The Sieve of Eratosthenes is a highly efficient way to generate a list of primes. It works by recognizing that all non-prime numbers are divisible by a prime number. We start with a list of all the numbers up through some value max. First, we cross off all numbers divisible by 2. Then, we look for the next prime (the next non-crossed off number) and cross off all numbers divisible by it. By crossing off all numbers divisible by 2, 3, 5, 7, 11, and so on, we wind up with a list of prime numbers from 2 through max. Write the code that implements the Sieve of Eratosthenes with the following prototype that returns an array of boolean values indicating if the corresponding integer is a prime number or not . Test the function with a main function . 用c++5.11寫出，且使用using namespace std;

以下是使用 C++ 实现 Sieve of Eratosthenes 的代码，返回一个布尔类型的数组，表示每个数是否为质数：

#include <iostream>
#include <vector>

using namespace std;

vector<bool> sieve_of_eratosthenes(int max) {
    vector<bool> is_prime(max + 1, true);
    is_prime[0] = is_prime[1] = false;

    for (int i = 2; i * i <= max; i++) {
        if (is_prime[i]) {
            for (int j = i * i; j <= max; j += i) {
                is_prime[j] = false;
            }
        }
    }

    return is_prime;
}

int main() {
    int max = 100;
    vector<bool> primes = sieve_of_eratosthenes(max);

    for (int i = 2; i <= max; i++) {
        if (primes[i]) {
            cout << i << " ";
        }
    }

    return 0;
}

在主函数中，我们调用 `sieve_of_eratosthenes` 函数，将 `max` 传递给它，并将返回的布尔类型向量存储在 `primes` 变量中。然后，我们遍历这个向量并打印所有值为 `true` 的索引，这些索引表示质数。