小编Nam*_*man的帖子

有没有办法在不使用math.pow或乘法运算符的情况下使用代码2 ^ power.至今,

我虽然使用了2个计数器和附加功能,但我的程序似乎没有用.到目前为止,这是我的工作.

int counter=0; // k 
int userNumber=0; // p 
int power=0;
int sum=0;

cout << "Enter a non-negative number: ";
cin >> userNumber;


while (userNumber > counter)
{
    power +=2;
    counter++;
    power++;
}

sum = power - 1;
// post-condition: Sum = 2^p -1
cout << "The output is " << sum << endl;
return 0;

我有三个名为"Guest","Guest_Address"和"Acutal_Address"的表.Guest_Address是guest和acutal_address之间的链接表.这就是我到目前为止所拥有的.

  SELECT GUEST_ADDRESS.ADDRESS_CODE,(GUEST_FNAME+' '+GUEST_LNAME) AS GUEST_NAMES
  FROM GUEST JOIN GUEST_ADDRESS 
  ON GUEST.ADDRESS_NUM = GUEST_ADDRESS.ADDRESS_NUM;

这只会加入Guest和Guest_address表,但我需要加入Guest和Acutal_Address.这是ERD.

我必须做一个掷骰子游戏,到最后,我必须做一些概率.到目前为止,这是我的代码.我想要它,以便循环重复1000次,并寻找用户输入的'probNumb'.我不确定这是否正确,但我可以说我输入了数字5.这就是我得到的.
"在1000次中,5次被滚动了1000次."

所以,不计算5次滚动的次数.我不允许使用break或continue语句,只允许循环和if else.

cout << "What number do you want the probability of ?";
cin >> probNumb;

while (probCount < 1000)
{
  ranNumb= 1 + (rand() % (5 + 1));
  ranNumb2= 1 + (rand() % (5 + 1));
  ranNumbFin = ranNumb + ranNumb2;
  probCount++;

  if (ranNumbFin = probNumb)
    probNumbCount++;
}

cout << "Out of 1000 times, " << probNumb << " was rolled " 
  << probNumbCount << "times." << endl;