给定一个非负整数N，任务是查找非负整数的数量
小于或等于N，其按位OR和与N相等。
例子 ：
Input : N = 3
Output : 1
0 is the only number in [0, 3]
that satisfies given property.
(0 + 3) = (0 | 3)
Input : 10
Output : 4
(0 + 10) = (0 | 10) (Both are 10)
(1 + 10) = (1 | 10) (Both are 11)
(4 + 10) = (4 | 10) (Both are 14)
(5 + 10) = (5 | 10) (Both are 15)
一个简单的解决方案是遍历从0到N的所有数字，如果两个都是相等的增量计数器，则对它们进行按位OR和SUM运算。
时间复杂度= O(N)。
一个有效的解决方案是遵循以下步骤。
1.在N中找到零位的计数。
2.返回pow(2，count)。
这个想法基于这样一个事实，当且仅当
x与N的按位与将为0
Let, N=10 =10102.
Bitwise AND of a number with N will be 0, if number
contains zero bit with all respective set bit(s) of
N and either zero bit or set bit with all respective
zero bit(s) of N (because, 0&0=1&0=0).
e.g.
bit : 1 0 1 0
position: 4 3 2 1
Bitwise AND of any number with N will be 0, if the number
has following bit pattern
1st position can be either 0 or 1 (2 ways)
2nd position can be 1 (1 way)
3rd position can be either 0 or 1 (2 ways)
4th position can be 1 (1 way)
Total count = 2*1*2*1 = 22 = 4.
C++
// C++ program to count numbers whose bitwise
// OR and sum with N are equal
#include
using namespace std;
// Function to find total 0 bit in a number
unsigned int CountZeroBit(int n)
{
unsigned int count = 0;
while(n)
{
if (!(n & 1))
count++;
n >>= 1;
}
return count;
}
// Function to find Count of non-negative numbers
// less than or equal to N, whose bitwise OR and
// SUM with N are equal.
int CountORandSumEqual(int N )
{
// count number of zero bit in N
int count = CountZeroBit(N);
// power of 2 to count
return (1 << count);
}
// Driver code
int main()
{
int N = 10;
cout << CountORandSumEqual(N);
return 0;
} Java
// Java program to count numbers whose bitwise
// OR and sum with N are equal
class GFG {
// Function to find total 0 bit in a number
static int CountZeroBit(int n)
{
int count = 0;
while(n > 0)
{
if ((n & 1) != 0)
count++;
n >>= 1;
}
return count;
}
// Function to find Count of non-negative
// numbers less than or equal to N, whose
// bitwise OR and SUM with N are equal.
static int CountORandSumEqual(int N )
{
// count number of zero bit in N
int count = CountZeroBit(N);
// power of 2 to count
return (1 << count);
}
//Driver code
public static void main (String[] args)
{
int N = 10;
System.out.print(CountORandSumEqual(N));
}
}
// This code is contributed by Anant Agarwal.Python3
# Python3 program to count numbers whose
# bitwise OR and sum with N are equal
# Function to find total 0 bit in a number
def CountZeroBit(n):
count = 0
while(n):
if (not(n & 1)):
count += 1
n >>= 1
return count
# Function to find Count of non-negative
# numbers less than or equal to N, whose
# bitwise OR and SUM with N are equal.
def CountORandSumEqual(N):
# count number of zero bit in N
count = CountZeroBit(N)
# power of 2 to count
return (1 << count)
# Driver code
N = 10
print(CountORandSumEqual(N))
# This code is contributed by Anant Agarwal.C#
// C# program to count numbers whose
// bitwise OR and sum with N are equal
using System;
class GFG
{
// Function to find total
// 0 bit in a number
static int CountZeroBit(int n)
{
int count = 0;
while(n>0)
{
if (n%2!=0)
count++;
n >>= 1;
}
return count;
}
// Function to find Count of non-negative
// numbers less than or equal to N, whose
// bitwise OR and SUM with N are equal.
static int CountORandSumEqual(int N )
{
// count number of zero bit in N
int count = CountZeroBit(N);
// power of 2 to count
return (1 << count);
}
//Driver code
public static void Main()
{
int N = 10;
Console.Write(CountORandSumEqual(N));
}
}
// This code is contributed by Anant Agarwal.PHP
>= 1;
}
return $count;
}
// Function to find Count of
// non-negative numbers less
// than or equal to N, whose
// bitwise OR and SUM with N
// are equal.
function CountORandSumEqual($N )
{
// count number of
// zero bit in N
$count = CountZeroBit($N);
// power of 2 to count
return (1 << $count);
}
// Driver code
$N = 10;
echo CountORandSumEqual($N);
// This code is contributed by Ajit
?>输出 ：
4上述解决方案的总时间复杂度将为O(log 2 (N))。