HDU 多校对抗第三场 L Visual Cube

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 2127    Accepted Submission(s): 984

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 c, please write a program to display the cube.

 

 

Input

The first line of the input contains an integer ), 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：给你长方体的长宽高，要你构造出（画出，一个按照题目要求给的长方体

题解：真tm变态呜呜呜，这个题，考你码力和找规律的能力，代码看了就懂了

　　一个面一个面的画，因为长方体在平面上只能展现三个面，所以循环六次就可以画出来了

代码如下：

#include <map>
#include <set>
#include <cmath>
#include <ctime>
#include <stack>
#include <queue>
#include <cstdio>
#include <cctype>
#include <bitset>
#include <string>
#include <vector>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <functional>
#define fuck(x) cout<<"["<<x<<"]";
#define FIN freopen("input.txt","r",stdin);
#define FOUT freopen("output.txt","w+",stdout);
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
typedef long long LL;
typedef pair<int, int> PII;
const int maxn = 1e5+5;

char mp[400][400];
int main(){
    #ifndef ONLINE_JUDGE
    FIN
    #endif
    int T;
    int a,b,c;
    scanf("%d",&T);
    while(T--){
        memset(mp,0,sizeof(mp));
        scanf("%d%d%d",&a,&b,&c);
        
        for(int i=a*2+3;i<=a*2+b*2+1;i+=2){
            for(int j=1;j<=b*2+c*2+1;j++){
                mp[j][i]=(j&1)?'+':'|';
            }
        }

        for(int i=a*2+2;i<=a*2+b*2+1;i+=2){
            for(int j=1;j<=b*2+c*2+1;j++){
                mp[j][i]=(j&1)?'.':'/';
            }
        }

        for(int i=1;i<=b*2;i+=2){
            for(int j=b*2+2-i;j<=a*2+b*2+2-i;j++){
                mp[i][j]=(j&1)?'+':'-';
            }
        }

        for(int i=2;i<=b*2;i+=2){
            for(int j=b*2+2-i;j<=a*2+b*2+2-i;j++){
                mp[i][j]=(j&1)?'.':'/';
            }
        }

        for(int i=2*b+1;i<=2*b+2*c+1;i+=2){
            for(int j=1;j<=2*a+1;j++){
                mp[i][j]=(j&1)?'+':'-';
            }
        }

        for(int i=2*b+2;i<=2*b+2*c;i+=2){
            for(int j=1;j<=2*a+1;j++){
                mp[i][j]=(j&1)?'|':'.';
            }
        }
        
        int x=c*2+b*2+1;
        int y=b*2+a*2+1;
        for(int i=1;i<=b*2;i++){
            for(int j=1;j<=2*b-i+1;j++){
                mp[x-i+1][y-j+1]='.';
            }
        }

        for(int i=1;i<=b*2;i++){
            for(int j=1;j<=2*b-i+1;j++){
                mp[i][j]='.';
            }
        }
        for(int i=1;i<=x;i++){
            for(int j=1;j<=y;j++){
                printf("%c",mp[i][j]);
            }
            printf("\n");
        }
    }

}

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