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

2

0

60

 

Sample Output

Case #1:

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

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

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

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

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

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

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

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

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

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

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

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

Case #2:

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

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

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

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

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

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

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

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

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

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

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

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

 

Source

简单题：

代码：

1 #include
       
         2 #include
        
          3 using namespace std; 4 int main() 5 { 6     int i,j,t,n,cnt; 7     scanf("%d",&t); 8     for(cnt=1;cnt<=t;cnt++) 9     {10       scanf("%d",&n);11       printf("Case #%d:\n",cnt);12       for(i=0;i<12;i++)13       {14         for(j=0;j<14;j++)15         {16             if(i==0||i==11)17             {18                 if(j==0||j==13)19                      printf("*");20                 else21                     printf("-");22             }23             else24                 if(j==0||j==13)25                     printf("|");26                 else27                    if(i<=10-n/10)28                        printf(".");29                 else30                       printf("-");31         }32          puts("");33       }34     }35     return 0;36 }