n = x+n⊕x 的解的数量
给定一个数 n，我们必须找到 x 的可能值的个数，使得 n = x + n ⊕ x。这里⊕表示 XOR
例:
Input : n = 3
Output : 4
The possible values of x are 0, 1, 2, and 3.
Input : n = 2
Output : 2
The possible values of x are 0 and 2.
蛮力逼近:我们可以看到 x 总是等于或小于 n，所以我们可以在[0，n]的范围内迭代，统计满足要求条件的值的个数。这种方法的时间复杂度是 O(n)。
c++
// C++ implementation of above approach
#include <bits/stdc++.h>
using namespace std;
// Function to find the number of
// solutions of n = n xor x
int numberOfSolutions(int n)
{
// Counter to store the number
// of solutions found
int c = 0;
for (int x = 0; x <= n; ++x)
if (n == x + n ^ x)
++c;
return c;
}
// Driver code
int main()
{
int n = 3;
cout << numberOfSolutions(n);
return 0;
}
Java
// Java implementation of above approach
import java.util.*;
import java.lang.*;
class GFG
{
// Function to find the number of
// solutions of n = n xor x
static int numberOfSolutions(int n)
{
// Counter to store the number
// of solutions found
int c = 0;
for (int x = 0; x <= n; ++x)
if (n == x + (n ^ x))
++c;
return c;
}
// Driver code
public static void main(String args[])
{
int n = 3;
System.out.print(numberOfSolutions(n));
}
}
// This code is contributed
// by Akanksha Rai(Abby_akku)
python 3
# Python 3 implementation
# of above approach
# Function to find the number of
# solutions of n = n xor x
def numberOfSolutions(n):
# Counter to store the number
# of solutions found
c = 0
for x in range(n + 1):
if (n ==( x +( n ^ x))):
c += 1
return c
# Driver code
if __name__ == "__main__":
n = 3
print(numberOfSolutions(n))
# This code is contributed
# by ChitraNayal
c
// C# implementation of above approach
using System;
class GFG
{
// Function to find the number of
// solutions of n = n xor x
static int numberOfSolutions(int n)
{
// Counter to store the number
// of solutions found
int c = 0;
for (int x = 0; x <= n; ++x)
if (n == x + (n ^ x))
++c;
return c;
}
// Driver code
public static void Main()
{
int n = 3;
Console.Write(numberOfSolutions(n));
}
}
// This code is contributed
// by Akanksha Rai(Abby_akku)
PHP
<?php
// PHP implementation of above approach
// Function to find the number of
// solutions of n = n xor x
function numberOfSolutions($n)
{
// Counter to store the number
// of solutions found
$c = 0;
for ($x = 0; $x <= $n; ++$x)
if ($n == $x + $n ^ $x)
++$c;
return $c;
}
// Driver code
$n = 3;
echo numberOfSolutions($n);
// This code is contributed
// by Akanksha Rai(Abby_akku)
Javascript
<script>
// Javascript implementation of above approach
// Function to find the number of
// solutions of n = n xor x
function numberOfSolutions(n)
{
// Counter to store the number
// of solutions found
let c = 0;
for(let x = 0; x <= n; ++x)
if (n == x + n ^ x)
++c;
return c;
}
let n = 3;
document.write(numberOfSolutions(n));
// This code is contributed by divyesh072019.
</script>
输出:
4
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