HDU6330 Visual Cube

3D立方体图形绘制

最新推荐文章于 2019-07-22 10:07:00 发布

小鱼爱吃火锅

最新推荐文章于 2019-07-22 10:07:00 发布

阅读量116

点赞数

CC 4.0 BY-SA版权

本文链接：https://blog.youkuaiyun.com/daoshen1314/article/details/87820857

思维专栏收录该内容

42 篇文章

订阅专栏

本文介绍了一种使用C++编程语言绘制3D长方体图形的方法，通过具体实例展示了如何根据给定的长宽高参数，生成相应的立体图形表示。代码详细解释了绘制过程，包括顶部、前部和侧面的构造。

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

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

题意：

给你一个长方体的长宽高，让你画出这个立体图形

代码：

#include<bits/stdc++.h>
using namespace std;
#define mem(a,b)memset(a,b,sizeof(a))
char p[200][200];
void draw(int x,int y)
{
    for(int i=1; i<=x; i++)
    {
        for(int j=1; j<=y; j++)
            printf("%c",p[i][j]);
        printf("\n");
    }
}
int main()
{
    int t,a,b,c;
    scanf("%d",&t);
    while(t--)
    {
        mem(p,'.');
        scanf("%d%d%d",&a,&b,&c);
        for(int i=1; i<=2*b; i++) //上面
            if(i&1)
            {
                for(int j=2*b+1-i+1,k=1; j<=2*a+2*b+1&k<=a+1; j+=2,k++)
                    p[i][j]='+';
                for(int j=2*b+1-i+2,k=1; j<=2*a+2*b+1&&k<=a; j+=2,k++)
                    p[i][j]='-';
            }
            else
                for(int j=2*b+1-i+1,k=1; j<=2*a+2*b+1&&k<=a+1; j+=2,k++)
                    p[i][j]='/';
        for(int i=2*b+1; i<=2*c+2*b+1; i++) //正面
            if(i&1)
            {
                for(int j=1,k=1; k<=a+1; j+=2,k++)
                    p[i][j]='+';
                for(int j=2,k=1; k<=a; j+=2,k++)
                    p[i][j]='-';
            }
            else
                for(int j=1,k=1; k<=a+1; j+=2,k++)
                    p[i][j]='|';
        for(int i=2*a+2*b+1,k=1; k<=2*b; i--,k++) //侧面
            if(k&1)
                for(int j=k,pos=1; j<=k+2*c; j++,pos++)
                    if(pos&1)
                        p[j][i]='+';
                    else
                        p[j][i]='|';
            else
                for(int j=k,pos=1; j<=k+2*c+1; j++,pos++)
                    if(pos&1)
                        p[j][i]='/';
        draw(2*c+2*b+1,2*a+2*b+1);// x=2*c+2*b+1 y=2*a+2*b+1 
    }
    return 0;
}