2018 Multi-University Training Contest 3
Problem Description
Little Q likes solving math problems very much. Unluckily, however, he does not have good spatial ability. Everytime he meets a 3D geometry problem, he will struggle to draw a picture.
Now he meets a 3D geometry problem again. This time, he doesn't want to struggle any more. As a result, he turns to you for help.
Given a cube with length a, width b and height c, please write a program to display the cube.
Input
The first line of the input contains an integer T(1≤T≤50), denoting the number of test cases.
In each test case, there are 3 integers a,b,c(1≤a,b,c≤20), denoting the size of the cube.
Output
For each test case, print several lines to display the cube. See the sample output for details.
Sample Input
2
1 1 1
6 2 4
Sample Output
..+-+
././|
+-+.+
|.|/.
+-+..
....+-+-+-+-+-+-+
.../././././././|
..+-+-+-+-+-+-+.+
./././././././|/|
+-+-+-+-+-+-+.+.+
|.|.|.|.|.|.|/|/|
+-+-+-+-+-+-+.+.+
|.|.|.|.|.|.|/|/|
+-+-+-+-+-+-+.+.+
|.|.|.|.|.|.|/|/.
+-+-+-+-+-+-+.+..
|.|.|.|.|.|.|/...
+-+-+-+-+-+-+....模拟题：个人水平有限，写的超级复杂，还是太蒻了
超霸说找规律so easy~ QAQ
AC代码：简单看看就好，代码权值几乎为0
#include<bits/stdc++.h>
using namespace std;
int main()
{
#ifdef LOCAL_FILE
freopen("in.txt","r",stdin);
#endif // LOCAL_FILE
ios_base::sync_with_stdio(0);
cin.tie(0),cout.tie(0);
int t;
int c,k,g;
cin>>t;
while(t--)
{
cin>>c>>k>>g;
for(int i=1;i<=2*(g+k)+1;i++)
{
for(int j=1;j<=2*(c+k)+1;j++)
{
if(i>=1 && i<=(2*k)+1-1)//上
{
if(i%2 == 1)
{
if(i <= 2*g+1)
{
if(j>=1 && j<=2*(c+k)+1-(2*c+i)+2*c+1+(i-1))
{
if((2*(c+k)+1)-j<(2*c+i))
{
if((j-(2*(c+k)+1-(2*c+i)))>=1 && (j-(2*(c+k)+1-(2*c+i)))<=2*c+1 && j%2 == 0)
{
cout<<"-";
}
else if(j%2 == 1)
cout<<"+";
else
cout<<".";
}
else
cout<<".";
}
}
else
{
if(j>=1 && j<=2*(c+k)+1-(2*c+i)+2*c+1+(2*g+1-1))
{
if((2*(c+k)+1)-j<(2*c+i))
{
if((j-(2*(c+k)+1-(2*c+i)))>=1 && (j-(2*(c+k)+1-(2*c+i)))<=2*c+1 && j%2 == 0)
{
cout<<"-";
}
else if(j%2 == 1)
cout<<"+";
else
cout<<".";
}
else
cout<<".";
}
else
cout<<".";
}
}
else if(i%2 == 0)
{
if(i <= 2*g+1)
{
if(j>=1 && j<=2*(c+k)+1-(2*c+i)+2*c+1+(i-1))
{
if((2*(c+k)+1)-j<(2*c+i))
{
if((j-(2*(c+k)+1-(2*c+i)))>=1 && (j-(2*(c+k)+1-(2*c+i)))<=2*c+1 && j%2 == 1)
{
cout<<".";
}
else if(j%2 == 0 && j!=2*(c+k)+1)
cout<<"/";
else
cout<<"|";
}
else
cout<<".";
}
}
else
{
if(j>=1 && j<=2*(c+k)+1-(2*c+i)+2*c+1+(2*g+1-1))
{
if((2*(c+k)+1)-j<(2*c+i))
{
if((j-(2*(c+k)+1-(2*c+i)))>=1 && (j-(2*(c+k)+1-(2*c+i)))<=2*c+1 && j%2 == 1)
{
cout<<".";
}
else if(j%2 == 0 && j!=2*(c+k)+1)
cout<<"/";
else
cout<<"|";
}
else
cout<<".";
}
else
cout<<".";
}
}
}
else if(i>=(2*k+1) && i<=(2*k+1)+(2*(g+k)+1-2*(2*k)-1))//中
{
if(i%2 == 1)
{
if(j>=1 && j<=(2*c+1) && j%2 == 0)
cout<<"-";
else if(j%2 == 1)
cout<<"+";
else
cout<<".";
}
else if(i%2 == 0)
{
if(j>=1 && j<=(2*c+1) && j%2 == 0)
cout<<".";
else if(j%2 == 1)
cout<<"|";
else
cout<<"/";
}
}
else if(i>=(2*k+1)+(2*(g+k)+1-2*(2*k)) && i<=2*(g+k)+1)//下
{
if(i%2 == 0)
{
if(j>=1 && j<=( (2*(c+k)+1)-((i-((2*k+1)+(2*(g+k)+1-2*(2*k)-1)))) ) )
{
if(j>=1 && j<=(2*c+1) && j%2 == 0)
cout<<".";
else if(j%2 == 1)
cout<<"|";
else
cout<<"/";
}
else
cout<<".";
}
else if(i%2 == 1)
{
if(j>=1 && j<=( (2*(c+k)+1)-((i-((2*k+1)+(2*(g+k)+1-2*(2*k)-1)))) ) )
{
if(j>=1 && j<=(2*c+1) && j%2 == 0)
cout<<"-";
else if(j%2 == 1)
cout<<"+";
else
cout<<".";
}
else
cout<<".";
}
}
}
cout<<endl;
}
}
return 0;
}