A Computer Graphics Problem

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 92 Accepted Submission(s): 80

Problem Description

In this problem we talk about the study of Computer Graphics. Of course, this is very, very hard. We have designed a new mobile phone, your task is to write a interface to display battery powers. Here we use '.' as empty grids. When the battery is empty, the interface will look like this:

*------------*
|............|
|............|
|............|
|............| 
|............| 
|............| 
|............| 
|............| 
|............| 
|............| 
*------------*

When the battery is 60% full, the interface will look like this:

*------------* 
|............| 
|............| 
|............| 
|............| 
|------------| 
|------------| 
|------------| 
|------------| 
|------------| 
|------------| 
*------------*

Each line there are 14 characters. Given the battery power the mobile phone left, say x%, your task is to output the corresponding interface. Here x will always be a multiple of 10, and never exceeds 100.

Input

The first line has a number T (T < 10) , indicating the number of test cases. For each test case there is a single line with a number x. (0 < x < 100, x is a multiple of 10)

Output

For test case X, output "Case #X:" at the first line. Then output the corresponding interface. See sample output for more details.

Sample Input

Sample Output

Case #1:

*------------*

|............|

*------------*

Case #2:

*------------*

|............|

|------------|

*------------*

Source

2013 ACM/ICPC Asia Regional Online —— Warmup2

简单题：

代码：

 #include<iostream>

 #include<cstdio>

 using namespace std;

 int main()

 {

     int i,j,t,n,cnt;

     scanf("%d",&t);

     for(cnt=;cnt<=t;cnt++)

     {

       scanf("%d",&n);

       printf("Case #%d:\n",cnt);

       for(i=;i<;i++)

       {

         for(j=;j<;j++)

         {

             if(i==||i==)

             {

                 if(j==||j==)

                      printf("*");

                 else

                     printf("-");

             }

             else

                 if(j==||j==)

                     printf("|");

                 else

                    if(i<=-n/)

                        printf(".");

                 else

                       printf("-");

         }

          puts("");

       }

     }

     return ;

 }

巴特西

HDUOJ----A Computer Graphics Problem

A Computer Graphics Problem

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