简单的嵌套循环问题.*形状*

Question

我现在脑震荡,所以我想问一个非常简单的问题.

Currenly,我试图打印出这样的开头

当输入为7时,输出为

* 
** 
* 
** 
* 
** 
*

在这里,我的代码是,它打印14次而不是7次,或者当我把N/2它打印出来时不打印奇数.

#include <iostream>
using namespace std;

int main () {


    int N;
    cout << " Please enter N " ;
    cin >> N;


    for (int i = 0; i < N ; i++) {

        cout << "*" << endl;

        for (int j = 0; j < 2; j++) {

            cout << "*" ;
        }
        cout << endl;
    }
}

Answer 1

对于每一个,N您打印两行,单行*,另一行,两行*.相反,只打印一行与任何基于该行一两个明星是奇数还是偶数.

#include <iostream>

int main ()
{
  unsigned int N;
  cout << " Please enter N " ;
  cin >> N;

  for(unsigned int i = 0; i < N; ++i)
  {
    if(i%2 == 0)
    {
      std::cout << "*" << std::endl;
    }
    else
    {
      std::cout << "**" << std::endl;
    }
  }
}

(未经测试的代码)