A Computer Graphics Problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

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

比较简单，思路可以自己找。

#include<cstdio>
int main()
{
  int t;
  int n,k=1;
  scanf("%d",&t);
  while(t--)
  {
    scanf("%d",&n);
	printf("Case #%d:\n",k++);
	printf("*------------*\n");
	for(int i=0;i<10-n/10;i++)
		printf("|............|\n");
	for(int j=0;j<n/10;j++)
	printf("|------------|\n");
		printf("*------------*\n");
  }
  return 0;
}

A Computer Graphics Problem 4176 2013上海邀请赛