Problem L. Visual Cube
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 1473 Accepted Submission(s): 694
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
解题思路：立体图分为：俯视图 + 正视图 + 右视图，此题先刷 俯视图 + 正视图，最后刷 右视图；一开始先记录每个视图有哪些规律图案，比如：“+-+-+-+”、“/././././”，然后再创建一个全是“点”的二维数组，把它描上去即可。
AC 代码
#include<bits/stdc++.h> #include<cmath> #define mem(a,b) memset(a,b,sizeof a); #define INF 0x3f3f3f3f using namespace std; typedef long long ll; int main() { int T; scanf("%d",&T); int a,b,c; while(T-- && ~scanf("%d%d%d",&a,&b,&c)) { string zf,xd,sd,dx,sz; for(int i=0;i<a;i++) { zf.append("+-"); xd.append("/."); sd.append("|."); } zf.append("+"); xd.append("/"); sd.append("|"); for(int i=0;i<c;i++) { sz.append("|+"); dx.append("./"); } // cout<<"+-："<<zf<<endl; // cout<<"/.："<<xd<<endl; // cout<<"|.："<<sd<<endl; // cout<<"./："<<dx<<endl; // cout<<"|+："<<sz<<endl; int rlen=2*c+1+2*b,clen=2*a+1+2*b; int con=2*b; char g[rlen+5][clen+5]; mem(g,'.'); for(int i=0;i<rlen;i++,con--) { for(int j=0;j<clen;j++) { if(con>0) { if(j+con<clen && i%2==0 && j<zf.length()) g[i][j+con]=zf[j]; else if(j+con<clen && i%2==1 && j<xd.length()) g[i][j+con]=xd[j]; } else { if(i%2==0 && j<zf.length()) g[i][j]=zf[j]; else if(i%2==1 && j<sd.length()) g[i][j]=sd[j]; } } } // 正视图和俯视图刷完，可以快速排错 // for(int i=0;i<rlen;i++) // { // for(int j=0;j<clen;j++) // { // printf("%c",g[i][j]); // } // puts(""); // } // puts("-----------------"); // 右视图刷完 con=2*b; int h=2*c,t=0; for(int j=clen-1;j>=0 && con--;j--,t++) { for(int i=1+t;i<=h+t;i++) { if(j%2==0 && i-1-t<sz.length()) g[i][j]=sz[i-1-t]; else if(j%2==1 && i-1-t<dx.length()) g[i][j]=dx[i-1-t]; } } for(int i=0;i<rlen;i++) { for(int j=0;j<clen;j++) { printf("%c",g[i][j]); } puts(""); } } return 0; }
