杭电暑期多校集训—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..XX..X....X.X....X.X..X......XX.....XX...XX.X..X.X....X....X.X..XX..X.X.........X.X..X.XX...XX.....XX...XX.

 

Sample Output

02:38
题意：读取ASCII码图像中的数字
#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;}