V*3
2 楼
现在大家都不缺分钟数吧? 那么用数据流量来节省分钟数意义何在?
J*9
3 楼
check whether it can be equal 1/2 + 1/4 + 1/8 + ...
see /// below
/**
* Given a real number between 0 and 1 (e.g., 0.72) that is passed in as a
double, print
* the binary representation. If the number c*annot be represented
accurately in binary
* with at most 32 characters, print "ERROR."
* hkBitsDoubleBits(): 0.720000 = 0.101110000101000111101011100001
* hkBitsDoubleBits(): 0.687500 = 0.1011
* hkBitsDoubleBits(): 0.500000 = 0.1
*/
char *hkBitsDoubleBits(double num, char *str, int maxSize)
{
if (!str)
return NULL;
if (num>=1.0 || num<0.0)
{
strcpy(str, "ERROR");
return str;
}
char *p = str;
*p++ = '0';
*p++ = '.';
maxSize -= 2;
while((num>0.0000000000001) && (maxSize>0))
{
double temp = 2.0 * num;
if (temp>=1.0)
{
*p++ = '1';
num = temp - 1.0;
/// if temp == 1.0, then can be represented exactly
}
else
{
*p++ = '0';
num = temp;
}
maxSize--;
}
*p = '\0';
return str;
}
see /// below
/**
* Given a real number between 0 and 1 (e.g., 0.72) that is passed in as a
double, print
* the binary representation. If the number c*annot be represented
accurately in binary
* with at most 32 characters, print "ERROR."
* hkBitsDoubleBits(): 0.720000 = 0.101110000101000111101011100001
* hkBitsDoubleBits(): 0.687500 = 0.1011
* hkBitsDoubleBits(): 0.500000 = 0.1
*/
char *hkBitsDoubleBits(double num, char *str, int maxSize)
{
if (!str)
return NULL;
if (num>=1.0 || num<0.0)
{
strcpy(str, "ERROR");
return str;
}
char *p = str;
*p++ = '0';
*p++ = '.';
maxSize -= 2;
while((num>0.0000000000001) && (maxSize>0))
{
double temp = 2.0 * num;
if (temp>=1.0)
{
*p++ = '1';
num = temp - 1.0;
/// if temp == 1.0, then can be represented exactly
}
else
{
*p++ = '0';
num = temp;
}
maxSize--;
}
*p = '\0';
return str;
}
S*2
4 楼
在室内信号很弱的时候,很多building内部信号都很弱(是的,我说的是t mobile),
但是wifi很强,这种时候wifi calling就特别好。
对已很多unlimited talk,用处不打。prepaid用处很大。
但是wifi很强,这种时候wifi calling就特别好。
对已很多unlimited talk,用处不打。prepaid用处很大。
d*x
5 楼
why don't you just check the denomanitor and see if it is 2^n?
for m/10^n, playing with gcd(m, 10^n) will give you the answer
【在 J**9 的大作中提到】
: check whether it can be equal 1/2 + 1/4 + 1/8 + ...
: see /// below
: /**
: * Given a real number between 0 and 1 (e.g., 0.72) that is passed in as a
: double, print
: * the binary representation. If the number c*annot be represented
: accurately in binary
: * with at most 32 characters, print "ERROR."
: * hkBitsDoubleBits(): 0.720000 = 0.101110000101000111101011100001
: * hkBitsDoubleBits(): 0.687500 = 0.1011
for m/10^n, playing with gcd(m, 10^n) will give you the answer
【在 J**9 的大作中提到】
: check whether it can be equal 1/2 + 1/4 + 1/8 + ...
: see /// below
: /**
: * Given a real number between 0 and 1 (e.g., 0.72) that is passed in as a
: double, print
: * the binary representation. If the number c*annot be represented
: accurately in binary
: * with at most 32 characters, print "ERROR."
: * hkBitsDoubleBits(): 0.720000 = 0.101110000101000111101011100001
: * hkBitsDoubleBits(): 0.687500 = 0.1011
j*2
7 楼
对用prepaid的人来说很有意义,有wifi的地方就不会扣分钟数
s*f
9 楼
本来你要买个 obi 的盒子吧,现在把个不用的板子放在家里冲着电,保持上着网也可
以当电话用了。:P
以当电话用了。:P
b*q
12 楼
意义重大啊,可以让各种pad成为可打进打出的电话。
b*g
20 楼
太有用了,家里一个obi当固话,再弄一个非私人用途
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