hdu-暑假集训-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): 0    Accepted Submission(s): 0

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

题意：

0~9  如下的表示方法：

.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.

这样相信都能看的懂题意了吧！打表暴力也能过

code 

#include<bits/stdc++.h>

using namespace std;
char m[7][22];
int fun(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 t;
    scanf("%d",&t);
    while(t--){
        for(int i=0;i<7;i++){
            scanf("%s",m[i]);
        }
        int t1,t2,t3,t4;
        t1=fun(1);
        t2=fun(6);
        t3=fun(13);
        t4=fun(18);
        printf("%d%d:%d%d\n",t1,t2,t3,t4);

    }
    return 0;
}