HDU6077Time To Get Up(模拟)

最新推荐文章于 2021-02-17 21:48:11 发布

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最新推荐文章于 2021-02-17 21:48:11 发布

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分类专栏：模拟文章标签：模拟

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本文介绍了一个时间解析问题，通过分析特定的ASCII图像来确定并输出所显示的时间。文章提供了一种解决方案，利用暴力模拟的方法逐个识别图像中的数字，并最终形成标准格式的时间字符串。

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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): 307 Accepted Submission(s): 248

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：给你时间的图像输出他的数

思路：观察特征直接暴力模拟

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <math.h>
using namespace std;
char a[10][30];
void judge(int n)
{
    int cot=0;
    for(int i=0+n;i<=3+n;i++)
            for(int j=0;j<7;j++)
            if(a[j][i]=='X')
            cot++;
        if(cot==4)
            printf("1");
        else if(cot==6)
            printf("7");
        else if(cot==8)
            printf("4");
        else if(cot==14)
            printf("8");
        else if(cot==12)
        {
            if(a[3][1+n]=='.'&&a[3][2+n]=='.')
                printf("0");
            else if(a[1][3+n]=='X'&&a[2][3+n]=='X')
                printf("9");
            else
                printf("6");
        }
        else if(cot==10)
        {
            if(a[1][3+n]=='X'&&a[5][3+n]=='.')
            printf("2");
            else if(a[1][3+n]=='X'&&a[5][3+n])
                printf("3");
            else printf("5");
        }
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        for(int i=0;i<7;i++)
            scanf("%s",a[i]);
        judge(0);
        judge(5);
        printf(":");
        judge(12);
        judge(17);
        printf("\n");
    }
    return 0;
}