題目：

Given two integers L and R, find the count of numbers in the range [L, R] (inclusive) 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.

理解：

就是找到範圍內的數中二進位制表示中置位的個數為質數的數目。由於範圍最大為10^6，最多需要20位數字表示，而20不是質數，因此最大為19即可。

實現：

class Solution {
public:
	int countPrimeSetBits(int l, int r) {
		set<int> primes = { 2, 3, 5, 7, 11, 13, 17, 19};
		int cnt = 0;
		for (int i = l; i <= r; i++) {
			int bits = 0;
			for (int n = i; n; n >>= 1)
				bits += n & 1;
			cnt += primes.count(bits);
		}
		return cnt;
	}
};
 

學到一個位操作n&(n-1)，可以每次把最末尾的1置為0，因此可以通過這個迴圈來判斷被置位的位元數。

int cnt=0;
while(n){
    n&=(n-1);
    ++cnt;
}