HDU4716：A Computer Graphics Problem-优快云博客

本文链接：https://blog.youkuaiyun.com/libin56842/article/details/11580761

本文介绍了一种简单的方法来设计手机电池电量的显示界面。通过使用特定字符来表示不同的电量状态，可以直观地展示出手机剩余电量的情况。文章提供了一个示例程序，展示了如何根据输入的百分比来生成相应的电池图标。

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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:
*------------*
|............|
|............|
|............|
|............|
|------------|
|------------|
|------------|
|------------|
|------------|
|------------|
*------------*

水题不解释

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;

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

    return 0;
}