HDUOJ----A Computer Graphics Problem

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Gxjun 2022-03-18 14:52:29 ©著作权

文章标签 基础题 HDUOJ java #include ios 文章分类 代码人生

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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：

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 }