counting
    // numbers that satisfy given property
    for (int i = 0; i <= x; i++)
        if ((x + i) == (x ^ i))
            count++;
 
    return count;
}
 
// Driver code
int main()
{
    int x = 10;
   
    // Function call
    cout << FindValues(x);
    return 0;
}

Java

// Java program to find count of values whose XOR
// with x is equal to the sum of value and x
// and values are smaller than equal to x
 
class Fib
{
    static int FindValues(int x)
    {
        // Initialize result
        int count = 0;
 
        // Traversing through all values between
        // 0 and x both inclusive and counting
        // numbers that satisfy given property
        for (int i = 0; i <= x; i++)
            if ((x + i) == (x ^ i))
                count++;
 
        return count;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int x = 10;
       
        // Function call
        System.out.println(FindValues(x));
    }
}

Python3

# Python3 program to find count of
# values whose XOR with x is equal
# to the sum of value and x and
# values are smaller than equal to x
 
 
def FindValues(x):
 
    # Initialize result
    count = 0
 
    # Traversing through all values
    # between 0 and x both inclusive
    # and counting numbers that
    # satisfy given property
    for i in range(0, x):
        if ((x + i) == (x ^ i)):
            count = count + 1
 
    return count
 
 
# Driver code
x = 10
 
# Function call
print(FindValues(x))
 
# This code is contributed
# by Shivi_Aggarwal

C#

// C# program to find count of values whose XOR
// with x is equal to the sum of value and x
// and values are smaller than equal to x
using System;
 
class Fib
{
    static int FindValues(int x)
    {
        // Initialize result
        int count = 0;
 
        // Traversing through all values between
        // 0 and x both inclusive and counting
        // numbers that satisfy given property
        for (int i = 0; i <= x; i++)
            if ((x + i) == (x ^ i))
                count++;
 
        return count;
    }
 
    // Driver code
    public static void Main()
    {
        int x = 10;
        
        // Function call
        Console.Write(FindValues(x));
    }
}
 
// This code is contributed by Nitin Mittal.

输出

4

时间复杂度： O(x)。

高效方法：
XOR模拟二进制加法，而不会结转到下一位。对于’a’的零位，我们可以加1或0而不会得到进位，这意味着xor = +，而如果’a’中的位为1，则x中的匹配位被强制为0，以避免携带。对于x中匹配数字中的’a’中的每个0，其总组合计数为2 ^(零个数)可以是1或0。因此，我们只需要以数字的二进制表示形式来计算0的数量，答案将为2 ^(零的数量)。

下面是上述想法的实现：

C++

// C++ program to count numbers whose bitwise
// XOR and sum with x are equal
#include 
using namespace std;
 
// Function to find total 0 bit in a number
long CountZeroBit(long x)
{
    unsigned int count = 0;
    while (x)
    {
       if (!(x & 1LL))
           count++;
       x >>= 1LL;
    }
    return count;
}
 
// Function to find Count of non-negative numbers
// less than or equal to x, whose bitwise XOR and
// SUM with x are equal.
long CountXORandSumEqual(long x)
{
    // count number of zero bit in x
    long count = CountZeroBit(x);
 
    // power of 2 to count
    return (1LL << count);
}
 
// Driver code
int main()
{
   long x = 10;
   
   // Function call
   cout << CountXORandSumEqual(x);
   return 0;
}

Java

// Java program to count
// numbers whose bitwise
// XOR and sum with x
// are equal
import java.io.*;
 
class GFG {
 
    // Function to find total
    // 0 bit in a number
    static long CountZeroBit(long x)
    {
        long count = 0;
        while (x > 0) {
            if ((x & 1L) == 0)
                count++;
            x >>= 1L;
        }
        return count;
    }
 
    // Function to find Count
    // of non-negative numbers
    // less than or equal to x,
    // whose bitwise XOR and
    // SUM with x are equal.
    static long CountXORandSumEqual(long x)
    {
        // count number of
        // zero bit in x
        long count = CountZeroBit(x);
 
        // power of 2 to count
        return (1L << count);
    }
 
    // Driver code
    public static void main(String[] args)
    {
 
        long x = 10;
 
        // Function call
        System.out.println(CountXORandSumEqual(x));
    }
}
 
// The code is contributed by ajit

Python3

# Python3 program to count numbers whose bitwise
# XOR and sum with x are equal
 
# Function to find total 0 bit in a number
 
 
def CountZeroBit(x):
 
    count = 0
    while (x):
 
        if ((x & 1) == 0):
            count += 1
        x >>= 1
 
    return count
 
# Function to find Count of non-negative numbers
# less than or equal to x, whose bitwise XOR and
# SUM with x are equal.
 
 
def CountXORandSumEqual(x):
 
    # count number of zero bit in x
    count = CountZeroBit(x)
 
    # power of 2 to count
    return (1 << count)
 
 
# Driver code
if __name__ == '__main__':
    x = 10
     
    # Function call
    print(CountXORandSumEqual(x))
 
# This code is contributed by 29AjayKumar

C＃

// C# program to count
// numbers whose bitwise
// XOR and sum with x
// are equal
using System;
 
class GFG {
 
    // Function to find total
    // 0 bit in a number
    static int CountZeroBit(int x)
    {
        int count = 0;
        while (x > 0) {
            if ((x & 1) == 0)
                count++;
            x >>= 1;
        }
        return count;
    }
 
    // Function to find Count
    // of non-negative numbers
    // less than or equal to x,
    // whose bitwise XOR and
    // SUM with x are equal.
    static int CountXORandSumEqual(int x)
    {
        // count number of
        // zero bit in x
        int count = CountZeroBit(x);
 
        // power of 2 to count
        return (1 << count);
    }
 
    // Driver code
    static public void Main()
    {
        int x = 10;
       
        // Function call
        Console.WriteLine(CountXORandSumEqual(x));
    }
}
 
// The code is contributed by ajit

的PHP

>= 1;
    }
    return $count;
}
 
// Function to find Count of
// non-negative numbers less
// than or equal to x, whose
// bitwise XOR and SUM with
// x are equal.
function CountXORandSumEqual($x)
{
    // count number of zero bit in x
    $count = CountZeroBit($x);
 
    // power of 2 to count
    return (1 << $count);
}
 
    // Driver code
    $x = 10;
 
    // Function call
    echo CountXORandSumEqual($x);
 
// This code is contributed by m_kit
?>

输出

4

时间复杂度： O(Log x)