;//二进制位逆序。int Grial(int x){int n = 32;int count = 0;while (n–){count|= ((x&0x1) << n);x = ((unsigned)x >> 1);}return count;}int main(){cout << Grial(1) << endl;return 0;};//二进制位中的奇偶位互换。int Grial(int x){int sum1 = x & 0x55555555;int sum2 = x & 0xaaaaaaaa;sum1 <<= 1;sum2 >>= 1;return sum1 | sum2;}int main(){cout << Grial(6) << endl;return 0;};//求两个数的和。int Grial(int x, int y){if (!y)return x;sum = x^y;int num = ((x&y)<<1);return Grial(sum,num);}int main(){cout << Grial(3, 5) << endl;return 0;};//判断一个数是不是2的n次方。bool Grial(int x){return (x&(x – 1)) == 0;}int main(){cout << Grial(4) << endl;return 0;};//求二进制中1的个数。int Grial(int x){int count = 0;while (x){x = x&(x – 1);count++;}return count;}int main(){cout << Grial(15) << endl;};//求平均值。int Grial(int x,int y){return (x&y) + ((x^y) >> 1);}int main(){cout << Grial(20, 48) << endl;return 0;};//求平均值。int Grial(int x,int y){return x – (x – y) / 2;}int main(){cout << Grial(3,9) << endl;return 0;}
，青春一经典当即永不再赎
