Time To Get Up
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 477 Accepted Submission(s): 381
题目链接：点击打开链接
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
题意：每个数字由四行七列组成，给你一个七行21列的图，让你识别改图的图案构成几点。（每个数字占4列，一共4个数字，再加上每个数字中间空一列，和冒号占一列，一共21列）
#include<stdio.h>
#include<string.h>
/*int s[7][45]=
{
.xx. .... .xx. .xx. .... .xx. .xx. .xx. .xx. .xx.
x..x ...x ...x ...x x..x x... x... ...x x..x x..x
x..x ...x ...x ...x x..x x... x... ...x x..x x..x
.... .... .xx. .xx. .xx. .xx. .xx. .... .xx. .xx.
x..x ...x x... ...x ...x ...x x..x ...x x..x ...x
x..x ...x x... ...x ...x ...x x..x ...x x..x ...x
.xx. .... .xx. .xx. .... .xx. .xx. .... .xx. .xx.
};*/
char h[7][21];
int book(int m)
{
if(h[0][m+1]=='X'&&h[0][m+2]=='X')
{
if(h[3][m+1]=='.'&&h[6][m+2]=='X')
return 0;
else if(h[3][m+2]=='.')
return 7;
else if(h[4][m]=='X'&&h[4][m+3]=='.')
return 2;
else if(h[4][m]=='X'&&h[1][m]=='X'&&h[1][m+3]=='X'&&h[4][m+3]=='X')
return 8;
else if(h[4][m]=='X'&&h[1][m+3]=='.')
return 6;
else if(h[4][m]=='.'&&h[1][m]=='.')
return 3;
else if(h[4][m]=='.'&&h[1][m]=='X'&&h[1][m+3]=='X')
return 9;
else if(h[4][m]=='.'&&h[1][m+3]=='.')
return 5;
}
else
{
if(h[3][m+1]=='.')
return 1;
else
return 4;
}
}
int main()
{
int T;
int a,b,c,d;
scanf("%d",&T);
getchar();
while(T--)
{
for(int i=0;i<7;i++)
{
for(int j=0;j<21;j++)
{
scanf("%c",&h[i][j]);
}
getchar();
}
a=book(0);
b=book(5);
c=book(12);
d=book(17);
printf("%d%d:%d%d\n",a,b,c,d);
}
return 0;
}
