Hdu4716 A Computer Graphics Problem (simulation), hdu4716graphics
Problem DescriptionIn 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.
InputThe 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)
OutputFor 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:*------------*|............||............||............||............||------------||------------||------------||------------||------------||------------|*------------*
Source2013 ACM/ICPC Asia Regional Online -- Warmup2
Simulate the remaining battery power. The input can only be an integer multiple of 10 in the range of 100. Analysis: Super Water Simulation questions can be directly output.
# 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 # define CL () 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 indicates that the power is used for (int I = 0; I <10; I ++) {if (n) {printf ("| ............ | \ n "); n --;} else {printf (" | ------------ | \ n ") ;}} printf (" * ------------ * \ n ");} return 0 ;}
Copyright Disclaimer: This article is an original article by the blogger and cannot be reproduced without the permission of the blogger.