int max1(int a,int b){
  return b & ( (a-b) >> 31) | a & (~(a-b) >> 31);
}

int max2(int x,int y){
  return x ^ ((x^y) & -(x < y));
}

int min1(int a,int b){
  return a & ((a-b) >> 31) | b & (~(a-b) >> 31);
}

int min2(int x,int y){
  return y ^ ((x^y) & -(x < y));
}

bool isSampSign(int x,int y){
  return (x^y) >= 0;
}

int getTwo(int n){
 // 2^n
 return 2 << (n-1);
}

bool isTwo(int n){
 return n > 0 ? (n&(n-1)) == 0:false;
}

int quyu(int m,int n){
  return m & (n-1);
}

int getAgv1(int n,int m){
 return (n+m) >> 1;
}

int getAgv2(int x,int y){
 return ((x^y) >> 1) + (x&y);
}

int getBit(int n,int m){  //从低位到高位 取n的第m位
 return (n >> (m-1)) & 1;
}

int setBit1(int n,int m){ //从低位到高位 将n的第m位置1
 return n | (1 << (m-1));
}

int setBit2(int n,int m){ //从低位到高位 将n的第m位置0
 return n & ~(1 << (m-1));
}

int getnum1(int n){
  //get n+1
  return -~n;
}

int getnum2(int n){
 return ~-n;
}

int getnum3(int n){
 return ~n + 1;
}

int main(int argc,char *argv[])
{
    int num1 = max1(3,5);
    int num2 = max2(3,5);
    int num3 = min1(3,5);
    int num4 = min2(3,5);
    cout<<getnum3(3)<<endl;
    //cout<<getBit2(4,1)<<endl;
    //cout<<getAgv1(3,5)<<” “<<getAgv2(3,5)<<endl;
    //cout<<quyu(2,4)<<” “<<quyu(5,3)<<endl;
    //cout<<isTwo(4)<<” “<<isTwo(5)<<endl;
    //cout<<getTwo(2)<<endl;
    //cout<<isSampSign(-1,2)<<endl;;
    //cout<<isSampSign(3,5)<<endl;
    //cout<<num3<<” “<<num4<<endl;
    //cout<<num1<<” “<<num2<<endl;
    return 0;
}