h*g
2 楼
看到战友show off的帖子,都是非常好的deal,自己看的话好象差远了,请问有什么心
得,想给儿子买一个电脑,可能中度游戏,谢谢!
得,想给儿子买一个电脑,可能中度游戏,谢谢!
y*g
3 楼
linkedin?
B*y
6 楼
想省更多的钱就得花更多的时间和精力去折腾,家里有娃的估计就得算了,差不多有点
deal就可以了,别羡慕别人的“超级”deal,你玩不起的。。。
deal就可以了,别羡慕别人的“超级”deal,你玩不起的。。。
w*e
7 楼
double power(double x, unsigned int n)
{
if (n == 0) return 1;
if (n == 1) return x;
int xsquare = x * x;
if (n % 2 == 1) // n is odd
{
return (x * power(xsquare, n / 2));
}
else // n is even
{
return power(xsquare, n / 2);
}
}
We can also write an iterative version, I think.
{
if (n == 0) return 1;
if (n == 1) return x;
int xsquare = x * x;
if (n % 2 == 1) // n is odd
{
return (x * power(xsquare, n / 2));
}
else // n is even
{
return power(xsquare, n / 2);
}
}
We can also write an iterative version, I think.
x*4
8 楼
打游戏就直接外星人吧,看S大神的dell帖子
g*y
9 楼
凑热闹
public double power(double x, int n) {
if (n < 0) return 1.0/power(x, -n);
double r = 1.0, pow = x;
while (n > 0) {
if ( (1 & n) > 0 ) r *= pow;
n >>= 1;
pow *= pow;
}
return r;
}
public double power(double x, int n) {
if (n < 0) return 1.0/power(x, -n);
double r = 1.0, pow = x;
while (n > 0) {
if ( (1 & n) > 0 ) r *= pow;
n >>= 1;
pow *= pow;
}
return r;
}
y*i
12 楼
牛人貌似使用机器人高频率刷,刷到好的自动下单。
p*a
14 楼
现在outlet系统还经常出问题,用filter搜会错过很多好机子,他们家码农太烂了
:牛人貌似使用机器人高频率刷,刷到好的自动下单。
:牛人貌似使用机器人高频率刷,刷到好的自动下单。
w*e
20 楼
If n = 4, it seems to return 1. Should we fix the last line like this?
return r * pow;
【在 g**********y 的大作中提到】
: 凑热闹
: public double power(double x, int n) {
: if (n < 0) return 1.0/power(x, -n);
: double r = 1.0, pow = x;
: while (n > 0) {
: if ( (1 & n) > 0 ) r *= pow;
: n >>= 1;
: pow *= pow;
: }
: return r;
return r * pow;
【在 g**********y 的大作中提到】
: 凑热闹
: public double power(double x, int n) {
: if (n < 0) return 1.0/power(x, -n);
: double r = 1.0, pow = x;
: while (n > 0) {
: if ( (1 & n) > 0 ) r *= pow;
: n >>= 1;
: pow *= pow;
: }
: return r;
D*e
23 楼
If can use recursion, this is simple. But the restriction of O(1) space
becomes meaningless.
becomes meaningless.
d*d
31 楼
电面小题.
double power(double x, int n);
要求时间lg(n), 空间O(1);
double power(double x, int n);
要求时间lg(n), 空间O(1);
y*g
32 楼
linkedin?
w*e
34 楼
double power(double x, unsigned int n)
{
if (n == 0) return 1;
if (n == 1) return x;
int xsquare = x * x;
if (n % 2 == 1) // n is odd
{
return (x * power(xsquare, n / 2));
}
else // n is even
{
return power(xsquare, n / 2);
}
}
We can also write an iterative version, I think.
{
if (n == 0) return 1;
if (n == 1) return x;
int xsquare = x * x;
if (n % 2 == 1) // n is odd
{
return (x * power(xsquare, n / 2));
}
else // n is even
{
return power(xsquare, n / 2);
}
}
We can also write an iterative version, I think.
g*y
35 楼
凑热闹
public double power(double x, int n) {
if (n < 0) return 1.0/power(x, -n);
double r = 1.0, pow = x;
while (n > 0) {
if ( (1 & n) > 0 ) r *= pow;
n >>= 1;
pow *= pow;
}
return r;
}
public double power(double x, int n) {
if (n < 0) return 1.0/power(x, -n);
double r = 1.0, pow = x;
while (n > 0) {
if ( (1 & n) > 0 ) r *= pow;
n >>= 1;
pow *= pow;
}
return r;
}
w*e
42 楼
If n = 4, it seems to return 1. Should we fix the last line like this?
return r * pow;
【在 g**********y 的大作中提到】
: 凑热闹
: public double power(double x, int n) {
: if (n < 0) return 1.0/power(x, -n);
: double r = 1.0, pow = x;
: while (n > 0) {
: if ( (1 & n) > 0 ) r *= pow;
: n >>= 1;
: pow *= pow;
: }
: return r;
return r * pow;
【在 g**********y 的大作中提到】
: 凑热闹
: public double power(double x, int n) {
: if (n < 0) return 1.0/power(x, -n);
: double r = 1.0, pow = x;
: while (n > 0) {
: if ( (1 & n) > 0 ) r *= pow;
: n >>= 1;
: pow *= pow;
: }
: return r;
D*e
45 楼
If can use recursion, this is simple. But the restriction of O(1) space
becomes meaningless.
becomes meaningless.
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