【多校训练】hdu 6077 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..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：

给出时间的图像，打印时间

思路：

模拟，判断特征

////  main.cpp//  1011////  Created by zc on 2017/8/3.//  Copyright © 2017年 zc. All rights reserved.//#include <iostream>#include<cstdio>#include<cstring>#include<cmath>using namespace std;char s[110][110];int d[10];int cal(int l){    memset(d,0,sizeof(d));    if(s[0][l+1]=='X')   d[1]=1;    if(s[1][l]=='X')    d[2]=1;    if(s[1][l+3]=='X')  d[3]=1;    if(s[3][l+1]=='X')  d[4]=1;    if(s[4][l]=='X')    d[5]=1;    if(s[4][l+3]=='X')   d[6]=1;    if(s[6][l+1]=='X')  d[7]=1;    int sum=0;    for(int i=1;i<8;i++)    sum+=d[i];    if(sum==2)  return 1;    if(sum==3)  return 7;    if(sum==4)  return 4;    if(sum==5)    {        if(!d[2]&&!d[6])    return 2;        if(!d[2]&&!d[5])    return 3;        if(!d[3]&&!d[5])    return 5;    }    if(sum==6)    {        if(!d[4])   return 0;        if(!d[3])   return 6;        if(!d[5])   return 9;    }    if(sum==7)  return 8;    return 0;}int main(int argc, const char * argv[]) {    int T;    scanf("%d",&T);    while(T--)    {        for(int i=0;i<7;i++)    scanf("%s",s[i]);        printf("%d%d:%d%d\n",cal(0),cal(5),cal(12),cal(17));    }}