给定一个正整数N ，任务是找到前N 个自然数的排列，使得相邻元素对的按位 AND( & ) 的乘积大于0 。如果没有找到这样的排列，则打印“不可能” 。
例子：
Input: N = 3
Output: 1 3 2
Explanation:
1 & 3 = 1
3 & 2 = 2
Product of bitwise AND of adjacent elements = 1 * 2 = 2(>0)
Therefore, the required output is 1 3 2
Input: 2
Output: Not Possible
Explanation:
Possible permutations of first N(= 2) natural numbers are: { { 1, 2 }, { 2, 1 } }
1 & 2 = 0
2 & 1 = 0
Therefore, the required output is Not Possible.
朴素方法：解决此问题的最简单方法是生成前 N 个自然数的所有可能排列。对于每个排列，检查其相邻元素的按位与的乘积是否大于0 。如果发现为真，则打印该排列。否则，打印“不可能” 。
时间复杂度： O(N * N!)
辅助空间： O(1)
有效的方法：可以根据以下观察来解决问题：
If N is a power of 2, then the bitwise AND of N with any number less than N must be 0.
If bitwise AND of adjacent elements is equal to 0, then the product of bitwise AND of its adjacent element must be 0.
请按照以下步骤解决问题：
- 如果N是2的幂，则打印“不可能” 。
- 初始化一个数组，比如arr[]来存储满足给定条件的前N 个自然数的排列。
- 迭代范围[1, N] ，并更新arr[i] = i
- 更新数组的前三个元素，使其相邻元素的按位与必须大于0 ，即arr[1] = 2 ， arr[2] = 3和arr[3] = 1
- 迭代范围[4, N] 。对于每个第i个值，检查i是否是2的幂。如果发现为真，则swap(arr[i], arr[i+1]) 。
- 最后，打印arr[] 。
下面是上述方法的实现：
C++
// C++ program to implement
// the above approach
#include
using namespace std;
// Function to check if a number
// is a power of 2 or not
bool isPowerOfTwo(int n)
{
if (n == 0)
return false;
return (ceil(log2(n))
== floor(log2(n)));
}
// Function to find the permutation of first N
// natural numbers such that the product of
// bitwise AND of adjacent elements is > 0
void findThePermutation(int N)
{
// Base Case, If N = 1, print 1
if (N == 1) {
cout << "1";
return;
}
// If N is a power of 2,
// print "Not Possible"
if (isPowerOfTwo(N)) {
cout << "Not Possible";
return;
}
// Stores permutation of first N
// natural numbers that satisfy
// the condition
int arr[N + 1];
// Iterate over the range [1, N]
for (int i = 1; i <= N; i++) {
// Update arr[i]
arr[i] = i;
}
// Update arr[1], arr[2] and arr[3]
arr[1] = 2, arr[2] = 3, arr[3] = 1;
// Iterate over the range [4, N]
for (int i = 4; i <= N; i++) {
// If i is a power of 2
if (isPowerOfTwo(i)) {
// Swap adjacent elements
swap(arr[i], arr[i + 1]);
// Update i
i++;
}
}
// Print the array
for (int i = 1; i <= N; i++)
cout << arr[i] << " ";
}
// Driver Code
int main()
{
// Input
int N = 5;
// Function Call
findThePermutation(N);
return 0;
} Java
// Java program to implement the above approach
class GFG
{
// Function to calculate the
// log base 2 of an integer
public static int log2(int N)
{
// calculate log2 N indirectly
// using log() method
int result = (int)(Math.log(N) / Math.log(2));
return result;
}
// Function to check if a number
// is a power of 2 or not
static boolean isPowerOfTwo(int n)
{
if (n == 0)
return false;
return Math.ceil(log2(n)) == Math.floor(log2(n));
}
// Function to find the permutation of first N
// natural numbers such that the product of
// bitwise AND of adjacent elements is > 0
static void findThePermutation(int N)
{
// Base Case, If N = 1, print 1
if (N == 1)
{
System.out.print("1");
return;
}
// If N is a power of 2,
// print "Not Possible"
if (isPowerOfTwo(N) == false)
{
System.out.print("Not Possible");
return;
}
// Stores permutation of first N
// natural numbers that satisfy
// the condition
int arr[] = new int[N + 1];
// Iterate over the range [1, N]
for (int i = 1; i <= N; i++)
{
// Update arr[i]
arr[i] = i;
}
// Update arr[1], arr[2] and arr[3]
arr[1] = 2; arr[2] = 3; arr[3] = 1;
int temp;
