2
i = 1;
while (i <= n)
j = i;
x = x+A[i];
while (j > 0)
y = x/(2*j);
j = j/2; // Assume here that this returns the floor of the quotient
i = 2*i;
return y;
我不知道我的答案,我得到了O(n )。什麼是這個僞代碼的運行時間