c++ - 钻石形状打印不正确

Question

好的，所以我必须创建这个菱形，它应该是菱形，但右上半部分缺失，左下半部分缺失。我有两个形状，但它们没有以菱形形式显示，而是打印出一个在另一个之上。

我假设某处。我必须打印出相同数量的空间才能将整个底部形状向右移动？

#include <iostream>

using namespace std;

int main()
{
   int row, col, siz;

   cout << "Enter the size of the diamond: " << endl;
   cin >> siz;

   cout << endl << endl;


   for(row = 1; row <= siz; row++){

      for(col = 1; col <= siz; col ++){
         if(col <= siz - row){
            cout << " ";
         }
         else{
                cout << "@";
             }
        }
        cout << "\n";
    }

    for(row = 1; row <= siz; row++){

        for(col = row; col <= siz; col++){
            cout << "@";
        }
        cout << "\n";
    }

    return 0;
}

Answer 1

这是我对你问题的解决方案:

#include <iostream>
using namespace std;

int main()
{
   int row, col, siz;

   cout << "Enter the size of the diamond: " << endl;
   cin >> siz;

   cout << endl << endl;

   // drawing top left side of diamon
   for(row = 1; row <= siz; row++){
      for(col = 1; col <= siz; col ++){
         if(col <= siz - row){
            cout << " ";
         }
         else{
                cout << "@";
             }
        }
        cout << "\n";
    }

    // drawing bottom right side of diamon
    for(row = 1; row <= siz; row++){
        for(col = 1; col <= siz*2-row; col++){
            if(col<siz){
               cout << " "; 
            }
            else{
                cout << "@";
            }
        }
        cout << "\n";
    }

    return 0;
}

解释: 下半场的第一列必须先绘制“siz”时间“”，然后再绘制“siz”时间“@”。所以它必须循环 siz*2;从那时起你总是有'siz'时间''然后每次都少一个'@'所以这就是为什么循环应该看起来像for(col = 1; col <= siz*2-row; col++)