题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=6330
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：
题意：输入T组测试样例，每组测试样例输入立方体的长宽高，然后打印出次立方体（例如样例输出）。
思路：这道题其实不难，就是一行一行的模拟，但要注意的是模拟的时候是通过模拟上面、前面、侧面的方法模拟的，而不是根据前面几行、后面几行和中间几行的方法模拟的，这样是错的，因为如果高很小，而宽很大时（即宽远大于高时）你就会发现这种方法的中间几行变成负的了，所以这种方法不可行，我一开始就是这样做的，还死活找不到错哪，可怜我快wa哭了（唉，弱鸡）
My DaiMa：
//#include <iostream>
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
//int n,a[55][55];
int main()
{
int t,a,b,c;
char s[100][100];
cin >> t;
while( t-- )
{
cin >> a >> b >> c;
for( int i = 0; i < 100; i++)
{
for( int j = 0; j < 100; j++)
s[i][j] = '.';
}
for(int i = 0; i < 2*b; i++)//上面
{
for( int j = 2*b-i ; j < 2*a+2*b-i; j++)
{
if(i%2 == 0)
{
if(j%2 == 0)
s[i][j] = '+';
else
s[i][j] = '-';
}
else
{
if(j%2 != 0)
s[i][j] = '/';
}
}
}
for( int i = 2*b; i < 2*c+1+2*b; i++)//前面
{
for( int j = 0; j < 2*a+1; j++)
{
if(i%2 == 0)
{
if(j%2 == 0)
s[i][j] = '+';
else
s[i][j] = '-';
}
else
{
if(j%2 == 0)
s[i][j] = '|';
}
}
}
for(int j = 2*a+1; j < 2*b+2*a+1; j++)//侧面
{
for( int i = 2*a+2*b-j; i < 2*a+2*b+1-j+2*c; i++)
{
if(j%2 == 0)
{
if(i%2 == 0)
s[i][j] = '+';
else
s[i][j] = '|';
}
else
{
if(i%2 != 0)
s[i][j] = '/';
}
}
}
for(int i = 0; i < 2*c+1+2*b; i++)
{
for(int j = 0; j < 2*a+1+2*b; j++)
printf("%c",s[i][j]);
printf("\n");
}
}
}