单选题

下述代码实现素数表的埃拉托斯特尼筛法,筛选出所有小于等于 n 的素数,则横线上应填的最佳代码是(    )。

void sieve_Eratosthenes(int n) {

      vector<bool> is_prime(n + 1, true);

      vector<int> primes;


      for (int i = 2; i * i <= n; i++) {

            if (is_prime[i]) {

                  primes.push_back(i);

                  _______________________ { // 在此处填入代码

                        is_prime[j] = false;

                  }

            }

      }

      for (int i = sqrt(n) + 1; i <= n; i++) {

            if (is_prime[i]) {

                  primes.push_back(i);

            }

      }

      return primes;

}

A

for (int j = i; j <= n; j++)

B

for (int j = i * i; j <= n; j++)

C

for (int j = i * i; j <= n; j += i)

D

for (int j = i; j <= n; j += i)

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