找出 N 位的可能排列
原文:https://www . geeksforgeeks . org/find-the-the-the-the-of-bit-of-n/
给定一个整数 N ，任务是找出 N 的位是否可以交替排列，即 0101… 或 10101… 。假设 N 表示为一个 32 位整数。 示例:
输入: N = 23 输出:否 “0000000000000000000000000000000000111”是 23 的二进制表示，所需的比特排列是不可能的。 输入: N = 524280 输出:是 二进制(524280)=“0000000000011111111111000”，可通过 重排为“01010101010101010101010101010101”。
方法:由于给定的整数必须用 32 位表示，并且 1 的数量必须等于其二进制表示中的 0 的数量，才能满足给定的条件。因此，N 中的设置位数必须为 16，这可以使用 __builtin_popcount() 轻松计算。以下是上述方法的实现:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
const int TOTAL_BITS = 32;
// Function that returns true if it is
// possible to arrange the bits of
// n in alternate fashion
bool isPossible(int n)
{
// To store the count of 1s in the
// binary representation of n
int cnt = __builtin_popcount(n);
// If the number set bits and the
// number of unset bits is equal
if (cnt == TOTAL_BITS / 2)
return true;
return false;
}
// Driver code
int main()
{
int n = 524280;
if (isPossible(n))
cout << "Yes";
else
cout << "No";
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation of the approach
import java.util.*;
class GFG
{
static int TOTAL_BITS = 32;
// Function that returns true if it is
// possible to arrange the bits of
// n in alternate fashion
static boolean isPossible(int n)
{
// To store the count of 1s in the
// binary representation of n
int cnt = Integer.bitCount(n);
// If the number set bits and the
// number of unset bits is equal
if (cnt == TOTAL_BITS / 2)
return true;
return false;
}
// Driver code
static public void main (String []arr)
{
int n = 524280;
if (isPossible(n))
System.out.println("Yes");
else
System.out.println("No");
}
}
// This code is contributed by 29AjayKumar
Python 3
# Python3 implementation of the approach
TOTAL_BITS = 32;
# Function that returns true if it is
# possible to arrange the bits of
# n in alternate fashion
def isPossible(n) :
# To store the count of 1s in the
# binary representation of n
cnt = bin(n).count('1');
# If the number set bits and the
# number of unset bits is equal
if (cnt == TOTAL_BITS // 2) :
return True;
return False;
# Driver code
if __name__ == "__main__" :
n = 524280;
if (isPossible(n)) :
print("Yes");
else :
print("No");
# This code is contributed by AnkitRai01
C
// C# implementation of the above approach
using System;
class GFG
{
static int TOTAL_BITS = 32;
static int CountBits(int value)
{
int count = 0;
while (value != 0)
{
count++;
value &= value - 1;
}
return count;
}
// Function that returns true if it is
// possible to arrange the bits of
// n in alternate fashion
static bool isPossible(int n)
{
// To store the count of 1s in the
// binary representation of n
int cnt = CountBits(n);
// If the number set bits and the
// number of unset bits is equal
if (cnt == TOTAL_BITS / 2)
return true;
return false;
}
// Driver code
public static void Main (String []arr)
{
int n = 524280;
if (isPossible(n))
Console.WriteLine("Yes");
else
Console.WriteLine("No");
}
}
// This code is contributed by Mohit kumar
java 描述语言
<script>
// Javascript implementation of the approach
const TOTAL_BITS = 32;
function CountBits(value)
{
let count = 0;
while (value != 0)
{
count++;
value &= value - 1;
}
return count;
}
// Function that returns true if it is
// possible to arrange the bits of
// n in alternate fashion
function isPossible(n)
{
// To store the count of 1s in the
// binary representation of n
let cnt = CountBits(n);
// If the number set bits and the
// number of unset bits is equal
if (cnt == parseInt(TOTAL_BITS / 2))
return true;
return false;
}
// Driver code
let n = 524280;
if (isPossible(n))
document.write("Yes");
else
document.write("No");
// This code is contributed by subhammahato348.
</script>
Output:
Yes
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