e······

#include <iostream>
using namespace std;
int n, t;
void f(int i)
{
	for (int j = 0 ; j < i ; j++)
		cout << ' ';
	cout << 'X';
	for (int j = 0 ; j < n - i * 2 - 2 ; j++)
		cout << ' ';
	cout << 'X' << endl ;
	
}
int main()
{
	cin >> n ;
	for (int i = 0 ; i < n / 2 ; i++)
		f(i);
	for (int i = 0 ; i < n / 2 ; i++)
		cout << ' ';
	cout << 'X' << endl;
	for (int i = n / 2 - 1; i >= 0 ; i--)
		f(i);
	return 0;
}

有点麻烦，但不多，来看一下本题的注意事项：

• 中间那一行要注意，只有一个“X”，不要打成了两个“X”
• 先掌握住空格的打印次数的规律，再确定循环变量的范围，千万不要急着下手

OK，话不多说，上代码！

已AC，请放心食用

#include <iostream>
using namespace std;
int main()
{
    int n;
    cin >> n;
    for (int i = 1; i <= n / 2; i++)
    {
        for (int j = 1; j <= i - 1; j++)
        {
            cout << " ";
        }
        cout << "X";
        for (int j = 1; j <= n - i * 2; j++)
        {
            cout << " ";
        }
        cout << "X" << endl;
    }
    for (int i = 1; i <= n / 2; i++)
    {
        cout << " ";
    }
    cout << "X" << endl;
    for (int i = 1; i <= n / 2; i++)
    {
        for (int j = 1; j <= n / 2 - i; j++)
        {
            cout << " ";
        }
        cout << "X";
        for (int j = 1; j <= i * 2 - 1; j++)
        {
            cout << " ";
        }
        cout << "X" << endl;
    }
    return 0;
}

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