h**o
发帖数: 548 | 1 given 1/10, 1/3.。。问能否 represented exactly in base 2?
怎么做？ | J**9
发帖数: 835 | 2 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;
} | d**********x
发帖数: 4083 | 3 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
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