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762.prime-number-of-set-bits-in-binary-representation

Statement

Metadata

给你两个整数 left 和 right ，在闭区间 [left, right] 范围内，统计并返回 计算置位位数为质数 的整数个数。

计算置位位数 就是二进制表示中 1 的个数。

  • 例如， 21 的二进制表示 10101 有 3 个计算置位。

 

示例 1：

输入：left = 6, right = 10
输出：4
解释：
6 -> 110 (2 个计算置位，2 是质数)
7 -> 111 (3 个计算置位，3 是质数)
9 -> 1001 (2 个计算置位，2 是质数)
10-> 1010 (2 个计算置位，2 是质数)
共计 4 个计算置位为质数的数字。

示例 2：

输入：left = 10, right = 15
输出：5
解释：
10 -> 1010 (2 个计算置位, 2 是质数)
11 -> 1011 (3 个计算置位, 3 是质数)
12 -> 1100 (2 个计算置位, 2 是质数)
13 -> 1101 (3 个计算置位, 3 是质数)
14 -> 1110 (3 个计算置位, 3 是质数)
15 -> 1111 (4 个计算置位, 4 不是质数)
共计 5 个计算置位为质数的数字。

 

提示：

  • 1 <= left <= right <= 106
  • 0 <= right - left <= 104

Metadata

Given two integers left and right, return the count of numbers in the inclusive range [left, right] 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.

 

Example 1:

Input: left = 6, right = 10
Output: 4
Explanation:
6  -> 110 (2 set bits, 2 is prime)
7  -> 111 (3 set bits, 3 is prime)
8  -> 1000 (1 set bit, 1 is not prime)
9  -> 1001 (2 set bits, 2 is prime)
10 -> 1010 (2 set bits, 2 is prime)
4 numbers have a prime number of set bits.

Example 2:

Input: left = 10, right = 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)
5 numbers have a prime number of set bits.

 

Constraints:

  • 1 <= left <= right <= 106
  • 0 <= right - left <= 104

Solution

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

#define endl "\n"
#define fi first
#define se second
#define all(x) begin(x), end(x)
#define rall rbegin(a), rend(a)
#define bitcnt(x) (__builtin_popcountll(x))
#define complete_unique(a) a.erase(unique(begin(a), end(a)), end(a))
#define mst(x, a) memset(x, a, sizeof(x))
#define MP make_pair

using ll = long long;
using ull = unsigned long long;
using db = double;
using ld = long double;
using VLL = std::vector<ll>;
using VI = std::vector<int>;
using PII = std::pair<int, int>;
using PLL = std::pair<ll, ll>;

using namespace __gnu_pbds;
using namespace std;
template <typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
const ll mod = 1e9 + 7;

template <typename T, typename S>
inline bool chmax(T &a, const S &b) {
    return a < b ? a = b, 1 : 0;
}

template <typename T, typename S>
inline bool chmin(T &a, const S &b) {
    return a > b ? a = b, 1 : 0;
}

#ifdef LOCAL
#include <debug.hpp>
#else
#define dbg(...)
#endif
// head

int is_prime[] = {0, 0, 1, 1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0};

class Solution {
public:
    int countPrimeSetBits(int left, int right) {
        int res = 0;

        for (int i = left; i <= right; i++) {
            res += is_prime[__builtin_popcount(i)];
        }

        return res;
    }
};

#ifdef LOCAL

int main() {
    return 0;
}

#endif
最后更新: October 11, 2023
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