// Iterate over the range [4, N]
for (int i = 4; i <= N; i++)
{
// If i is a power of 2
if (isPowerOfTwo(i) == true)
{
// Swap adjacent elements
temp = arr[i];
arr[i] = arr[i + 1];
arr[i + 1] = temp ;
// Update i
i++;
}
}
// Print the array
for (int i = 1; i <= N; i++)
System.out.print(arr[i] + " ");
}
// Driver Code
public static void main(String[] args)
{
// Input
int N = 5;
// Function Call
findThePermutation(N);
}
}
// This code is contributed by AnkThonPython3
# Python3 program to implement
# the above approach
from math import ceil, floor, log2
# Function to check if a number
# is a power of 2 or not
def isPowerOfTwo(n):
if (n == 0):
return False
return (ceil(log2(n)) == floor(log2(n)))
# Function to find the permutation of first N
# natural numbers such that the product of
# bitwise AND of adjacent elements is > 0
def findThePermutation(N):
# Base Case, If N = 1, pr1
if (N == 1):
print("1")
return
# If N is a power of 2,
# print "Not Possible"
if (isPowerOfTwo(N)):
print("Not Possible")
return
# Stores permutation of first N
# natural numbers that satisfy
# the condition
arr = [i for i in range(N + 1)]
# Iterate over the range [1, N]
for i in range(1, N + 1):
# Update arr[i]
arr[i] = i
# Update arr[1], arr[2] and arr[3]
arr[1], arr[2], arr[3] = 2, 3, 1
# Iterate over the range [4, N]
for i in range(4, N + 1):
# If i is a power of 2
if (isPowerOfTwo(i)):
# Swap adjacent elements
arr[i], arr[i + 1] = arr[i + 1], arr[i]
# Update i
i += 1
# Print the array
for i in range(1, N + 1):
print(arr[i], end = " ")
# Driver Code
if __name__ == '__main__':
# Input
N = 5
# Function Call
findThePermutation(N)
# This code is contributed by mohit kumar 29C#
// C# program to implement the above approach
using System;
class GFG
{
// Function to calculate the
// log base 2 of an integer
public static int log2(int N)
{
// calculate log2 N indirectly
// using log() method
int result = (int)(Math.Log(N) / Math.Log(2));
return result;
}
// Function to check if a number
// is a power of 2 or not
static bool isPowerOfTwo(int n)
{
if (n == 0)
return false;
return Math.Ceiling((double)log2(n)) ==
Math.Floor((double)log2(n));
}
// Function to find the permutation of first N
// natural numbers such that the product of
// bitwise AND of adjacent elements is > 0
static void findThePermutation(int N)
{
// Base Case, If N = 1, print 1
if (N == 1)
{
Console.Write("1");
return;
}
// If N is a power of 2,
// print "Not Possible"
if (isPowerOfTwo(N) == false)
{
Console.Write("Not Possible");
return;
}
// Stores permutation of first N
// natural numbers that satisfy
// the condition
int []arr = new int[N + 1];
// Iterate over the range [1, N]
for (int i = 1; i <= N; i++)
{
// Update arr[i]
arr[i] = i;
}
// Update arr[1], arr[2] and arr[3]
arr[1] = 2; arr[2] = 3; arr[3] = 1;
int temp;
// Iterate over the range [4, N]
for (int i = 4; i <= N; i++)
{
// If i is a power of 2
if (isPowerOfTwo(i) == true)
{
// Swap adjacent elements
temp = arr[i];
arr[i] = arr[i + 1];
arr[i + 1] = temp ;
// Update i
i++;
}
}
// Print the array
for (int i = 1; i <= N; i++)
Console.Write(arr[i] + " ");
}
// Driver Code
public static void Main(String[] args)
{
// Input
int N = 5;
// Function Call
findThePermutation(N);
}
}
// This code is contributed by 29AjayKumarJavascript
输出：
2 3 1 5 4时间复杂度： O(N)
辅助空间： O(N)
如果您想与行业专家一起参加直播课程，请参阅Geeks Classes Live