使得 a = b + (a^b)的 b 的值的数量
给定一个整数
。求
的解数，满足方程:
a = b + (a^b)
示例:
Input: a = 4
Output: 2
The only values of b are 0 and 4 itself.
Input: a = 3
Output: 4
一个天真的解决方案是从 0 迭代到
，并计算满足给定方程的值的数量。我们只需要遍历到，因为任何大于的值都会给出异或值> ，因此不能满足等式。
下面是上述方法的实现:
C++
// C++ program to find the number of values
// of b such that a = b + (a^b)
#include <bits/stdc++.h>
using namespace std;
// function to return the number of solutions
int countSolutions(int a)
{
int count = 0;
// check for every possible value
for (int i = 0; i <= a; i++) {
if (a == (i + (a ^ i)))
count++;
}
return count;
}
// Driver Code
int main()
{
int a = 3;
cout << countSolutions(a);
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to find the number of values
// of b such that a = b + (a^b)
import java.io.*;
class GFG {
// function to return the number of solutions
static int countSolutions(int a)
{
int count = 0;
// check for every possible value
for (int i = 0; i <= a; i++) {
if (a == (i + (a ^ i)))
count++;
}
return count;
}
// Driver Code
public static void main (String[] args) {
int a = 3;
System.out.println( countSolutions(a));
}
}
// This code is contributed by inder_verma
Python 3
# Python 3 program to find
# the number of values of b
# such that a = b + (a^b)
# function to return the
# number of solutions
def countSolutions(a):
count = 0
# check for every possible value
for i in range(a + 1):
if (a == (i + (a ^ i))):
count += 1
return count
# Driver Code
if __name__ == "__main__":
a = 3
print(countSolutions(a))
# This code is contributed
# by ChitraNayal
C
// C# program to find the number of
// values of b such that a = b + (a^b)
using System;
class GFG
{
// function to return the
// number of solutions
static int countSolutions(int a)
{
int count = 0;
// check for every possible value
for (int i = 0; i <= a; i++)
{
if (a == (i + (a ^ i)))
count++;
}
return count;
}
// Driver Code
public static void Main ()
{
int a = 3;
Console.WriteLine(countSolutions(a));
}
}
// This code is contributed by inder_verma
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP program to find the number of
// values of b such that a = b + (a^b)
// function to return the
// number of solutions
function countSolutions($a)
{
$count = 0;
// check for every possible value
for ($i = 0; $i <= $a; $i++)
{
if ($a == ($i + ($a ^ $i)))
$count++;
}
return $count;
}
// Driver Code
$a = 3;
echo countSolutions($a);
// This code is contributed
// by inder_verma
?>
java 描述语言
<script>
// Javascript program to find the number of values
// of b such that a = b + (a^b)
// function to return the number of solutions
function countSolutions(a)
{
let count = 0;
// check for every possible value
for (let i = 0; i <= a; i++) {
if (a == (i + (a ^ i)))
count++;
}
return count;
}
// Driver Code
let a = 3;
document.write(countSolutions(a));
</script>
Output:
4
时间复杂度 : O(a)
一个有效的方法是当我们为不同的 T2 值写可能的答案时，观察答案的模式。只有设置位用于确定可能答案的数量。这个问题的答案永远是 2^(number，它可以通过观察来确定。
下面是上述方法的实现:
C++
// C++ program to find the number of values
// of b such that a = b + (a^b)
#include <bits/stdc++.h>
using namespace std;
// function to return the number of solutions
int countSolutions(int a)
{
int count = __builtin_popcount(a);
count = pow(2, count);
return count;
}
// Driver Code
int main()
{
int a = 3;
cout << countSolutions(a);
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to find the number of values
// of b such that a = b + (a^b)
import java.io.*;
class GFG
{
// function to return the number of solutions
static int countSolutions(int a)
{
int count = Integer.bitCount(a);
count =(int) Math.pow(2, count);
return count;
}
// Driver Code
public static void main (String[] args)
{
int a = 3;
System.out.println(countSolutions(a));
}
}
// This code is contributed by Raj
Python 3
# Python3 program to find the number
# of values of b such that a = b + (a^b)
# function to return the number
# of solutions
def countSolutions(a):
count = bin(a).count('1')
return 2 ** count
# Driver Code
if __name__ == "__main__":
a = 3
print(countSolutions(a))
# This code is contributed by
# Rituraj Jain
C
// C# program to find the number of
// values of b such that a = b + (a^b)
class GFG
{
// function to return the number
// of solutions
static int countSolutions(int a)
{
int count = bitCount(a);
count =(int) System.Math.Pow(2, count);
return count;
}
static int bitCount(int n)
{
int count = 0;
while (n != 0)
{
count++;
n &= (n - 1);
}
return count;
}
// Driver Code
public static void Main()
{
int a = 3;
System.Console.WriteLine(countSolutions(a));
}
}
// This code is contributed by mits
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP program to find the number of
// values of b such that a = b + (a^b)
// function to return the number
// of solutions
function countSolutions($a)
{
$count = bitCount($a);
$count = (int)pow(2, $count);
return $count;
}
function bitCount($n)
{
$count = 0;
while ($n != 0)
{
$count++;
$n &= ($n - 1);
}
return $count;
}
// Driver Code
$a = 3;
echo (countSolutions($a));
// This code is contributed by ajit
?>
java 描述语言
<script>
// Javascript program to find the number
// of values of b such that a = b + (a^b)
function bitCount(n)
{
let count = 0;
while (n != 0)
{
count++;
n &= (n - 1);
}
return count;
}
// Function to return the number of solutions
function countSolutions(a)
{
let count = bitCount(a);
count = Math.pow(2, count);
return count;
}
// Driver Code
let a = 3;
document.write(countSolutions(a));
// This code is contributed by subhammahato348
</script>
Output:
4
时间复杂度: O(log N)
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