#include <iostream> using namespace std; //二進制位逆序。 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; } #include <iostream> using namespace std; //二進制位中的奇偶位互換。 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; } #include <iostream> using namespace std; //求兩個數的和。
int Grial(int x, int y) { if (!y)return x; //int ret = x^y; //-1%2=-1; int sum = x^y; int num = ((x&y)<<1); return Grial(sum,num); } int main() { cout << Grial(3, 5) << endl; return 0; } #include <iostream> using namespace std; //推斷一個數是不是2的n次方。 bool Grial(int x) { return (x&(x - 1)) == 0; } int main() { cout << Grial(4) << endl; return 0; } #include <iostream> using namespace std; //求二進制中1的個數。 int Grial(int x) { int count = 0; while (x) { x = x&(x - 1); count++; } return count; } int main() { cout << Grial(15) << endl; } #include <iostream> using namespace std; //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; } #include <iostream> using namespace std; //求平均值。 int Grial(int x,int y) { return x - (x - y) / 2; } int main() { cout << Grial(3,9) << endl; return 0; }