求助这个题怎么做啊

当输入为6的时候，输出打印都要几秒

尽力了,最开始想使用c语言的数组,指针太麻烦了,使用vector更简单.
这题还是有点难度,尤其要打印到屏幕上,cout只能一行一行输出,但将X替换为5个X,不只有简单的替换,还有其行列都增大了3倍的问题,没法一个一个X输出.所以用二维数组将结果保存起来,每次替换都要增大数组的size.
#include <iostream>
#include<stdio.h>
#include <vector>
using namespace std;
const int nScale = 3; //放大倍数
void fun(vector<vector<char>>& vec,int &nCycles)
{
if (nCycles == 1)
return;
int nSrcSize = vec.size();
vector<vector<char>> vecNew;;
vecNew.resize(nSrcSize*nScale);
for (int i = 0; i < vecNew.size(); ++i)
vecNew[i].resize(nSrcSize * nScale);
for(int i = 0; i< nSrcSize; i++)
{
for(int j = 0; j < nSrcSize; j++)
{
char nTemp = vec[i][j];
if(nTemp == 'X')
{
vecNew[nScale*i+0][nScale*j + 0] = 'X';
vecNew[nScale*i+0][nScale*j + 1] = ' ';
vecNew[nScale*i+0][nScale*j + 2] = 'X';
vecNew[nScale*i+1][nScale*j + 0] = ' ';
vecNew[nScale*i+1][nScale*j + 1] = 'X';
vecNew[nScale*i+1][nScale*j + 2] = ' ';
vecNew[nScale*i+2][nScale*j + 0] = 'X';
vecNew[nScale*i+2][nScale*j + 1] = ' ';
vecNew[nScale*i+2][nScale*j + 2] = 'X';
}
else if(nTemp == ' ')
{
vecNew[nScale*i + 0][nScale*j + 0] = ' ';
vecNew[nScale*i + 0][nScale*j + 1] = ' ';
vecNew[nScale*i + 0][nScale*j + 2] = ' ';
vecNew[nScale*i + 1][nScale*j + 0] = ' ';
vecNew[nScale*i + 1][nScale*j + 1] = ' ';
vecNew[nScale*i + 1][nScale*j + 2] = ' ';
vecNew[nScale*i + 2][nScale*j + 0] = ' ';
vecNew[nScale*i + 2][nScale*j + 1] = ' ';
vecNew[nScale*i + 2][nScale*j + 2] = ' ';
}
}
}
vec.swap(vecNew);
fun(vec, --nCycles);
}
int main()
{
vector<vector<char>> vec = {{'X'}};
int nCycles = 1;
scanf_s("%d", &nCycles);
fun(vec, nCycles);
cout << "result:" << endl;
for (int i = 0; i < vec.size(); i++)
{
for (int j = 0; j < vec[i].size(); j++)
{
char nTemp = vec[i][j];
cout << nTemp;
}
cout << endl;
}
return 0;
}