A Computer Graphics Problem 4176 2013上海邀請賽

標籤：演算法   水題   

A Computer Graphics Problem Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 968    Accepted Submission(s): 688


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

比較簡單，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;}