摘要： 4 int clp2(unsigned int n) 5 { 6 --n; 7 n = n | (n >> 1); 8 n = n | (n >> 2); 9 n = n | (n >> 4); 10 n = n | (n >> 8); 11 n = n | (n >> 16); 12 return n + 1; 13 }unsigned flp2 (unsigned x){ x = x | (x >> 1) x = x | (x >> 2) x = x | (x >> 4); x = x | (x 阅读全文

摘要： 4 int bitcount(unsigned int n) 5 { 6 n = n - ((n>>1)&033333333333) - ((n>>2)&011111111111); 7 n = (n + (n>>3))&030707070707; 8 n = n % 0x3f; 9 return n; 10 } 17 int bitcount(unsigned int n) 18 { 19 n = n - ((n>>1)&0x77777777) - ((n>>2)&0x33333333)-(( 阅读全文