HDUOJ----A Computer Graphics Problem

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

2013 ACM/ICPC Asia Regional Online —— Warmup2

简单题：

代码：

 1 #include<iostream>

 2 #include<cstdio>

 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                 else

21                     printf("-");

22             }

23             else

24                 if(j==0||j==13)

25                     printf("|");

26                 else

27                    if(i<=10-n/10)

28                        printf(".");

29                 else

30                       printf("-");

31         }

32          puts("");

33       }

34     }

35     return 0;

36 }

View Code