HDU 2017 多校联赛4 1011 Time To Get Up
Problem Description
Little Q’s clock is alarming! It’s time to get up now! However, after reading the time on the clock, Little Q lies down and starts sleeping again. Well, he has 5 alarms, and it’s just the first one, he can continue sleeping for a while.
Little Q’s clock uses a standard 7-segment LCD display for all digits, plus two small segments for the ”:”, and shows all times in a 24-hour format. The ”:” segments are on at all times.
Your job is to help Little Q read the time shown on his clock.
Input
The first line of the input contains an integer T(1≤T≤1440), denoting the number of test cases.
In each test case, there is an 7×21 ASCII image of the clock screen.
All digit segments are represented by two characters, and each colon segment is represented by one character. The character ”X” indicates a segment that is on while ”.” indicates anything else. See the sample input for details.
Output
For each test case, print a single line containing a string t in the format of HH:MM, where t(00:00≤t≤23:59), denoting the time shown on the clock.
Sample Input
1
.XX…XX…..XX…XX.
X..X….X……X.X..X
X..X….X.X….X.X..X
……XX…..XX…XX.
X..X.X….X….X.X..X
X..X.X………X.X..X
.XX…XX…..XX…XX.
Sample Output
02:38
题目大意：
这就是给你一个7*21的字符矩阵，看它打印的时间是多少。下面是0—9的打印方法。每个数字占4个字符，数字间空有一格，方便区分。
.xx. …. .xx. .xx. …. .xx. .xx. .xx. .xx. .xx.
x..x …x …x …x x..x x… x… …x x..x x..x
x..x …x …x …x x..x x… x… …x x..x x..x
…. …. .xx. .xx. .xx. .xx. .xx. …. .xx. .xx.
x..x …x x… …x …x …x x..x …x x..x …x
x..x …x x… …x …x …x x..x …x x..x …x
.xx. …. .xx. .xx. …. .xx. .xx. …. .xx. .xx.
c++
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
char m[8][25];
int sz(int x)
{
if(m[3][x]=='X'&&m[3][x+1]=='X')
{
if(m[2][x-1]=='.'&&m[1][x-1]=='.')
{
if(m[4][x-1]=='.'&&m[5][x-1]=='.')
return(3);
else return(2);
}
else
{
if(m[2][x+2]=='.'&&m[1][x+2]=='.')
{
if(m[4][x-1]=='.'&&m[5][x-1]=='.')
return(5);
else return(6);
}
else
{
if(m[4][x-1]=='.'&&m[5][x-1]=='.')
{
if(m[6][x]=='.'&&m[6][x+1]=='.')
return(4);
else return(9);
}
else return(8);
}
}
}
else
{
if(m[6][x]=='.'&&m[6][x+1]=='.')
{
if(m[0][x]=='.'&&m[0][x+1]=='.')
return(1);
else return(7);
}
else return(0);
}
}
int main()
{
int a,b,c,d,e;
cin>>a;
getchar();
while(a--)
{
for(b=0;b<7;b++)
{
for(c=0;c<21;c++)
scanf("%c",&m[b][c]);
getchar();
}
b=sz(1);
c=sz(6);
d=sz(13);
e=sz(18);
printf("%d%d:%d%d\n",b,c,d,e);
}
return 0;
}
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