第2题

bool move(int b[], int n)
{
    for (int i = 0; i < n; ++i) {
        if (b[i] == 1) {
            b[i] = 0;
        }
        else {
            b[i] = 1;
            return true;
        }
    }
    return false;
}
void print(int n)
{
    int b[n];
    for (int i = 0; i < n; ++i) b[i] = 0;
    do {
        for (int i = 0; i < n; ++i) std::cout << b[i];
        std::cout << "\n";
    }
    while (move(b, n));
}
int count(int n)
{
    int b[100];
    for (int i = 0; i < n; ++i) b[i] = 0;
    int c = 0;
    do {
        c++;
    }
    while (move(b, n));
    return c;
}

print(n) 与 count(n) 的时间复杂度为（　）。

{{ select(1) }}

• $\Theta(2^n)$，$\Theta(2^n)$
• $\Theta(n\cdot 2^n)$，$\Theta(2^n)$
• $\Theta(n\cdot 2^n)$，$\Theta(n\cdot 2^n)$
• $\Theta(4^n)$，$\Theta(2^n)$