For this problem, you are to output the Hilbert Curve up to a certain recursion depth, using just 'X'.Draw the smallest Hilbert curve (that is not divided any further) as follows:
XXX X XXX To see how to draw larger Hilbert curve, take a look at the sample output. InputThe input contains several testcases. Each is specified by an integer n (1 < n < 9). OutputFor each test case, output the Hilbert curve with the certain recursion depth n.The output must not contain any trailing blanks. Print an empty line after each test case. Sample Input
2 3 Sample Output
XXXXX X
X X X
XXX XXX
X
XXX XXX
X X X
XXXXX X
XXXXX XXXXX XXX
X X X X X
XXX XXX XXX XXX
X X X
XXX XXX X XXX X
X X X X X X X
XXXXX X XXX XXX
X
XXXXX X XXX XXX
X X X X X X X
XXX XXX X XXX X
X X X
XXX XXX XXX XXX
X X X X X
XXXXX XXXXX XXX
#include<iostream>
using namespace std;
void main()
{
int lines[9];
lines[0]=1;
for(int i=1;i<9;i++)
lines[i]=lines[i-1]*2+1;
char Hilbert[520][520];
Hilbert[0][0]='X';
int r=1;
for(int i=1;i<9;i++)
{
for(int j=0;j<r;j++)
for(int k=0;k<r;k++)
{
Hilbert[j][r+1+k]=Hilbert[k][r-1-j];
Hilbert[r+1+j][k]=Hilbert[j][k];
Hilbert[r+1+j][r+1+k]=Hilbert[r-1-k][j];
}
Hilbert[0][r]='X';
Hilbert[r][r-1]='X';
Hilbert[2*r][r]='X';
r=r*2+1;
}
int n;
while(cin>>n)
{
for(int i=0;i<lines[n];i++)
{
int len=lines[n]-1;
while(Hilbert[i][len]!='X')
len--;
for(int j=0;j<=len;j++)
if(Hilbert[i][j]=='X')
cout<<Hilbert[i][j];
else
cout<<' ';
cout<<endl;
}
cout<<endl;
}
}
没有对数组进行初始化，如果全部先赋为空格反而会费时间 在打印时判断更好
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