HDU 多校对抗第三场 L Visual Cube

Problem 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 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：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("
");
        }
    }

}