给定一个整数 。查找解决方案的数量满足等式:
a = b + (a^b)
例子:
Input: a = 4
Output: 2
The only values of b are 0 and 4 itself.
Input: a = 3
Output: 4
一个幼稚的解决方案是从0迭代到并计算满足给定方程的值的数量。我们只需要走到因为任何大于将给出XOR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
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
// 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
Python3
# 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
PHP
Javascript
C++
// C++ program to find the number of values
// of b such that a = b + (a^b)
#include
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
// 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
Python3
# 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
PHP
Javascript
输出:
4
时间复杂度:O(a)
一种有效的方法是,当我们针对不同的值编写可能的解决方案时,观察答案的模式。 。仅将设置的位用于确定可能的答案的数量。问题的答案将始终是2 ^(设置位数),可以通过观察确定。
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
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
// 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
Python3
# 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
的PHP
Java脚本
输出:
4
时间复杂度: O(log N)
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