题解 | 求素数

求素数

https://www.nowcoder.com/practice/ea1c82be0c144d74925a044973fe9a5c

解题思路

这是一个区间素数统计问题。需要高效地判断一个数是否为素数，并统计区间内的素数个数。

关键点：

高效的素数判断方法
避免对每个数都进行完整的素数判断
处理大数据范围（可达1000000）

算法步骤：

使用埃氏筛法预处理素数表
统计区间 $[M,N]$ 内的素数个数

代码

cpp
java
python

#include <bits/stdc++.h>
using namespace std;

class Solution {
private:
    vector<bool> isPrime;
    
    void sieve(int n) {
        isPrime.resize(n + 1, true);
        isPrime[0] = isPrime[1] = false;
        
        for (int i = 2; i * i <= n; i++) {
            if (isPrime[i]) {
                for (int j = i * i; j <= n; j += i) {
                    isPrime[j] = false;
                }
            }
        }
    }
    
public:
    int countPrimes(int m, int n) {
        sieve(n);
        
        int count = 0;
        for (int i = m; i <= n; i++) {
            if (isPrime[i]) {
                count++;
            }
        }
        
        return count;
    }
};

int main() {
    int m, n;
    cin >> m >> n;
    
    Solution solution;
    cout << solution.countPrimes(m, n) << endl;
    
    return 0;
}

import java.util.*;

public class Main {
    static class Solution {
        private boolean[] isPrime;
        
        private void sieve(int n) {
            isPrime = new boolean[n + 1];
            Arrays.fill(isPrime, true);
            isPrime[0] = isPrime[1] = false;
            
            for (int i = 2; i * i <= n; i++) {
                if (isPrime[i]) {
                    for (int j = i * i; j <= n; j += i) {
                        isPrime[j] = false;
                    }
                }
            }
        }
        
        public int countPrimes(int m, int n) {
            sieve(n);
            
            int count = 0;
            for (int i = m; i <= n; i++) {
                if (isPrime[i]) {
                    count++;
                }
            }
            
            return count;
        }
    }
    
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int m = sc.nextInt();
        int n = sc.nextInt();
        
        Solution solution = new Solution();
        System.out.println(solution.countPrimes(m, n));
        
        sc.close();
    }
}

class Solution:
    def __init__(self):
        self.is_prime = []
    
    def sieve(self, n: int):
        self.is_prime = [True] * (n + 1)
        self.is_prime[0] = self.is_prime[1] = False
        
        i = 2
        while i * i <= n:
            if self.is_prime[i]:
                for j in range(i * i, n + 1, i):
                    self.is_prime[j] = False
            i += 1
    
    def count_primes(self, m: int, n: int) -> int:
        self.sieve(n)
        return sum(1 for i in range(m, n + 1) if self.is_prime[i])

# 读取输入
m, n = map(int, input().split())

solution = Solution()
print(solution.count_primes(m, n))