求二进制数中1的个数之平行算法

#include <cstdio> /* int ans[1000000]; int main(){ freopen("1.out", "w", stdout); int idc = 0; long long n = 0xffff; for(long long i=1; i<n; i<<=1){ if(n & i) ans[++idc] = 1; else ans[++idc] = 0; } for(int i=idc; i; i--){ printf("%d", ans[i]); } } */ int func(unsigned int i)// { unsigned int temp = i; temp = (temp & 0x55555555) + ((temp>> 1) & 0x55555555); //temp相邻位相加 0x55555555 = 1010101010101010101010101010101 temp = (temp & 0x33333333) + ((temp >> 2) & 0x33333333); //temp相邻（以2为单位）相加 0x33333333 = 110011001100110011001100110011 temp = (temp & 0x0f0f0f0f) + ((temp>> 4) & 0x0f0f0f0f); //temp相邻（以4为单位）相加 0x0f0f0f0f = 1111000011110000111100001111 temp = (temp & 0xff00ff) + ((temp>> 8) & 0xff00ff); //temp相邻（以8为单位）相加 0xff00ff = 111111110000000011111111 temp = (temp & 0xffff) + ((temp>> 16) & 0xffff) ; //temp相邻（以16为单位）相加 0xffff = 1111111111111111 return temp; } /* 先将n写成二进制形式，然后相邻位相加，重复这个过程，直到只剩下一位。 */

膜大佬