【多校训练】hdu 6077 Time To Get Up

最新推荐文章于 2019-05-15 21:06:32 发布

弱菜zc

最新推荐文章于 2019-05-15 21:06:32 发布

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分类专栏：多校联合训练文章标签： 2017 多校训练

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本文介绍了一种通过解析特定格式的时间图像来读取时间的方法。针对每个数字段落的特征进行模拟判断，实现了从输入的ASCII图像中准确识别并输出24小时制的时间。

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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：

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));
    }
}