Given two integers L and R, find the count of numbers in the range L, R having a prime number of set bits in their binary representation.

(Recall that the number of set bits an integer has is the number of 1s present when written in binary. For example, 21 written in binary is 10101 which has 3 set bits. Also, 1 is not a prime.)

Example 1:

Input: L = 6, R = 10
Output: 4
Explanation:
6 -> 110 (2 set bits, 2 is prime)
7 -> 111 (3 set bits, 3 is prime)
9 -> 1001 (2 set bits , 2 is prime)
10->1010 (2 set bits , 2 is prime)
Example 2:

Input: L = 10, R = 15
Output: 5
Explanation:
10 -> 1010 (2 set bits, 2 is prime)
11 -> 1011 (3 set bits, 3 is prime)
12 -> 1100 (2 set bits, 2 is prime)
13 -> 1101 (3 set bits, 3 is prime)
14 -> 1110 (3 set bits, 3 is prime)
15 -> 1111 (4 set bits, 4 is not prime)
Note:

L, R will be integers L <= R in the range [1, 10^6].
R - L will be at most 10000.

```/*判断整型数字n是否是素数*/
bool isPrime(int n)
{
/*1不是素数*/
if(1 == n)
return false;

for(int i = 2;i < n/2 + 1;i++)
{
if(n % i == 0)
return false;
}

return true;
}

/*统计整型数字n中二进制形式中1的个数*/
int countbits(int n)
{
int count = 0;
while(n)
{
n = n&(n-1);
count += 1;
}

return count;
}

int countPrimeSetBits(int L, int R)
{
int n = 0;
int count = 0;

for(int i = L;i <= R;i++)
{
n = countbits(i);

if(isPrime(n))
count += 1;
}

return count;
}
```

```/*统计整型数字n中二进制形式中1的个数*/
int countbits(int n)
{
int count = 0;
while(n)
{
n = n&(n-1);
count += 1;
}

return count;
}

int countPrimeSetBits(int L, int R)
{
int n = 0;
int count = 0;
int primes = {2,3,5,7,11,13,17,19};

for(int i = L;i <= R;i++)
{
n = countbits(i);

/*在数组中查找是否存在，二分查找*/
int low = 0;int high = 7;
while(low <= high)
{
int mid = (low+high)/2;

if (primes[mid] == n)
{
count++;
break;
}
else if (primes[mid]>n)
{
high = mid-1;
}
else
{
low = mid+1;
}
}

}

return count;
}
```