给定一个正整数N，找出有多少个严格小于N的正整数具有与N相同的置位位数。
例子 ：
Input : 8
Output :3
Explanation: Binary representation of
8 : 1000, so number of set bits in 8 is 1.
So the integers less than 8 with same number
of set bits are : 4, 2, 1
Input :1
Output :0
Input :4
Output :2
方法：
1. Using __builtin_popcount() inbuilt function, count set bits in N and store into a
temp variable
2. Iterate from n-1 to 1 and also count set bits in i using __builtin_popcount()
function
3. Now, compare temp with __builtin_popcount(i)
4. If both are equal then increment counter variable
5. Return counter
下面是上述方法的实现。
C++
// CPP program to find numbers less than N
// that have same Number Of Set Bits As N
#include
using namespace std;
int smallerNumsWithSameSetBits(int n)
{
// __builtin_popcount function that count
// set bits in n
int temp = __builtin_popcount(n);
// Iterate from n-1 to 1
int count = 0;
for (int i = n - 1; i > 0; i--) {
// check if the number of set bits
// equals to temp increment count
if (temp == __builtin_popcount(i))
count++;
}
return count;
}
// Driver Code
int main()
{
int n = 4;
cout << smallerNumsWithSameSetBits(n);
return 0;
} Java
// Java program to find numbers less than N
// that have same Number Of Set Bits As N
class GFG {
// returns number of set bits in a number
static int __builtin_popcount(int n)
{
int d, t = 0;
while(n > 0)
{
d = n % 2;
n = n / 2;
if(d == 1)
t++;
}
return t;
}
static int smallerNumsWithSameSetBits(int n)
{
// __builtin_popcount function that count
// set bits in n
int temp = __builtin_popcount(n);
// Iterate from n-1 to 1
int count = 0;
for (int i = n - 1; i > 0; i--) {
// check if the number of set bits
// equals to temp increment count
if (temp == __builtin_popcount(i))
count++;
}
return count;
}
// Driver Code
public static void main(String[] args)
{
int n = 4;
System.out.println(
smallerNumsWithSameSetBits(n));
}
}
// This code is contributed by Arnab Kundu.Python3
# Python3 program to find numbers
# less than N that have same
# Number Of Set Bits As N
def __builtin_popcount(n) :
t = 0
while(n > 0) :
d = n % 2
n = int(n / 2)
if(d == 1) :
t = t + 1
return t
def smallerNumsWithSameSetBits(n) :
# __builtin_popcount function
# that count set bits in n
temp = __builtin_popcount(n)
# Iterate from n-1 to 1
count = 0
for i in range(n-1,0,-1) :
# check if the number of
# set bits equals to temp
# increment count
if (temp == __builtin_popcount(i)) :
count = count + 1
return count
# Driver Code
n = 4
print (smallerNumsWithSameSetBits(n))
# This code is contributed by
# Manish Shaw(manishshaw1)C#
// C# program to find numbers less than N
// that have same Number Of Set Bits As N
using System;
class GFG {
// returns number of set bits in a number
static int __builtin_popcount(int n)
{
int d, t = 0;
while(n > 0)
{
d = n % 2;
n = n / 2;
if(d == 1)
t++;
}
return t;
}
static int smallerNumsWithSameSetBits(int n)
{
// __builtin_popcount function that count
// set bits in n
int temp = __builtin_popcount(n);
// Iterate from n-1 to 1
int count = 0;
for (int i = n - 1; i > 0; i--) {
// check if the number of set bits
// equals to temp increment count
if (temp == __builtin_popcount(i))
count++;
}
return count;
}
// Driver Code
static public void Main(String []args)
{
int n = 4;
Console.WriteLine(
smallerNumsWithSameSetBits(n));
}
}
// This code is contributed by Arnab Kundu.PHP
0)
{
$d = $n % 2;
$n = intval($n / 2);
if($d == 1)
$t++;
}
return $t;
}
function smallerNumsWithSameSetBits($n)
{
// __builtin_popcount function
// that count set bits in n
$temp = __builtin_popcount($n);
// Iterate from n-1 to 1
$count = 0;
for ($i = $n - 1; $i > 0; $i--)
{
// check if the number of
// set bits equals to temp
// increment count
if ($temp == __builtin_popcount($i))
$count++;
}
return $count;
}
// Driver Code
$n = 4;
echo (smallerNumsWithSameSetBits($n));
// This code is contributed by
// Manish Shaw(manishshaw1)
?>输出：
2