2018暑期HDU多校第三场—— L Visual Cube

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

..+-+
././|
+-+.+
|.|/.
+-+..
....+-+-+-+-+-+-+
.../././././././|
..+-+-+-+-+-+-+.+
./././././././|/|
+-+-+-+-+-+-+.+.+
|.|.|.|.|.|.|/|/|
+-+-+-+-+-+-+.+.+
|.|.|.|.|.|.|/|/|
+-+-+-+-+-+-+.+.+
|.|.|.|.|.|.|/|/.
+-+-+-+-+-+-+.+..
|.|.|.|.|.|.|/...
+-+-+-+-+-+-+....

题解：

就是简单的模拟输出图像，不过如果是一行一行输出的要注意长宽较大高较小的情况，建议把图像拆成一个矩形和两个菱形来输出，这样可以回避很多麻烦。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

char board[105][105];

int main(){
    
    int T;
    int A,B,C;
    scanf("%d",&T);
    while(T--){
        memset(board,'.',sizeof board);
        scanf("%d %d %d",&A,&B,&C);
        int L = 2*C+1+2*B;
        int LL = 2*A+2*B+1;
        
        for(int i=1 ; i<=L ; ++i){//填侧面的菱形 
            int l1 = 0,l2 = 0;
            if(i<=B*2){
                l1 = 2*B-i+1+1+2*A;
            }
            if(L-i<B*2){
                l2 = 2*B-(L-i);
            }
            for(int j=l1+1 ; j<=LL-l2 ; ++j){
                if(i&1){
                    if(j&1)board[i][j] = '+';
                    else board[i][j] = '.';
                }
                else {
                    if(j&1)board[i][j] = '|';
                    else board[i][j] = '/';
                }
            }
        }
        
        for(int i=1 ; i<=B*2 ; ++i){//填上边的菱形 
            int t = 2*B-i+1+1;
            for(int j=t ; j<=t+2*A-1 ; j+=2){
                if(i&1){
                    board[i][j] = '+';
                    board[i][j+1] = '-';
                }
                else{
                    board[i][j] = '/';
                    board[i][j+1] = '.';    
                }
            }
            if(i&1)board[i][t+2*A] = '+';
            else board[i][t+2*A] = '/';
        }
        
        for(int i=L ; i>2*B ; i--){//填左下角的大矩形 
            if(i&1){
                for(int j=1 ; j<=2*A ; j+=2){
                    board[i][j] = '+';
                    board[i][j+1] = '-';
                }
                board[i][2*A+1] = '+';
            }
            else {
                for(int j=1 ; j<=2*A ; j+=2){
                    board[i][j] = '|';
                    board[i][j+1] = '.';
                }
                board[i][2*A+1] = '|';
            }
        }
        
        for(int i=1 ; i<=L ; ++i){//输出 
            for(int j=1 ; j<=LL ; ++j){
                printf("%c",board[i][j]);
            }
            printf("\n");
        }
    }
    
    return 0;
}