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
解题思路：
通过枚举出0--9个数，然后分析输入的与枚举出来的关系编写代码！
具体代码：
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
char str[8][22];
int shu[10][7][4]=
{
0,1,1,0,
1,0,0,1,
1,0,0,1,
0,0,0,0,
1,0,0,1,
1,0,0,1,
0,1,1,0,
0,0,0,0,
0,0,0,1,
0,0,0,1,
0,0,0,0,
0,0,0,1,
0,0,0,1,
0,0,0,0,
0,1,1,0,
0,0,0,1,
0,0,0,1,
0,1,1,0,
1,0,0,0,
1,0,0,0,
0,1,1,0,
0,1,1,0,
0,0,0,1,
0,0,0,1,
0,1,1,0,
0,0,0,1,
0,0,0,1,
0,1,1,0,
0,0,0,0,
1,0,0,1,
1,0,0,1,
0,1,1,0,
0,0,0,1,
0,0,0,1,
0,0,0,0,
0,1,1,0,
1,0,0,0,
1,0,0,0,
0,1,1,0,
0,0,0,1,
0,0,0,1,
0,1,1,0,
0,1,1,0,
1,0,0,0,
1,0,0,0,
0,1,1,0,
1,0,0,1,
1,0,0,1,
0,1,1,0,
0,1,1,0,
0,0,0,1,
0,0,0,1,
0,0,0,0,
0,0,0,1,
0,0,0,1,
0,0,0,0,
0,1,1,0,
1,0,0,1,
1,0,0,1,
0,1,1,0,
1,0,0,1,
1,0,0,1,
0,1,1,0,
0,1,1,0,
1,0,0,1,
1,0,0,1,
0,1,1,0,
0,0,0,1,
0,0,0,1,
0,1,1,0
};
void ton(int n)
{
for(int k=0;k<10;k++)
{
int kk=0;//检测数是否被找到
for(int i=0;i<7;i++)
{
for(int j=n;j<n+4;j++)
{
if((shu[k][i][j-n]==0&&str[i][j]=='.')||(shu[k][i][j-n]==1&&str[i][j]=='X')) continue;
kk=1;break;
}
if(kk==1) break ;
}
if(kk==0) {printf("%d",k);return ;}
}
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
for(int i=0;i<7;i++)
scanf("%s",str[i]);
ton(0);
ton(5);
printf(":");
ton(12);
ton(17);
printf("\n");
}
return 0;
}
