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
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;}
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，“人”的结构就是相互支撑，“众”人的事业需要每个人的参与。
