给定一个nrr个正整数的数组arr [] 。任务是检查数组的任何子集的按位与是2的幂(即1、2、4、8、16…)。
例子：
Input : n = 3, arr[] = { 12, 13, 7 }
Output : Yes
Subset { 12, 7 } has Biwise AND value 4, which
is power of 2.
Input : n = 2, arr[] = { 10, 20 }
Output : No
注意，要使数字为2的幂，它应该只有1个置1的位。
如果n为1，那么我们只需检查数字是否只有唯一的一位。
对于n大于1，我们的任务是从数组中选择那些按位与导致仅单个位集编号的数字。为此，我们搜索一个位置，在该位置集合中的所有元素都在该位置设置了一位。例如，对于集合{4(100)，6(110)，7(111)}，在位置2(从右到左，从0开始的索引)中，为所有元素设置了位。因此，按位与运算得到4，即2的幂。
以下是此方法的实现：
C++
// CPP Program to check if Bitwise AND of any
// subset is power of two
#include
using namespace std;
const int NUM_BITS = 32;
// Check for power of 2 or not
bool isPowerOf2(int num)
{
return (num && !(num & (num - 1)));
}
// Check if there exist a subset whose bitwise AND
// is power of 2.
bool checkSubsequence(int arr[], int n)
{
// if there is only one element in the set.
if (n == 1)
return isPowerOf2(arr[0]);
// Finding a number with all bit sets.
int total = 0;
for (int i = 0; i < NUM_BITS; i++)
total = total | (1 << i);
// check all the positions at which the bit is set.
for (int i = 0; i < NUM_BITS; i++) {
int ans = total;
for (int j = 0; j < n; j++) {
// include all those elements whose
// i-th bit is set
if (arr[j] & (1 << i))
ans = ans & arr[j];
}
// check for the set contains elements
// make a power of 2 or not
if (isPowerOf2(ans))
return true;
}
return false;
}
// Driver Program
int main()
{
int arr[] = { 12, 13, 7 };
int n = sizeof(arr) / sizeof(arr[0]);
if (checkSubsequence(arr, n))
printf("YES\n");
else
printf("NO\n");
return 0;
} Java
// Java Program to check if Bitwise AND of any
// subset is power of two
import java.io.*;
import java.util.*;
public class GFG {
static int NUM_BITS = 32;
// Check for power of 2 or not
static boolean isPowerOf2(int num)
{
if(num != 0 && (num & (num - 1)) == 0)
return true;
return false;
}
// Check if there exist a
// subset whose bitwise AND
// is power of 2.
static boolean checkSubsequence(int []arr, int n)
{
// if there is only one
// element in the set.
if (n == 1)
return isPowerOf2(arr[0]);
// Finding a number with
// all bit sets.
int total = 0;
for (int i = 0; i < NUM_BITS; i++)
total = total | (1 << i);
// check all the positions
// at which the bit is set.
for (int i = 0; i < NUM_BITS; i++)
{
int ans = total;
for (int j = 0; j < n; j++)
{
// include all those
// elements whose
// i-th bit is set
int p = arr[j] & (1 << i);
if (p == 0)
ans = ans & arr[j];
}
// check for the set
// contains elements
// make a power of 2
// or not
if (isPowerOf2(ans))
return true;
}
return false;
}
// Driver Code
public static void main(String args[])
{
int []arr = {12, 13, 7};
int n = arr.length;
if (checkSubsequence(arr, n))
System.out.println("YES");
else
System.out.println("NO");
}
}
// This code is contributed by
// Manish Shaw (manishshaw1)Python3
# Python3 Program to check if Bitwise AND of any
# subset is power of two
NUM_BITS = 32
# Check for power of 2 or not
def isPowerOf2(num):
return (num and (num & (num - 1)) == 0)
# Check if there exist a subset whose bitwise AND
# is power of 2.
def checkSubsequence(arr, n):
# if there is only one element in the set.
if (n == 1):
return isPowerOf2(arr[0])
# Finding a number with all bit sets.
total = 0
for i in range(0, NUM_BITS):
total = total | (1 << i)
# check all the positions at which the bit is set.
for i in range(0, NUM_BITS):
ans = total
for j in range(0, n):
# include all those elements whose
# i-th bit is set
if (arr[j] & (1 << i)):
ans = ans & arr[j]
# check for the set contains elements
# make a power of 2 or not
if (isPowerOf2(ans)):
return True
return False
# Driver Program
arr = [ 12, 13, 7 ]
n = len(arr)
if (checkSubsequence(arr, n)):
print ("YES\n")
else:
print ("NO\n")
# This code is contributed by Manish Shaw
# (manishshaw1)C#
// C# Program to check if Bitwise AND of any
// subset is power of two
using System;
using System.Collections.Generic;
class GFG {
static int NUM_BITS = 32;
// Check for power of 2 or not
static bool isPowerOf2(int num)
{
if(num != 0 && (num & (num - 1)) == 0)
return true;
return false;
}
// Check if there exist a
// subset whose bitwise AND
// is power of 2.
static bool checkSubsequence(int []arr, int n)
{
// if there is only one
// element in the set.
if (n == 1)
return isPowerOf2(arr[0]);
// Finding a number with
// all bit sets.
int total = 0;
for (int i = 0; i < NUM_BITS; i++)
total = total | (1 << i);
// check all the positions
// at which the bit is set.
for (int i = 0; i < NUM_BITS; i++)
{
int ans = total;
for (int j = 0; j < n; j++)
{
// include all those
// elements whose
// i-th bit is set
int p = arr[j] & (1 << i);
if (p == 0)
ans = ans & arr[j];
}
// check for the set
// contains elements
// make a power of 2
// or not
if (isPowerOf2(ans))
return true;
}
return false;
}
// Driver Code
public static void Main()
{
int []arr = {12, 13, 7};
int n = arr.Length;
if (checkSubsequence(arr, n))
Console.Write("YES\n");
else
Console.Write("NO\n");
}
}
// This code is contributed by
// Manish Shaw (manishshaw1)PHP
输出：
YES
参考：
https://stackoverflow.com/questions/35990794/subset-of-array-a-in-which-if-we-do-and-of-all-elements-of-that-subset-then-outp