从 1 到 N 的所有奇数的按位“与”
给定一个整数 N ，任务是从范围【1，N】中找到所有奇数的位“与”(&)。
示例:
输入: N = 7 输出: 1 (1 & 3 & 5 & 7) = 1
输入:N = 1 T3】输出: 1
天真的做法:从 1 开始，对所有奇数 ≤ N 进行逐位 AND 运算。
下面是上述方法的实现:
C++
// C++ implementation of the approach
#include <iostream>
using namespace std;
// Function to return the bitwise AND
// of all the odd integers from
// the range [1, n]
int bitwiseAndOdd(int n)
{
// Initialize result to 1
int result = 1;
// Starting from 3, bitwise AND
// all the odd integers less
// than or equal to n
for (int i = 3; i <= n; i = i + 2) {
result = (result & i);
}
return result;
}
// Driver code
int main()
{
int n = 10;
cout << bitwiseAndOdd(n);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation of the approach
class GFG
{
// Function to return the bitwise AND
// of all the odd integers from
// the range [1, n]
static int bitwiseAndOdd(int n)
{
// Initialize result to 1
int result = 1;
// Starting from 3, bitwise AND
// all the odd integers less
// than or equal to n
for (int i = 3; i <= n; i = i + 2)
{
result = (result & i);
}
return result;
}
// Driver code
public static void main (String[] args)
{
int n = 10;
System.out.println(bitwiseAndOdd(n));
}
}
// This code is contributed by AnkitRai01
Python 3
# Python3 implementation of the approach
# Function to return the bitwise AND
# of all the odd integers from
# the range [1, n]
def bitwiseAndOdd(n) :
# Initialize result to 1
result = 1;
# Starting from 3, bitwise AND
# all the odd integers less
# than or equal to n
for i in range(3, n + 1, 2) :
result = (result & i);
return result;
# Driver code
if __name__ == "__main__" :
n = 10;
print(bitwiseAndOdd(n));
# This code is contributed by AnkitRai01
C
// C# implementation of the approach
using System;
class GFG
{
// Function to return the bitwise AND
// of all the odd integers from
// the range [1, n]
static int bitwiseAndOdd(int n)
{
// Initialize result to 1
int result = 1;
// Starting from 3, bitwise AND
// all the odd integers less
// than or equal to n
for (int i = 3; i <= n; i = i + 2)
{
result = (result & i);
}
return result;
}
// Driver code
public static void Main()
{
int n = 10;
Console.WriteLine(bitwiseAndOdd(n));
}
}
// This code is contributed by AnkitRai01
java 描述语言
<script>
// Javascript implementation of the approach
// Function to return the bitwise AND
// of all the odd integers from
// the range [1, n]
function bitwiseAndOdd(n)
{
// Initialize result to 1
var result = 1;
// Starting from 3, bitwise AND
// all the odd integers less
// than or equal to n
for(var i = 3; i <= n; i = i + 2)
{
result = (result & i);
}
return result;
}
// Driver code
var n = 10;
document.write(bitwiseAndOdd(n));
// This code is contributed by noob2000
</script>
Output:
1
时间复杂度: O(n)
辅助空间: O(1)
有效方法:如果整数的最低有效位为 1(在这种情况下是所有奇数)，则按位与 1 的“与”运算结果总是为 1。所以，所有情况下的结果都是 1。
下面是上述方法的实现:
C++
// C++ implementation of the approach
#include <iostream>
using namespace std;
// Function to return the bitwise AND
// of all the odd integers from
// the range [1, n]
int bitwiseAndOdd(int n)
{
return 1;
}
// Driver code
int main()
{
int n = 10;
cout << bitwiseAndOdd(n);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation of the approach
class GFG
{
// Function to return the bitwise AND
// of all the odd integers from
// the range [1, n]
static int bitwiseAndOdd(int n)
{
return 1;
}
// Driver code
public static void main (String[] args)
{
int n = 10;
System.out.println(bitwiseAndOdd(n));
}
}
// This code is contributed by AnkitRai01
Python 3
# Python 3 implementation of the approach
# Function to return the bitwise AND
# of all the odd integers from
# the range [1, n]
def bitwiseAndOdd(n):
return 1
# Driver code
n = 10
print(bitwiseAndOdd(n))
# This code is contributed by ApurvaRaj
C
// C# implementation of the approach
using System;
class GFG
{
// Function to return the bitwise AND
// of all the odd integers from
// the range [1, n]
static int bitwiseAndOdd(int n)
{
return 1;
}
// Driver code
public static void Main()
{
int n = 10;
Console.WriteLine(bitwiseAndOdd(n));
}
}
// This code is contributed by AnkitRai01
java 描述语言
<script>
// Javascript implementation of the approach
// Function to return the bitwise AND
// of all the odd integers from
// the range [1, n]
function bitwiseAndOdd(n)
{
return 1;
}
// Driver code
var n = 10;
document.write( bitwiseAndOdd(n));
//This code is contributed by SoumikMondal
</script>
Output:
1
时间复杂度: O(1)
辅助空间: O(1)
版权属于：月萌API www.moonapi.com，转载请注明出处
