A Computer Graphics Problem
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 43 Accepted Submission(s): 40
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:
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)
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.
See sample output for more details.
Sample Input
2
0
60
0
60
Sample Output
Case #1:
*------------*
|............|
|............|
|............|
|............|
|............|
|............|
|............|
|............|
|............|
|............|
*------------*
Case #2:
*------------*
|............|
|............|
|............|
|............|
|------------|
|------------|
|------------|
|------------|
|------------|
|------------|
*------------*
*------------*
|............|
|............|
|............|
|............|
|............|
|............|
|............|
|............|
|............|
|............|
*------------*
Case #2:
*------------*
|............|
|............|
|............|
|............|
|------------|
|------------|
|------------|
|------------|
|------------|
|------------|
*------------*
Source
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/* ***********************************************
Author :kuangbin
Created Time :2013-9-11 12:01:07
File Name :2013-9-11\1001.cpp
************************************************ */ #include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;
int main()
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
int iCase = ;
int T;
int n;
scanf("%d",&T);
while(T--)
{
iCase++;
printf("Case #%d:\n",iCase);
scanf("%d",&n);
printf("*");
for(int i = ;i < ;i++)printf("-");
printf("*\n");
int t = n/;
for(int i = ;i < -t;i++)
{
printf("|");
for(int j = ;j < ;j++)printf(".");
printf("|\n");
}
for(int i = ;i < t;i++)
{
printf("|");
for(int j = ;j < ;j++)printf("-");
printf("|\n");
}
printf("*");
for(int i = ;i < ;i++)printf("-");
printf("*\n"); }
return ;
}