原题：http://acm.hdu.edu.cn/showproblem.php?pid=6330
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
..+-+ ././| +-+.+ |.|/. +-+.. ....+-+-+-+-+-+-+ .../././././././| ..+-+-+-+-+-+-+.+ ./././././././|/| +-+-+-+-+-+-+.+.+ |.|.|.|.|.|.|/|/| +-+-+-+-+-+-+.+.+ |.|.|.|.|.|.|/|/| +-+-+-+-+-+-+.+.+ |.|.|.|.|.|.|/|/. +-+-+-+-+-+-+.+.. |.|.|.|.|.|.|/... +-+-+-+-+-+-+....分析：模拟题，题目意思是输入a, b, c代表一个长方体的长宽高，要求画出这个长方体的3D图。做法很容易，只不过要注意细节，做的时候因为一个等号的原因，挖了两发。
#include<bits/stdc++.h> using namespace std; const int maxn = 500; char pic[maxn][maxn]; int t, n, m; int a, b, c; int x, y; int main(){ scanf("%d", &t); while(t--){ memset(pic, '.', sizeof(pic)); scanf("%d%d%d", &a, &b, &c); x = 2*(c+b) + 1; y = 2*(a+b) + 1; for(int j=x; j>=2*b+1; j-=2){ for(int i=1; i<=2*a+1; i++){ if(i&1) pic[j][i] = '+'; else pic[j][i] = '-'; } } for(int j=x-1; j>2*b+1; j-=2){ for(int i=1; i<=2*a+1; i+=2){ if(i&1) pic[j][i] = '|'; } } int pos = 2; for(int i=2*b; i>=1; i--){ for(int j=pos; j<2*a+1+pos; j++){ if(i&1){ if(j&1) pic[i][j] = '+'; else pic[i][j] = '-'; } else{ if(j%2 == 0) pic[i][j] = '/'; } } pos++; } pos = x-1; for(int j=2*a+2; j<=y; j++){ for(int i=pos; i>pos-2*c; i--){ if(j%2 == 0){ if(i%2 == 0) pic[i][j] = '/'; } else { if(i&1) pic[i][j] = '+'; else pic[i][j] = '|'; } } pos--; } for(int i=1; i<=x; i++){ for(int j=1; j<=y; j++){ printf("%c", pic[i][j]); } printf("\n"); } } }
