hdu 多校联赛 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：看图说数、
题目并不难 大多数参赛的队伍都做出来了 就是一个简单看图说数 可是我还是因为
格式问题错了好几次。
ac代码（代码可能写的有点麻烦 但易于理解）：
#include<bits/stdc++.h>using namespace std;#define maxn 10#define maxm 24char p[maxn][maxm];//开一个矩阵列表 用于存储图int main(){    int t;    int a,b,c,d;    scanf("%d",&t);     int i=0;     int h=t;    while(t--)    {        i++;        memset(p,0,sizeof(p));        if(t==h-1)        getchar();        a=0,b=0,c=0,d=0;        for(int i = 1; i<=7;i++)        {            for(int j=1;j<=21;j++)                scanf("%c",&p[i][j]);//读入图               if(i!=7)               getchar();//坑在这里 最后一行的输入不能再让getchar吃掉回车 否则第二张图读的时候会出错        }    if(p[1][2]=='.')//判断第一个字        a=1;    else if(p[2][1]=='.')        a=2;    else if(p[4][2]=='.'&&p[1][4]=='X')        a=0;    /*---------------------*///判断第三个字    if(p[2][16]=='.')        c=5;    else if(p[5][16]=='.')        c=2;    else if(p[1][14]=='.'&&p[2][13]=='X')        c=4;    else if(p[7][14]=='.')        c=1;    else if(p[4][14]=='.')        c=0;    else if(p[4][14]=='X'&&p[2][13]=='.')        c=3;    /*---------------------*///判断第二个字    if(p[2][9]=='.'&&p[5][6]=='.')        b=5;    else if(p[2][9]=='.'&&p[5][6]=='X')        b=6;    else if(p[1][7]=='.'&&p[4][7]=='.')        b=1;    else if(p[1][7]=='.'&&p[4][7]=='X')        b=4;    else if(p[7][7]=='.')        b=7;    else if(p[4][7]=='.')        b=0;    else if(p[5][9]=='.')        b=2;    else if(p[2][6]=='.'&&p[5][6]=='.')        b=3;    else if(p[5][6]=='.')        b=9;    else if (p[2][6]=='X'&&p[5][6]=='X')        b=8;    /*-----------------------*///判断第四个字    if(p[2][21]=='.'&&p[5][18]=='.')        d=5;    else if(p[2][21]=='.'&&p[5][18]=='X')        d=6;    else if(p[1][19]=='.'&&p[4][19]=='.')        d=1;    else if(p[1][19]=='.'&&p[4][19]=='X')        d=4;    else if(p[7][19]=='.')        d=7;    else if(p[4][19]=='.')        d=0;    else if(p[5][21]=='.')        d=2;    else if(p[2][18]=='.'&&p[5][18]=='.')        d=3;    else if(p[5][18]=='.')        d=9;    else if (p[2][18]=='X'&&p[5][18]=='X')        d=8;    printf("%d%d:%d%d\n",a,b,c,d);    getchar();//吃掉回车 防止影响下次的读入    }    return 0;}