HDU 6077 Time To Get Up【】

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..XX..X....X.X....X.X..X......XX.....XX...XX.X..X.X....X....X.X..XX..X.X.........X.X..X.XX...XX.....XX...XX.

 

Sample Output

02:38

通过观察可以发现每个字符所占的位置的特征是不同的，其中0，1，和7是中间是空的。。。。。

#include<iostream>#include<cstdio>#include<cstring>#include<string>#include<cmath>#include<queue>#include<stack>#include<vector>#include<map>#include<set>#include<algorithm>using namespace std;#define ll long long#define ms(a,b)  memset(a,b,sizeof(a))#define maxn 510const int M=1e7+10;const int inf=0x3f3f3f3f;const int mod=1e9+7;const double eps=1e-10;int n,m,k;char s[100][190];int check(int x,int y){    if(s[3][x+1]=='.'&&s[3][x+2]=='.')    {        if(s[0][x+1]=='.')return 1;        if(s[6][x+1]=='.')return 7;        else return 0;    }    else {        if(s[0][x+1]=='.'&&s[6][x+1]=='.')return 4;        if(s[1][x]=='.'&&s[4][x]=='.')return 3;        if(s[1][x]=='.'&&s[4][y]=='.')return 2;        if(s[1][y]=='.'&&s[4][x]=='.')return 5;        if(s[1][y]=='.')return 6;        if(s[4][x]=='.')return 9;       return 8;    }}int main(){  int t;int j;  scanf("%d",&t);  getchar();  while(t--){    for(int i=0;i<7;i++){        for(j=0;j<10;j++)          scanf("%c",&s[i][j]);        char a1,a2;        scanf("%c%c",&a1,&a2);        for(j=10;j<21-2;j++)          scanf("%c",&s[i][j]);        getchar();    }    int a1=check(0,3);    int a2=check(5,8);    int a3=check(10,13);    int a4=check(15,18);    printf("%d%d:%d%d\n",a1,a2,a3,a4);  }  return 0;}