给定数字N,任务是找到X的计数,以使N XOR X == N OR X ,其中0 <= X <= N
例子:
Input: N = 5
Output: 2
For N = 5,
5 XOR 2 == 5 OR 2
5 XOR 0 == 5 OR 0
Thus, count is 2.
Input: N = 7
Output: 1
For N = 7,
7 XOR 0 == 7 OR 0
Thus, count is 1.
方法:想法是将给定的数字转换为二进制,然后计算其中的未设置位。 2 ^ count给我们X的数量,使得N XOR X == N OR X。
下面是上述方法的实现:
C++
// C++ program to find
// the XOR equals OR count
#include
#include
using namespace std;
class gfg {
// Function to calculate count
// of numbers with XOR equals OR
public:
int xorEqualsOrCount(int N)
{
// variable to store count of unset bits
int count = 0;
int bit;
while (N > 0) {
bit = N % 2;
if (bit == 0)
count++;
N = N / 2;
}
return (int)pow(2, count);
} };
// Driver code
int main()
{
gfg g ;
int N = 7;
cout<
Java
// Java program to find the XOR equals OR count
import java.io.*;
import java.util.*;
class GFG {
// Function to calculate count of numbers with XOR equals OR
static int xorEqualsOrCount(int N)
{
// variable to store count of unset bits
int count = 0;
int bit;
while (N > 0) {
bit = N % 2;
if (bit == 0)
count++;
N = N / 2;
}
return (int)Math.pow(2, count);
}
// Driver code
public static void main(String args[])
{
int N = 7;
System.out.println(xorEqualsOrCount(N));
}
}
Python3
# Python3 program to find
# the XOR equals OR count
# Function to calculate count
# of numbers with XOR equals OR
def xorEqualsOrCount(N) :
# variable to store
# count of unset bits
count = 0
while(N > 0) :
bit = N % 2
if bit == 0 :
count += 1
N //= 2
return int(pow(2, count))
# Driver code
if __name__ == "__main__" :
N = 7
print(xorEqualsOrCount(N))
# This code is contributed by
# ANKITRAI1
C#
// C# program to find
// the XOR equals OR count
using System;
class GFG {
// Function to calculate count
// of numbers with XOR equals OR
static int xorEqualsOrCount(int N)
{
// variable to store count of unset bits
int count = 0;
int bit;
while (N > 0) {
bit = N % 2;
if (bit == 0)
count++;
N = N / 2;
}
return (int)Math.Pow(2, count);
}
// Driver code
public static void Main()
{
int N = 7;
Console.WriteLine(xorEqualsOrCount(N));
}
}
// This code is contributed by inder_verma..
PHP
0)
{
$bit = $N % 2;
if ($bit == 0)
$count++;
$N = intval($N / 2);
}
return pow(2, $count);
}
// Driver code
$N = 7;
echo xorEqualsOrCount($N);
// This code is contributed
// by ChitraNayal
?>
Javascript
输出:
1
如果您希望与行业专家一起参加现场课程,请参阅《 Geeks现场课程》和《 Geeks现场课程美国》。