(3*4096+15*256+3*16+3). How many 1's are there in the binary
representation of the result.
Is there a quick way to count the number of set bits in this number
manually???
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either count it with a for loop or if you are using C, use
__builtin_popcount(int x)
or
x is given
int count = 0;
for(int i=0; i32;i++) count += (x (1i)) != 0;
On Tue, Aug 3, 2010 at 12:04 PM, amit amitjaspal...@gmail.com wrote:
(3*4096+15*256+3*16+3). How many 1's are there in the binary
it is
3*2^12 + 15*2^8 + 3*2^4 + 3
so it becomes
312 + 158 + 34 +3.. let this equal to n
take a int count=0;
while(n){
n=n(n-1);
count++;
}
return count;
2010/8/4 Seçkin Can Şahin seckincansa...@gmail.com
either count it with a for loop or if you are using C, use
it is
3*2^12 + 15*2^8 + 3*2^4 + 3
so it becomes
312 + 158 + 34 +3.. let this equal to n
take
while(n){
2010/8/4 Seçkin Can Şahin seckincansa...@gmail.com
either count it with a for loop or if you are using C, use
__builtin_popcount(int x)
or
x is given
int count = 0;
for(int i=0;
Hi Amit,
If you are answer this question orally then the format you have given is
just a hexadecimal number. Its value would be 0x3f33. So the number of
1's would be simply 2+4+2+2 = 10.
Programatically whatever solution Sahin has given seems to be good enough.
2010/8/4 Seçkin Can Şahin
