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：

#include<iostream>
#include<cstdio>
using namespace std;
int main()
{
    int i,j,t,n,cnt;
    scanf("%d",&t);
    for(cnt=1;cnt<=t;cnt++)
    {
      scanf("%d",&n);
      printf("Case #%d:\n",cnt);
      for(i=0;i<12;i++)
      {
        for(j=0;j<14;j++)
        {
            if(i==0||i==11)
            {
                if(j==0||j==13)
                     printf("*");
                else
                    printf("-");
            }
            else
                if(j==0||j==13)
                    printf("|");
                else
                   if(i<=10-n/10)
                       printf(".");
                else
                      printf("-");
        }
         puts("");
      }
    }
    return 0;
}