题目:
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 1
s 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 integersL <= R
in the range[1, 10^6]
.R - L
will be at most 10000.
题意思路:
输入L和R形成闭区间[L, R],区间中的任意数字,其二进制形式中包含的1的数量为质数个,则计数器加1,最终输出该区间中所有符合这个条件的总计数,也就是计数器的值。
题目中提示输入的L,R的差别值不超过10^6,L-R不超过10000,计数器不用考虑超大的情况,10^6的范围说明可能存在的数字中包含1的最大数量,如输入10^6,则log_2(10^6) < 30,那么即使是该范围中包含1最多的数,也不会超过30个,则初始化只需要列出30之内的质数作为备选即可。
代码:
int countPrimeSetBits(int L, int R) {
set<int> primes = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29};
int count = 0;
for(int cur = L; cur <= R; cur++)
{
bitset<30> tmp(cur);
if(primes.count(tmp.count()))
count++;
}
return count;
}
其中,使用了bitset这个类,让所有初始化的值转换为二进制格式,用自带的count函数就可以计算出1的个数。