所有偶数到 N 的按位“与”
给定一个整数 N ，任务是找到从 1 到 N 的所有偶数的按位 and ( &)
示例:
输入:2 T3】输出: 2
输入: 10 输出: 0 解释:2、4、6、8、10 的位与为 0。
天真方法:将结果初始化为 2。从 4 到 n 迭代循环(对于所有偶数)并通过按位“与”(&)更新结果。 以下是实施办法:
C++
// C++ implementation of the above approach
#include <iostream>
using namespace std;
// Function to return the bitwise &
// of all the even numbers upto N
int bitwiseAndTillN(int n)
{
// Initialize result as 2
int result = 2;
for (int i = 4; i <= n; i = i + 2) {
result = result & i;
}
return result;
}
// Driver code
int main()
{
int n = 2;
cout << bitwiseAndTillN(n);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation of the above approach
class GFG
{
// Function to return the bitwise &
// of all the even numbers upto N
static int bitwiseAndTillN(int n)
{
// Initialize result as 2
int result = 2;
for (int i = 4; i <= n; i = i + 2)
{
result = result & i;
}
return result;
}
// Driver code
public static void main (String[] args)
{
int n = 2;
System.out.println(bitwiseAndTillN(n));
}
}
// This code is contributed by AnkitRai01
Python 3
# Python3 implementation of the above approach
# Function to return the bitwise &
# of all the even numbers upto N
def bitwiseAndTillN(n) :
# Initialize result as 2
result = 2;
for i in range(4, n + 1, 2) :
result = result & i;
return result;
# Driver code
if __name__ == "__main__" :
n = 2;
print(bitwiseAndTillN(n));
# This code is contributed by AnkitRai01
C
// C# implementation of the above approach
using System;
class GFG
{
// Function to return the bitwise &
// of all the even numbers upto N
static int bitwiseAndTillN(int n)
{
// Initialize result as 2
int result = 2;
for (int i = 4; i <= n; i = i + 2)
{
result = result & i;
}
return result;
}
// Driver code
public static void Main()
{
int n = 2;
Console.WriteLine(bitwiseAndTillN(n));
}
}
// This code is contributed by AnkitRai01
java 描述语言
<script>
// Javascript implementation of the above approach
// Function to return the bitwise &
// of all the even numbers upto N
function bitwiseAndTillN(n) {
// Initialize result as 2
let result = 2;
for (let i = 4; i <= n; i = i + 2) {
result = result & i;
}
return result;
}
// Driver code
let n = 2;
document.write(bitwiseAndTillN(n));
</script>
Output:
2
高效方式:高效方式是 N 小于 4 返回 2，N 全部返回 0>= 4，因为 2 和 4 的按位 and 为 0，0 与任意数字的按位 and 为 0。 以下是实施办法:
C++
// C++ implementation of the above approach
#include <iostream>
using namespace std;
// Function to return the bitwise &
// of all the numbers upto N
int bitwiseAndTillN(int n)
{
if (n < 4)
return 2;
else
return 0;
}
int main()
{
int n = 2;
cout << bitwiseAndTillN(n);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation of the above approach
class GFG
{
// Function to return the bitwise &
// of all the numbers upto N
static int bitwiseAndTillN(int n)
{
if (n < 4)
return 2;
else
return 0;
}
// Driver code
public static void main (String[] args)
{
int n = 2;
System.out.println(bitwiseAndTillN(n));
}
}
// This code is contributed by AnkitRai01
Python 3
# Python3 implementation of the above approach
# Function to return the bitwise &
# of all the numbers upto N
def bitwiseAndTillN( n):
if (n < 4):
return 2
else:
return 0
# Driver code
n = 2
print(bitwiseAndTillN(n))
# This code is contributed by ANKITKUMAR34
C
// C# implementation of the above approach
using System;
class GFG
{
// Function to return the bitwise &
// of all the numbers upto N
static int bitwiseAndTillN(int n)
{
if (n < 4)
return 2;
else
return 0;
}
// Driver code
public static void Main()
{
int n = 2;
Console.WriteLine(bitwiseAndTillN(n));
}
}
// This code is contributed by AnkitRai01
java 描述语言
<script>
// JavaScript implementation of the above approach
// Function to return the bitwise &
// of all the numbers upto N
function bitwiseAndTillN(n)
{
if (n < 4)
return 2;
else
return 0;
}
// driver code
let n = 2;
document.write (bitwiseAndTillN(n));
// this code is contributed by shivanisinghss2110
</script>
Output:
2
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