Lintcode1046: Prime Number of Set Bits in Binary Reresenation

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本文探讨了一种算法，用于计算指定范围内整数的二进制表示中具有素数个数的1位的数量。通过遍历范围内的每个数，检查其二进制表示中1的个数是否为素数，从而得出最终计数。

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Prime Number of Set Bits in Binary Representation
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
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)
Notice
1.L, R will be integers L <= R in the range [1, 10^6].
2.R - L will be at most 10000.

求Prime数的练习题。

class Solution {
public:
    /**
     * @param L: an integer
     * @param R: an integer
     * @return: the count of numbers in the range [L, R] having a prime number of set bits in their binary representation
     */
    int countPrimeSetBits(int L, int R) {
        int result = 0;
        for (int i = L; i <= R; ++i) {
            if (isPrime(numOfSetBits(i))) {
                result++;
            }
        }
        return result;
    }

private:
    inline int numOfSetBits(int n) {
        int count = 0;
        while (n > 0) {
            if (n & 1) {
                count++;
            }
            n >>= 1;
        }
        return count;
    }

    inline int isPrime(int n) {
        if (n <= 1) return false;
        if (n == 2) return true;
        if ((n & 0x1) == 0) return false;
        int sqrtN = sqrt(n);
        for (int i = 2; i <= sqrtN; ++i) {
            if ((n % i) == 0) return false;
        }
        return true;
    }    
};