HDU 6077 Time To Get Up【模拟题】【水题】

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..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

 

Source

2017 Multi-University Training Contest - Team 4

对数字的特征进行一下判断即可。

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#include <string>#include <map>#include <cstring>#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;}