hdu4716 A Computer Graphics Problem（模拟）

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

2060

Sample Output

Case #1:*------------*|............||............||............||............||............||............||............||............||............||............|*------------*Case #2:*------------*|............||............||............||............||------------||------------||------------||------------||------------||------------|*------------*

Source

2013 ACM/ICPC Asia Regional Online —— Warmup2 

题意：模拟电池剩余电量，输入只能是100以内10的整数倍。

分析：超级水的模拟题，直接输出即可。

#include <iostream>#include <cstdio>#include <cstring>#include <stack>#include <queue>#include <map>#include <set>#include <vector>#include <cmath>#include <algorithm>using namespace std;const double eps = 1e-6;const double pi = acos(-1.0);const int INF = 0x3f3f3f3f;const int MOD = 1000000007;#define ll long long#define CL(a) memset(a,0,sizeof(a))int main (){    int T,ii,n;    scanf ("%d",&T);    for (ii=1; ii<=T; ii++)    {        scanf ("%d",&n);        printf ("Case #%d:\n",ii);        printf ("*------------*\n");        n=10-n/10;//n表示已使用电量        for (int i=0; i<10; i++)        {            if (n)            {                printf ("|............|\n");                n--;            }            else            {                printf ("|------------|\n");            }        }        printf ("*------------*\n");    }    return 0;}