2018HDU多校赛-Problem L. Visual Cube 【模拟】

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：不用多少都是理解的，就是打印立方体

题解：这题就是模拟，但是在模拟的时候容易出错。用一个二维数组保存，分为最上面，正面和侧面来模拟，这题需要耐心和要保持头脑清晰。具体代码如下：

#include<bits/stdc++.h>

using namespace std;
char cub[100][100];
int main(void){
    int T;
    scanf("%d",&T);
    while(T--){
        int a,b,c;
        scanf("%d%d%d",&a,&b,&c);
        for(int i=1;i<100;i++)
        for(int j=1;j<100;j++)
        cub[i][j]='.';
        
        int d=2*b+1;
        int f=2*b;
        
        //上面 
        for(int i=1;i<=f;i++){
            if(i%2){
                for(int j=d;j<=d+2*a;j++){
                    if(j%2) cub[i][j]='+';
                    else cub[i][j]='-';
                }
            }
            else{
                for(int j=d;j<=d+2*a;j++){
                    if(j%2)    cub[i][j]='.';
                    else    cub[i][j]='/';
                }
           }
           d--; 
        } 
        
        //正面 
            int l=2*b+2*c+1;
            int m=2*a+1;
            for(int i=f+1;i<=l;i++){ // f=2*b
                if((i-f)%2){
                    for(int j=1;j<=m;j++){
                        if(j%2) cub[i][j]='+';
                        else cub[i][j]='-';    
                    }
                }
                else{
                    for(int j=1;j<=m;j++){
                        if(j%2) cub[i][j]='|';
                        else cub[i][j]='.';    
                    }
                }
    
            }
            
        //侧面 
        int indexi=2*b+1;
        for(int j=m; j<=m+b*2; j++)
        {
            if((j-a*2)%2)
            {
                for(int i=indexi; i<=indexi+c*2; i++)
                {
                    if((i-indexi)%2==0)
                        cub[i][j]='+';
                    else cub[i][j]='|';
                }
            }
            else
            {
                for(int i=indexi; i<=indexi+c*2; i++)
                {
                    if((i-indexi)%2==0)
                        cub[i][j]='/';
                    else cub[i][j]='.';
                }

            }
            indexi--;
        } 
    
    int row=b*2+1+c*2,col= a*2+b*2+1;
    
    for(int i=1; i<=row; i++)
    {
        for(int j=1; j<=col; j++)
            printf("%c",cub[i][j]);
        puts("");
    }
    }
    return 0;
}