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 aa, width bb and height cc, please write a program to display the cube.
Input
The first line of the input contains an integer T(1≤T≤50)T(1≤T≤50), denoting the number of test cases.
In each test case, there are 33 integers a,b,c(1≤a,b,c≤20)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
..+-+ ././| +-+.+ |.|/. +-+.. ....+-+-+-+-+-+-+ .../././././././| ..+-+-+-+-+-+-+.+ ./././././././|/| +-+-+-+-+-+-+.+.+ |.|.|.|.|.|.|/|/| +-+-+-+-+-+-+.+.+ |.|.|.|.|.|.|/|/| +-+-+-+-+-+-+.+.+ |.|.|.|.|.|.|/|/. +-+-+-+-+-+-+.+.. |.|.|.|.|.|.|/... +-+-+-+-+-+-+....
【解析】
题意:给你长a宽b高c,打印出这个长方体。
模拟完了真的心累。
#include <bits/stdc++.h>
using namespace std;
int main()
{
char cub[100][100];
int a, b, c, t;
scanf("%d", &t);
while (t--)
{
memset(cub, '.', sizeof(cub));
scanf("%d%d%d", &a, &b, &c);
for (int j = 2*b+2*c+1; j >= 2 * b + 1; j -= 2) //正面
for (int i = 1; i <= 2 * a + 1; i++)
{
if (i & 1) cub[j][i] = '+';
else cub[j][i] = '-';
}
for (int j = 2 * b + 2 * c; j>2 * b + 1; j -= 2)//正面
for (int i = 1; i <= 2 * a + 1; i += 2)
cub[j][i] = '|';
int beg = 2;
for (int i = 2 * b; i >= 1; i--)//上面
{
for (int j = beg; j < 2 * a + 1 + beg; j ++)
{
if (i & 1)
{
if (j & 1) cub[i][j] = '+';
else cub[i][j] = '-';
}
else
if (j % 2 == 0) cub[i][j] = '/';
}
beg++;
}
beg = 2 * b + 2 * c;
for (int j = 2 * a + 2; j <= 2*a+2*b+1; j++)//侧面
{
for (int i = beg; i > beg - 2 * c; i--)
{
if (j &1 )
{
if (i & 1) cub[i][j] = '+';
else cub[i][j] = '|';
}
else
if (i % 2 == 0) cub[i][j] = '/';
}
beg--;
}
for (int i = 1; i <= 2 * b + 2 * c + 1; i++)
{
for (int j = 1; j <= 2*a+2*b+1; j++)
printf("%c", cub[i][j]);
printf("\n");
}
}
return 0;
}