17暑假多校联赛4.11 HDU 6077 Time To Get Up

Time To Get Up

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 524288/524288 K (Java/Others)  

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

  题目网址：http://acm.hdu.edu.cn/showproblem.php?pid=6077  

分析

题意：当时没翻译题目，直接看到了图片还有样例输入输出，就明白是给出矩阵，输出矩阵显示的时间 因为他的输入都是7x21的矩阵，所以每个数的位置是固定的，只用推出是哪个数即可 根据中心两点是否为X可以分为两堆，再根据左上，右上，左下，右下将各个数分辨出来，这种方法直接将4个位置一起考虑，可以直接写一个函数调用4次就行了 因为是时间第一个位置只能为0，1，2，第三个位置只能为0，1，2，3，4，5，6，所以也可以少几次判断 也可以直接将每个数的点给列出，判断满足哪个数的情况就输出谁，不过这样会使代码很长很乱  

代码

#include <bits/stdc++.h> using namespace std; char alarm[8][22]; int change(int x) { if(alarm[3][x]=='X') { if(alarm[2][x-1]=='.') { if(alarm[4][x-1]=='.')return(3); return(2); } if(alarm[2][x+2]=='.') { if(alarm[4][x-1]=='.')return(5); return(6); } if(alarm[4][x-1]=='.') { if(alarm[6][x]=='.')return(4); return(9); } return(8); } if(alarm[6][x]=='.') { if(alarm[0][x]=='.')return(1); return(7); } return(0); } int main() { int T; scanf("%d",&T); getchar(); while(T--) { for(int i=0; i<7; i++) { for(int j=0; j<21; j++)scanf("%c",&alarm[i][j]); getchar(); } printf("%d%d:%d%d\n",change(1),change(6),change(13),change(18)); } }