2017 Multi-University Training Contest No.4

Time To Get Up

题面

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
5alarms, 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 Out

02:38

上图是CLOCK的模型

这道模拟题我的思路是检测出四个数字就行，那么对于每个数字如上图，如何确定它的值？
可以这样模拟：
LED上有七个显示条，我们定义从上到下三个横杠为1,2,3；左到右是4,5,6,7；
那么0~9的识别码是可以这样表示

1. 1011111 0
2. 0000011 1
3. 1110110 2
4. 1110011 3
5. 0101011 4
6. 1111001 5
7. 1111101 6
8. 1000011 7
9. 1111111 8
10. 1111011 9

#include<bits/stdc++.h>using namespace std;char rec[7][21];//0,1 3,1 6,1 ; 1,0; 4,0 1,3 4,3//1011111 0//0000011 1//1110110 2//1110011 3//0101011 4//1111001 5//1111101 6//1000011 7//1111111 8//1111011 9int chc[7][2]={{0,1},{3,1},{6,1 } ,{1,0 } ,{4,0} , {1,3} ,{4,3}};//七个检测点相对于顶点的位置int pt[4][2]={{0,0}, {0,5} ,{0,12} ,{0,17} };  //四个顶点的位置int screen[4];  //转换的数字int num(int x,int y)   //获得识别码{    int sum=0;    int cnt=1000000;    for(int i=0;i<7;i++)    {        int ansx=chc[i][0];        int ansy=chc[i][1];        if(rec[x+ansx][y+ansy]=='X') sum+=cnt;        cnt/=10;    }    return sum;}int true_time(int ans)  //转换识别码{    if(ans==1011111) return 0;    if(ans==11) return 1;    if(ans==1110110) return 2;    if(ans==1110011) return 3;    if(ans==101011) return 4;    if(ans==1111001) return 5;    if(ans==1111101) return 6;    if(ans==1000011) return 7;    if(ans==1111111) return 8;    if(ans==1111011) return 9;    return -1;}int get_screen()   //开始转换并输出{    for(int i=0;i<4;i++)    {        int x=pt[i][0];        int y=pt[i][1];        screen[i]=true_time(num(x,y));    }    cout<<screen[0]<<screen[1]<<':'<<screen[2]<<screen[3]<<endl;}int main(){    //freopen("fk1.txt","w",stdout);    int t;    cin>>t;    while(t--)    {        for(int i=0;i<7;i++)        {            for(int j=0;j<21;j++)            {                cin>>rec[i][j];            }        }        get_screen();    }    return 0;}