HDU6077-Time To Get Up

Time To Get Up

                                                                  Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)
                                                                                            Total Submission(s): 548    Accepted Submission(s): 438


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(1T1440), 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(00:00t23:59), 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
 

Source
2017 Multi-University Training Contest - Team 4
 

题意:将给出的字符转化为数字

解题思路:每个数字找出一些特殊点进行判断即可


#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 

using namespace std;

#define LL long long
const int INF = 0x3f3f3f3f;
const LL mod = 998244353;

char ch[10][50];

int main()
{
	int t;
	scanf("%d", &t);
	while (t--)
	{
		for (int i = 0; i<7; i++) scanf("%s", ch[i]);
		int a[5],p=1;
		for (int i = 0; i < 4; i++)
		{
			if (ch[0][p] == 'X')
			{
				if (ch[1][p-1] == 'X')
				{
					if (ch[1][p+2] == 'X')
					{
						if (ch[3][p] == '.') a[i] = 0;
						else if (ch[4][p-1] == 'X') a[i] = 8;
						else a[i] = 9;
					}
					else if (ch[4][p-1] == 'X') a[i] = 6;
					else a[i] = 5;
				}
				else if (ch[3][p] == '.') a[i] = 7;
				else if (ch[4][p-1] == 'X') a[i] = 2;
				else a[i] = 3;
			}
			else if (ch[3][p] == 'X') a[i] = 4;
			else a[i] = 1;
			if (i == 1) p += 7;
			else p += 5;
		}
		printf("%d%d:%d%d\n", a[0], a[1], a[2], a[3]);
	}
	return 0;
}

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