（hdu6077）2017杭电多校联赛第四场-Time To Get Up 模拟题

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

题目大意：在矩阵里有。和X，用X构成了数字，我们需要根据题目给出的矩阵集输出表示的数字。

解题思路：由于题目中给出了0-9的所有矩阵表示，所以我们直接使用if条件判断就可以了，每个位置的每个数都判断一下，注意细节。

ac代码：
#include<cstdio>
#include<iostream>
#include<cstring>
using namespace std;
int f[10][9]={
{1,1,1,1,1,1,0},
{0,0,1,1,0,0,0},
{0,1,1,0,1,1,1},
{0,1,1,1,1,0,1},
{1,0,1,1,0,0,1},
{1,1,0,1,1,0,1},
{1,1,0,1,1,1,1},
{0,1,1,1,0,0,0},
{1,1,1,1,1,1,1},
{1,1,1,1,1,0,1}};
char c[10][25];
int pan(int a[]){
    for(int i=0;i<10;i++){
        int flag=0;
        for(int j=0;j<7;j++){
            //printf("a[%d]=%d f[%d][%d]=%d\n",j,a[j],i,j,f[i][j]);
            if(a[j]!=f[i][j]){
                flag=1;
                break;
            }
        }
        if(flag==0)return i;
    }
    return 10;
}
int main(){
    int t;
    scanf("%d",&t);
    while(t--){
        for(int i=0;i<7;i++){
            scanf("%s",c[i]);
        }
        int h1,h2,m1,m2;
        int a[10];
        memset(a,0,sizeof(a));
        if(c[0][1]=='X'&&c[0][2]=='X')a[1]=1;
        if(c[3][1]=='X'&&c[3][2]=='X')a[6]=1;
        if(c[6][1]=='X'&&c[6][2]=='X')a[4]=1;
        if(c[1][0]=='X'&&c[2][0]=='X')a[0]=1;
        if(c[4][0]=='X'&&c[5][0]=='X')a[5]=1;
        if(c[1][3]=='X'&&c[2][3]=='X')a[2]=1;
        if(c[4][3]=='X'&&c[5][3]=='X')a[3]=1;
        //for(int i=0;i<7;i++)cout<<a[i]<<" ";
        h1=pan(a);//cout<<h1<<endl;
        memset(a,0,sizeof(a));
        if(c[0][6]=='X'&&c[0][7]=='X')a[1]=1;
        if(c[3][6]=='X'&&c[3][7]=='X')a[6]=1;
        if(c[6][6]=='X'&&c[6][7]=='X')a[4]=1;
        if(c[1][5]=='X'&&c[2][5]=='X')a[0]=1;
        if(c[4][5]=='X'&&c[5][5]=='X')a[5]=1;
        if(c[1][8]=='X'&&c[2][8]=='X')a[2]=1;
        if(c[4][8]=='X'&&c[5][8]=='X')a[3]=1;
        //for(int i=0;i<7;i++)cout<<a[i]<<" ";
        h2=pan(a);//cout<<h2<<endl;
        memset(a,0,sizeof(a));
        if(c[0][13]=='X'&&c[0][14]=='X')a[1]=1;
        if(c[3][13]=='X'&&c[3][14]=='X')a[6]=1;
        if(c[6][13]=='X'&&c[6][14]=='X')a[4]=1;
        if(c[1][12]=='X'&&c[2][12]=='X')a[0]=1;
        if(c[4][12]=='X'&&c[5][12]=='X')a[5]=1;
        if(c[1][15]=='X'&&c[2][15]=='X')a[2]=1;
        if(c[4][15]=='X'&&c[5][15]=='X')a[3]=1;
            //for(int i=0;i<7;i++)cout<<a[i]<<" ";
        m1=pan(a);//cout<<m1<<endl;
        memset(a,0,sizeof(a));
        if(c[0][18]=='X'&&c[0][19]=='X')a[1]=1;
        if(c[3][18]=='X'&&c[3][19]=='X')a[6]=1;
        if(c[6][18]=='X'&&c[6][19]=='X')a[4]=1;
        if(c[1][17]=='X'&&c[2][17]=='X')a[0]=1;
        if(c[4][17]=='X'&&c[5][17]=='X')a[5]=1;
        if(c[1][20]=='X'&&c[2][20]=='X')a[2]=1;
        if(c[4][20]=='X'&&c[5][20]=='X')a[3]=1;
        m2=pan(a);
        printf("%d%d:%d%d\n",h1,h2,m1,m2);
    }
    return 0;
}