Time To Get Up

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 364 Accepted Submission(s): 288

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 

Source

对数字的特征进行一下判断即可。

#include 
    
     #include 
     
      #include 
      
       #include 
       
        #include 
        
         #include 
         
          #include 
          #include 
           
            #define INF 0x3f3f3f3f#define ms(x,y) memset(x,y,sizeof(x))using namespace std;typedef long long ll;const double pi = acos(-1.0);const double eps = 1e-8;const int maxn = 1e6 + 10;char a[10][30] = { 0 };int solve(int p){ if (a[3][p + 1] == '.') { if (a[6][p + 1] == 'X') return 0; if (a[0][p + 1] == 'X') return 7; return 1; } if (a[1][p] == '.') { if (a[4][p] == 'X') return 2; return 3; } if (a[1][p + 3] == '.') { if (a[4][p] == '.') return 5; return 6; } if (a[4][p] == '.') { if (a[6][p + 1] == '.') return 4; return 9; } return 8;}int main(){ int t; scanf("%d", &t); while (t--) { ms(a, 0); for (int i = 0; i < 7; i++) { for (int j = 0; j < 21; j++) { scanf(" %c", &a[i][j]); } } int h1, h2, m1, m2; for (int i = 0; i < 4; i++) { int p; if (i == 0) p = 0; else if (i == 1) p = 5; else if (i == 2) p = 12; else p = 17; if (i == 0) h1 = solve(p); else if (i == 1) h2 = solve(p); else if (i == 2) m1 = solve(p); else m2 = solve(p); } printf("%d%d:%d%d\n", h1, h2, m1, m2); } return 0;}

转载于:https://www.cnblogs.com/Archger/p/8451614.html

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