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工作学习 / 学科技术讨论 / BitCounting ---- An interesting solution#define BITCOUNT(x) (((BX_(x)+(BX_(x)>>4)) & 0x0F0F0F0F) % 255)
#define BX_(x) ((x) - (((x)>>1)&0x77777777) - (((x)>>2)&0x33333333) - (((x)>>3)&0x11111111))
-simali(0);
2008-3-11
{158}
(#4325290@0)
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Anyone understand these lines. I don't.
-simali(0);
2008-3-19
(#4341815@0)