HDU 6077 Time To Get Up -暴力枚举-2017多校联盟4 第11题

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最新推荐文章于 2019-03-16 23:33:52 发布

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本文介绍了一种通过解析特殊格式的7段数码管显示来读取时间的方法。主要针对24小时制的HH:MM格式的时间显示，并提供了解题思路及C++实现代码。

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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：

读出用x组成的7字数码管的显示值，格式是HH:MM

解题思路

一共四位，每位有0~9 这10中可能，枚举对比一下就好。

代码

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;

const char num[10][7][5] = {".XX.", "X..X", "X..X", "....", "X..X", "X..X", ".XX.",
                            "....", "...X", "...X", "....", "...X", "...X", "....",
                            ".XX.", "...X", "...X", ".XX.", "X...", "X...", ".XX.",
                            ".XX.", "...X", "...X", ".XX.", "...X", "...X", ".XX.",
                            "....", "X..X", "X..X", ".XX.", "...X", "...X", "....",
                            ".XX.", "X...", "X...", ".XX.", "...X", "...X", ".XX.",
                            ".XX.", "X...", "X...", ".XX.", "X..X", "X..X", ".XX.",
                            ".XX.", "...X", "...X", "....", "...X", "...X", "....",
                            ".XX.", "X..X", "X..X", ".XX.", "X..X", "X..X", ".XX.",
                            ".XX.", "X..X", "X..X", ".XX.", "...X", "...X", ".XX."};

char tm[10][25];
char n[7][5];

int check(){
    for (int i=0; i<10; i++){
        bool flag = true;
        for (int j=0; j<7; j++){
            if (strcmp(n[j], num[i][j]) != 0){
                flag = false;
                break;
            }
        }
        if (flag){
            return i;
        }
    }
    return -1;
}

int main()
{
    int t;

    scanf("%d", &t);
    while (t > 0){
        t--;

        for (int i=0; i<7; i++) scanf("%s", tm[i]);

        int a, b, c, d;
        for (int i=0; i<7; i++){
            strncpy(n[i], tm[i], 4);
        }
        a = check();
        for (int i=0; i<7; i++){
            strncpy(n[i], tm[i]+5, 4);
        }
        b = check();
        for (int i=0; i<7; i++){
            strncpy(n[i], tm[i]+12, 4);
        }
        c = check();
        for (int i=0; i<7; i++){
            strncpy(n[i], tm[i]+17, 4);
        }
        d = check();
        printf("%d%d:%d%d\n", a,b,c,d);
    }
    return 0;
}